Rings. Chapter Definitions and Examples


 Jonas Anderson
 2 years ago
 Views:
Transcription
1 Chapter 5 Rings Nothing proves more clearly that the mind seeks truth, and nothing reflects more glory upon it, than the delight it takes, sometimes in spite of itself, in the driest and thorniest researches of algebra.  Bernard de Fontenelle This chapter introduces what is in some ways the next logical structure in modern or abstract algebra: the ring. The ring we know best at this point is the integers under addition and multiplication. It is in many ways the quintessential ring, in the sense that the strangeness of other rings is measured by the properties they fail to share with the integers. 5.1 Definitions and Examples Definition 5.1 A ring is a set R with two operations, called addition, denoted +, and multiplication, denoted or, which obey the following rules: (R, +) is a commutative group. Multiplication is associative. The left and right distributive laws: Definition 5.3 If for a ring (R, +, ) multiplication is commutative then we say R is a commutative ring. Definition 5.4 If for a ring (R, +, ) there is an identity element for multiplication we call R a ring with identity or a ring with one. The multiplicative identity of the ring is also called the one of the ring, if it exists. Example 5.1 Both (Z, +, ) and (Z n, +, ) are examples of commutative rings with one. Notice that all of the ring properties are properties we are already familiar with for the integers and that the integers (mod n) inherit these properties by applying Proposition Example 5.2 A ring consisting only of the element zero is called the trivial ring. Example 5.3 The n n matrices over R form a noncommutative ring with one using matrix addition and multiplication. The one of this ring is the identity matrix I n. This ring is denoted R n n. Proposition 5.1 Let F(R, R) = {f : R R} a (b + c) = (a b) + (a c) be the set of functions from the real numbers to the real numbers. We know how to add and multiply such (b + c) a = (b a) + (c a) functions from our experience in introductory calculus hold. classes (the result is simply the standard multipli cation and addition of R on the output values of the Definition 5.2 For a ring (R, +, ) the group functions at the point at which they are evaluated). (R, +) is called the additive group of R. The identity of this group is called the zero of the ring. It is usually denoted 0. Under these forms of addition and multiplication the functions from R to itself form a commutative ring with one. 123
2 124 CHAPTER 5. RINGS We know that addition and multiplication of functions is associative by noticing that at each value of x R we are performing associative addition and multiplication of real numbers; the function in total follows these individual associative operations. The additive identity is the function f(x) = 0 and the additive inverse of an element f(x) is f(x) and so we have assembled the additive group. The distributive law for functions follows from the distributive law for each individual x R. We now know that F is a ring. The one is f(x) = 1 and commutativity results from commutativity of multiplication of functions at each x R. Proposition 5.2 Suppose that (R, +, ) is a ring. Then for any r R 0 r = r 0 = 0 Let a, r R and compute: (a + 0) r = (a r) + (0 r) (a r) = (a r) + (0 r) (a r) (a r) = (a r) (a r) + (0 r) 0 = 0 r Which gives us half of the proposition. The other half is proved in the same manner. The following proposition demonstrates that the additive inverse in a ring behaves as we would expect it to with respect to multiplication. Proposition 5.3 Suppose that (R, +, ) is a ring. Then for any a, b R (a b) = ( a) b = a ( b) Using the left distributive law and proposition 5.2 we have: a 0 = 0 a (b + ( b)) = 0 a b + a ( b) = 0 a ( b) = (a b) Which gives us half of the proposition, the other half is proved in a similar manner using the right distributive law. Small rings can be efficiently summarized by giving the addition and multiplication tables. Example 5.4 Below are both the addition and multiplication tables for (Z 3, +, ) We have already encountered zero divisors in Chapter 1, Definition 3.22 defined them for the integers (mod n) and we now repeat the definition generally for rings. Definition 5.5 Suppose that (R, +, ) is a ring and that x, y R. If x y = 0 and we have that x = 0 and y = 0 then we call x a left zero divisor and y a rightzero divisor. If an element of the ring is both a left zero divisor and a right zero divisor it is said to be a zero divisor. Zero itself is not a zero divisor. Note that for a commutative ring, an element is a left zero divisor iff it is a right zero divisor, as such a commutative ring that has no zero divisors has no left and no right zero divisors and vice versa. This is not true in general for noncommutative rings. We now define a type of ring that is extremely well behaved and typified by the integers. Definition 5.6 An integral domain is a commutative ring with one that has no zero divisors. Example 5.5 The ring (Z, +, ) (or the ring Z) is an integral domain. Proposition 5.4 The Left cancellation law If (I, +, ) is an integral domain and a = 0 is an element of I then when a b = a c we have that b = c.
3 5.1. DEFINITIONS AND EXAMPLES 125 First note that: (a c) + (a ( c)) = a (c c) (a c) + (a ( c)) = 0 So (a c) = a ( c). Then compute: a b = a c (a b) (a c) = 0 a (b c) = 0 Since I is an integral domain, there are no zero divisors. Since a = 0 this forces b c = 0 which in turn tells us that b = c. There is also a right cancellation law, the proof of which is left as Exercise 5.1. Example 5.6 The three familiar arithmetic systems Q, R and C are all examples of integral domains. Henceforth we will say the ring R rather than the ring (R, +, ) and will write a b as ab in order to enhance the clarity of presentation of the material. Proposition 5.5 The ring Z n is an integral domain iff n is prime. the proof of this proposition is an exercise. Definition 5.7 Suppose that R is a ring and the S R. Then we say S is a subring of R if S is a ring using the same operations as R. Notice that every ring is a subring of itself. Proposition 5.6 A nonempty subset S of a ring R is a subring if (i) S is closed under the addition of R. (ii) S is closed under the multiplication of R. (iii) S is closed under negation. the proof of this proposition is an exercise If S is a subring of R then we write S R. It is implicit in Proposition 5.6 that negation, which associates an element of the ring with its inverse in the additive group of the ring, is a unary operation on the ring. Example 5.7 Notice that the even integers are closed under addition, multiplication, and negation. The even integers are thus a subring of Z. They form an example of a commutative ring without one. Proposition 5.7 Suppose that R is a ring with one. Then the one is unique. Suppose that there exist two elements e and f that both act as multiplicative identities. This means that an element of R multiplied by either of these elements returns that same element of R. Then we have e e = e and e f = e. Subtracting these two equations we get: e e e f = e e e (e f) = 0 e f = 0 e = f Definition 5.8 Suppose that R is a commutative ring with one. We define R to be the subset of R so that for each a R there exists b R such that ab = 1. The subset R is called the units of R. Proposition 5.8 If R is a commutative ring with 1 then the set of units R form a commutative group under the multiplication operation of the ring. The first step is to prove that multiplication is even an operation on the putative group of units. This requires that we prove they are closed. Let a, b be units. Then there exist c, d so that ac = 1 and bd = 1. Notice that (ab)(dc) = a(bd)c = a(1)c = ac = 1 so ab is also a unit. We have that the multiplication operation is commutative since we are dealing with a commutative ring. We now check the group axioms. G1 is inherited from the definition of a ring: ring multiplication is associative. G2 is satisfied by 1. G3 is satisfied by the definition of a unit, since a R implies that there exists some b R such that ab = 1, since R is commutative ab = ba = 1, and hence b R.
4 126 CHAPTER 5. RINGS Recall that in Definition 4.7 we defined how to multiply an element of an additive group by a natural number or its negative. There is a very similar definition for multiplying a natural number or its negative by an element of a ring. Definition 5.9 If n is a natural number and r is an element of a ring then we define: 0 r = 0, that is the natural number zero multiplied by r is equal to the zero of the ring. n r = r + (n 1) r. ( n) r = n ( r). This definition permits us to make a definition that places rings into different categories. Definition 5.10 The characteristic of a ring R is the smallest positive natural number n so that n r = 0 for all r R. If no such n exists then we say the ring has characteristic zero. The notation for the characteristic of a ring R is char(r). Example 5.8 The ring Z n is an example of a ring of characteristic n while Z is an example of a ring of characteristic zero. Definition 5.11 Suppose R is a commutative ring with one such that the nonzero elements of R form a group under multiplication. Then we call R a field. We insist that there be at least one nonzero element and so fields have size at least 2. Example 5.9 The rings Q, R, C, and Z p when p is prime are all examples of fields. Notice that another way to define a field is to say a commutative ring with one in which all nonzero elements are units. you are asked to prove that this is an equivalent way to define a field in Problem Proposition 5.9 Every field is an integral domain. Suppose that F is a field and that ab = 0. If a is zero we are done. If a = 0 then a 1 ab = 1 b = b = 0 and we see that b is zero. We deduce that F has no zero divisors and so is an integral domain. Proposition 5.10 Every finite integral domain is a field. An integral domain is a commutative ring with 1. This means that it is a field if its nonzero elements form a group. Proposition 5.8 tells us that this can be done by proving that all nonzero elements are units. Let I be a finite integral domain. Proposition 5.4 tells us that, for nonzero a, if ab = ac then b = c in I; but this means that left multiplication by a = 0 is a 1 to 1 map from I to I. Since I is finite this means that left multiplication by a is a bijection of I with itself. From this we may deduce that for some d I, ad = 1 and so a is a unit. The proposition follows. Proposition 5.10 demonstrates the assertion in Example 5.9 that Z p is a field when p is prime, assuming that the student has done Problem 5.6. Definition 5.12 Let S R be a subset of a ring R. We define S to be a subring of R that contains S, such that for all subrings T R where S T, we have that S T. S is called the subring of R generated by S. When R is a finite ring the subring generated by a subset S may be thought of as the result of closing S under the operations addition, multiplication and negation. One of the most efficient methods of generating more groups from the groups we already had was the direct product of groups. Rings have a similar notion. Definition 5.13 If (R, +, ) and (S,, ) are rings then the set with the operations R S = {(r, s) : r R, s S} (r, s) + (r, s ) = (r + r, s s ) (r, s) (r, s ) = (r r, s s ) is the direct product of R and S. Proposition 5.11 The direct product of rings is a ring.
5 5.1. DEFINITIONS AND EXAMPLES 127 Note that the direct product of the additive groups is a commutative group and so there is no problem with the addition operator. Associativity of multiplication is inherited componentwise. Likewise the distributive laws work in each component of the direct product and so work overall. The direct product of rings lets us build a fairly substantial variety of rings. The next construction permits us to make rings with an extraordinary variety of properties. Warning: vast amounts of fiddly calculation lie ahead. Definition 5.14 Let G be a group and R be a ring. We define the group ring of R over G to be the set of formal sums : R[G] = r g g : r g R g G The ring R is called the coefficient ring of the group ring. Proposition 5.12 Let and x = g G r g g y = g G s g g be members of R[G]. Define: x + y = g G(r g + s g ) g x y = g G Then (R[G], +, ) is a ring. (r g s h )gh h G Addition in R and the formal addition of R[G] are both associative and so the addition given is associative. Let z = g G 0 g. Then it is easy to see that z + x = x + z = x and so z is the zero of R[G]. Notice that g G r g g + g G r g g = z so every element has an additive inverse and we see (R[G], +) is a group. That it is a commutative group follows from the fact that R is commutative. Let x = g G r g g, and Then y = g G s g g, w = g G t g g x (y w) = x (s h t k )hk = h G k G (r g s h t k )ghk = g G h G k G (r g s h )gh w = g G h G (x y) w and we see that the multiplication is associative. Now Compute: g G x (y + w) = (r g (s h + t h ))gh = g G h G h G(r g s h )gh + g G x y + x w (r g t h )gh = h G so the left distributive law holds. The right distributive law is proved in a very similar manner. It should be noted that during group ring multiplication when group multiplication yields a group element which does not explicitly have a ring coefficient we assign the zero of the ring to be that group elements coefficient. The following example illustrates how group ring addition and multiplication operations work.
6 128 CHAPTER 5. RINGS Example 5.10 Give the addition and multiplication table for the group ring of the ring integers (mod 2) over the group Z 2. In other words R = Z 2 [Z 2 ] Let the group Z 2 = {e, α}. Then the elements of R are: z = 0 e + 0 α o = 1 e + 0 α a = 0 e + 1 α b = 1 e + 1 α Following the definition of addition and multiplication in the group ring we get: + z o a b z z o a b o o z b a a a b z o b b a o z z o a b z z z z z o z o a b a z a o b b z b b z The group ring is a somewhat bizarre construction, but it provides a vast library of examples of rings. In the problems we will demonstrate that some of the properties of R can be inherited by R[G], depending on the choice of group. When computing products in a group ring there are some conventions (besides the coefficient zero convention mentioned above). First of all, elements that are multiplied by the zero of the ring are not displayed so that 0 α+r β is written simply as r β. The element with all coefficients zero is written 0 and is the zero of the group ring. Second, if R is a ring with one then we write elements or parts of elements of the form 1 α as just α, suppressing the writing of the one of R. Third, elements of the form r e where e is the identity of G are written r with the element 1 e written as 1; it is the one of the group ring. Fourth, all the coefficients of a given element of the group are added. So, for example, if we have Z[G] with α G, the element 2 α + 5 α is written 7 α. Example 5.11 Compute the multiplication table of the group ring Z 3 [Z 2 ]. Lets let the elements of the group Z 2 be {e, a}. Using the conventions above the elements are 0, 1, 2, a, 1 + a, 2 + a, 2a, 1 + 2a, and 2 + 2a. The table is shown in Figure 5.1. Problems Problem 5.1 Review Proposition 5.4 and then state and prove the right cancellation law. Problem 5.2 In the same fashion as Example 5.4 give the addition and multiplication tables for the ring (Z 4, +, ) Problem 5.3 Give an example of a finite noncommutative ring with one. Problem 5.4 Give an example of a noncommutative ring without one. Problem 5.5 Let (G, +) be a commutative group with identity 0. If we define g h = 0 for all g, h G then is the resulting object a ring? Problem 5.6 Prove Proposition 5.5. Hint: most of this problem can be done by citing an example. Problem 5.7 Examine the object below, given by an addition and multiplication table on the set R = {0, 1, q, v}: q v q v v q q q v 0 1 v v q q v q v q 0 q v 1 v 0 v 1 q Prove that this structure is a ring using the following steps: (i) Demonstrate that (R, +) is a group by finding which group it is isomorphic to. (ii) Prove that the nonzero elements of R form a group under multiplication and identify that group. (iii) Using (ii) prove that multiplication is associative. This is not too hard if you deal with zero sensibly.
7 5.1. DEFINITIONS AND EXAMPLES a 1+a 2+a 2a 1+2a 2+2a a 1+a 2+a 2a 1+2a 2+2a a 2+2a 1+2a a 2+a 1+a a 0 a 2a 1 1+a 1+2a 2 2+a 2+2a 1+a 0 1+a 2+2a 1+a 2+2a 0 2+2a 0 1+a 2+a 0 2+a 1+2a 1+2a 0 2+a 2+a 1+2a 0 2a 0 2a a 2 2+2a 2+a 1 1+2a 1+a 1+2a 0 1+2a 2+a 2+a 0 1+2a 1+2a 2+a 0 2+2a 0 2+2a 1+a 2+2a 1+a 0 1+a 0 2+2a Figure 5.1: The multiplication table of Z 3 [Z 2 ] (iv) Remembering to use Proposition 5.2 liberally, prove that the left distributive law holds: a (b + c) = a b + a c. (v) Prove that the multiplication is commutative and from this deduce the right distributive law. Problem 5.8 Prove that the ring in Problem 5.7 is an integral domain. Problem 5.9 Prove Proposition 5.6. Hint: remember that you already have propositions for subgroups. Problem 5.10 State and prove a result that lists all possible subrings of the integers. Start with example 5.7. Problem 5.11 Prove that every subring of the integers is generated by a single integer. Problem 5.12 Suppose we take the matrix 0 1 M = 1 1 with entries from Z 2. Find the set of matrices {M} generated by closing the set {M} under both matrix addition and matrix multiplication. Make addition and multiplication tables. Is the resulting subring of the ring Z of 2 2 matrices over Z 2 : (i) a commutative ring with one? (ii) an integral domain? (iii) a field? Prove your answers. Problem 5.13 Suppose we take the matrix 0 1 M = 2 0 with entries from Z 3. Find the set of matrices {M} generated by closing the set {M} under both matrix addition and matrix multiplication (there are 9 of them). Make addition and multiplication tables. Is the resulting subring of the ring Z of 2 2 matrices over Z 3 : (i) a commutative ring with one? (ii) an integral domain? (iii) a field? Prove your answers.
8 130 CHAPTER 5. RINGS Problem 5.14 Find all subrings of Z 8. Problem 5.15 Suppose that p is prime. subrings of the ring Z p. Find all Problem 5.16 Prove that if R is a commutative ring with one for which all nonzero elements are units then R is a field. Problem 5.17 Suppose that F is a field of characteristic p for some prime integer p. Show that F contains a subring that is identical to the ring Z p. Problem 5.18 Prove that the direct product of two commutative rings is commutative. Problem 5.19 Prove that the direct product of two rings with one is a ring with one. Problem 5.20 Suppose that R = S T is a direct product of rings with S and T each having size in excess of 2. Prove that R has zero divisors. Problem 5.21 Compute the characteristic of Z n Z m. Problem 5.22 Suppose that G is a group with n elements. How many elements does the group ring Z 2 [G] contain? Problem 5.23 Prove that every ring is isomorphic to a group ring. Hint: this is really easy. Problem 5.24 Prove that if R is a commutative ring and G is a commutative group then the group ring R[G] is a commutative ring. Problem 5.25 Prove that if R is a ring with one and G is a group then the group ring R[G] is a ring with one. Problem 5.26 Prove that if R is a ring with at least one nonzero element and G is a noncommutative group then R[G] is a noncommutative ring. Problem 5.27 Let G be a group. Prove that if R is a ring of characteristic n then so is R[G]. Problem 5.28 Make an addition and multiplication table for the group ring Z 2 [Z 3 ]. Note that this group ring has 8 elements. Problem 5.29 Prove or disprove Z 2 [Z 3 ] is an integral domain. Problem 5.30 Find the group of units of Z 3 [Z 2 ]. Problem 5.31 Prove that Z 3 [Z 2 ] = Z 3 Z Homomorphisms and Ideals Definition 5.15 If R and S are rings and f : R S is a function so that (i) f(r + r ) = f(r) + f(r ), and (ii) f(r r ) = f(r) f(r ). then we say that f is a ring homomorphism. Definition 5.16 If g : R S is a ring homomorphism then we call the image of g a homomorphic image of R. Proposition 5.13 Any homomorphic image of a ring is a ring. Let R and S be rings and let g : R S be a ring homomorphism. Note that (R, +) and (S, +) are both commutative groups and g is a group homomorphism. Thus image(g) is a commutative group and the remaining ring properties (associative multiplication, and left and right distributive laws) are inherited from R via the ring homomorphism properties: g(ab) = g(a)g(b) and g(a + b) = g(a) + g(b). Proposition 5.14 Any homomorphic image of a ring with 1 is a ring with 1. Let R be a ring with 1 and let g : R S be a ring homomorphism. Set Q = image(g), the image of g. By the previous proposition Q is a ring. Let e = g(1). Let a Q. Then there must be some b R so that g(b) = a. Note that: 1 b = b = b 1 g(1 b) = g(b) = g(b 1) g(1) g(b) = g(b) = g(b) g(1) e a = a = e a Since a was chosen arbitrarily, it follows that e is a multiplicative identity of Q.
9 5.2. HOMOMORPHISMS AND IDEALS 131 A point that may make the above proposition a little confusing is the way we proved e was an identity of Q when g was a function from R to S. A natural question for those not yet fully Zen with abstract algebra is what happened to S? The problem solved by using Q is that g need not be a surjection and so Q = image(g) S. It is possible to construct examples so that e is the identity of Q, Q is a proper subset of S, and e is not an identity of S. In fact we ask you to find this example in an exercise. Corollary 5.1 If R is a ring with 1 and π : R S is a surjective ring homomorphism then S is a ring with one and π(1) is the identity of S. Making π surjective forces S = Q = image(π) in the proof of Proposition 5.14 and the corollary follows from the proof of Proposition Proposition 5.15 The homomorphic image of a commutative ring is a commutative ring. this proof is left as an exercise. A group homomorphism was a map between groups that preserved the group operation. A ring homomorphism, similarly, is a map between rings that preserves the ring operations. Notice that if f : R S is a ring homomorphism then f is also a homomorphism of the additive groups of R and S Example 5.12 The map π n : Z Z n, given by i [i] n is a ring homomorphism. This follows from Proposition One use we made of group homomorphisms was defining the idea of isomorphism that told us when two algebraic structures were the same as groups. A similar notion is available for rings. Definition 5.17 A bijective ring homomorphism is called a ring isomorphism. If f : R S is a ring isomorphism we say the rings R and S are isomorphic. This is denoted R = S. Definition 5.18 Suppose that R and S are rings and g : R S is a ring homomorphism. Then I g = {r R : g(r) = 0 S } is called the kernel of g. We denote the kernel of a ring homomorphism g by ker(g). Example 5.13 From Example 5.12 we know that π 2 : Z Z 2, is a ring homomorphism. The kernel of this map consists of the even integers. Proposition 5.16 Let F be the ring of functions specified in Proposition 5.1. Let r R be a real number. Then the map r : F R given by f(x) f(r) is a ring homomorphism. Recall that addition and multiplication of functions are defined by (f +g)(x) = f(x)+g(x) and (f g)(x) = f(x)g(x). Given this: r (f + g) = f(r) + g(r) = r (f) + r (g) r (f g) = f(r)g(r) = r (f) r (g) and we have that r is a ring homomorphism. Many of the facts we proved about groups imply properties of rings. Proposition 5.17 If f : R S is a ring homomorphism then f(0 R ) = 0 S, where 0 R and 0 S are the zeros of R and S, respectively. Apply proposition 4.14 to the additive groups of R and S. Definition 5.19 Suppose that I is a nonempty subset of a ring R so that: (i) (I, +) is a subgroup of (R, +). (ii) For all x R and y I we have xy I and yx I.
10 132 CHAPTER 5. RINGS Then we say I is an ideal of the ring. We write I R when I is an ideal of R. Proposition 5.18 If R is a ring and I R then 0 I. Since I is nonempty by definition there is some a I. Let b R. Then b a I and b a I by the second ideal property. The sum of these two elements is zero and so 0 I. Proposition 5.19 Every ring R contains the two ideals R and {0}. the proof that these are ideals is an exercise. The next proposition gives an important family of ideals. Proposition 5.20 Let R be a commutative ring and for r R let I = {rx : x R}. Then I R. Check the two ideal properties. If a, b I, a = rx and b = ry so a + b = rx + ry = r(x + y) I and I is closed under addition, if a = rx I then r( x) I, noting that rx + r( x) = we have that I is closed under negation so (I, +) is a subgroup of (R, +) and we have the first ideal property. Suppose a I and c R then a = rx and ca = c(rx) = r(cx) I and the proposition follows. Definition 5.20 For a commutative ring R and an element r R, the ideal I = {rx : x R} consisting of all multiples of r is called a principal ideal. The element r R is called the principle of the ideal. Proposition 5.20 demonstrates I is an ideal. We also call the principle of a principle ideal its generator. Definition 5.21 If R is a ring and S R then the ideal generated by S is the ideal of R denoted by S such that S S, and for any ideal I of R such that S I we have that S I. If S is a singleton set then this ideal is a principle ideal. Definition 5.22 A principle ideal domain is an integral domain in which all the ideals are principle. Example 5.14 The ring of integers, Z is a principle ideal domain. To see this note that an ideal of the integers is closed under greatest common divisors, forcing any ideal to be generated by its least positive element (or zero if the ideal contains no other elements). As we will see in the next proposition, ideals in rings play a very similar role to normal subgroups in group theory. Proposition 5.21 If g : R S is a ring homomorphism then ker(g) R. Suppose that x, y ker(g) and that z R. Then g(x + y) = g(x) + g(y) = = 0 showing that x + y ker(g) and so we have property (i) required for an ideal. g(zx) = g(z)g(x) = g(z) 0 = 0 so that zx ker(g), similarly xz ker(g), giving us property (ii). As with groups, in the domain of rings it turns out that any ideal is the kernel of some homomorphism. Proposition 5.22 Let R be a ring and let I R. Define a relation on R by r s if r s I. Then is an equivalence relation. this proof is left as an exercise. Now that we know that differing by an element of an ideal is an equivalence relation we may define a ring structure on the equivalence classes. Definition 5.23 If R is a ring and I R let R/I be the set of equivalence classes on R induced by the equivalence relation given in Proposition We denote the equivalence classes by r+i for each r R. These equivalence classes are also called cosets of I. Notice that the cosets of an ideal I are also cosets of I as a subgroup of the additive group of R. Proposition 5.23 Let R be a ring and let I R. Define operations on R/I by (i) (r + I) + (s + I) = (r + s) + I, and (ii) (r + I) (s + I) = (r s) + I.
11 5.2. HOMOMORPHISMS AND IDEALS 133 Then (R/I, +, ) is a ring called the factor ring of R by I. Note that if R is a ring with one then (R/I, +, ) is a ring with one. Notice that I is a subgroup of (R, +). Since this group is commutative it follows that I is a normal subgroup and so R/I is a factor group and all of the needed properties for addition in R/I follow. Associativity of multiplication is inherited from R, as are the right and left distributive laws. The identity of R/I is 1 + I, if the ring R has an identity to start with. Corollary 5.2 Let R be a ring. The map π I : R R/I given by r r + I is a ring homomorphism. This homomorphism is called the canonical homomorphism or natural quotient map. Once it has been established that R/I is a ring this follows directly from the definitions of + and on R/I. Note that the kernel of a canonical homomorphism is an ideal. We now know that every ring homomorphism has an ideal as its kernel and that all ideals are kernels of ring homomorphisms. There is thus a matching of kernels and homomorphisms with each uniquely specifying the other. Next we will look at some of the properties of ideals. Definition 5.24 Suppose that R is a ring and that I and J are ideals of R. Then the sum of I and J is: I + J = {a + b : a I, b J} Proposition 5.24 If R is a ring and I and J are ideals of R then I + J is also an ideal of R. Suppose that x, y I + J then for some a, b I and c, d J, x = a + c and y = b + d. This means that x + y = (a + c) + (b + d) = (a + b) + (c + d) Notice that a + b I and c + d J meaning x + y I + J and so we have shown I + J is closed under addition. If x I + J, with x = a + c where a I and c J, then since I and J are subgroups of R (with respect to addition), we have that a I and c J and hence ( a) + ( c) = x I + J. Thus I + J is closed under negation, and hence I + J is a subgroup of R. Let r R. Then: r x = r(a + c) = r a + r c Since a I, ra I and since c J, rc J and so we see that rx I + J. In a similar manner it can be shown that xr I + J. This completes the proof that I + J is an ideal. Definition 5.25 Suppose that R is a ring and that I and J are ideals of R. Then we define the product of I and J to be the set of all finite sums of the form i 1 j 1 + i 2 j i n j n for i k I and j m J. The product of I and J is denoted IJ. Proposition 5.25 If R is a ring and I and J are ideals of R then IJ is also an ideal of R. The sum of two finite sums of the given form is a finite sum of the given form and so IJ is closed under addition. Suppose that r R and i 1 j 1 + i 2 j i n j n IJ. Then r(i 1 j 1 + i 2 j i n j n ) = (ri 1 )j 1 + (ri 2 )j (ri n )j n Each ri k I because I R and so the result of multiplying R by any element of IJ is an element of IJ (and vice versa). Finally note that for i 1 j 1 +i 2 j i n j n IJ, with i k I and j m J we have that i k I for each i k and so ( i 1 )j 1 + ( i 2 )j ( i n )j n IJ. Thus we have that IJ is an ideal. When an ideal is generated by a set {x} we often say, for the sake of simplicity, that the ideal is generated by x. Definition 5.26 An ideal I of a commutative ring R is prime if for all a, b R, if ab I then a I or b I. Proposition 5.26 If P is a prime ideal of a commutative ring with 1, R, then R/P is an integral domain.
12 134 CHAPTER 5. RINGS Let R be a commutative ring with 1 and let P R be a prime ideal. Set Q = R/P and let π P : R Q be the canonical homomorphism. The definition of factor rings ensures that π P is surjective which permits us to apply Corollary 5.1 to deduce Q is a ring with one. Proposition 5.15 tells us that Q is commutative. Recall that P is the equivalence class of elements of R that is the zero of Q. Suppose that a + P, b + P Q. Then if (a + P ) (b + P ) = 0 it follows that ab P. Since P is a prime ideal we see that a P or b P and so a + P = P or b + P = P. But this means that if any two elements of Q have zero as a product then one of these elements must be zero. This means that Q has no zero divisors. We have shown that Q is a commutative ring with one that has no zero divisors. This is the definition of an integral domain and the proposition follows. Notice that the notion of being a prime ideal generalizes, in a sense, the notion of being a prime number. The prime ideals of the integers are, for example, exactly those that are principle ideals generated by prime numbers. You are asked to verify this fact in an exercise. Definition 5.27 An ideal I of a ring is said to be maximal if it is a proper ideal and is contained in no other proper ideal. Proposition 5.27 If R is a commutative ring with 1 and M R is a maximal ideal of R then M is also a prime ideal of R. Let ab M. If a M or b M then we are done so assume neither is in M. Let A = a and B = b be the principal ideals generated by a and b. Since a, b / M the ideals M +A and M +B contain M and so must be R since M is maximal. Since R is a ring with 1, R R = R meaning that (M +A)(M +B) = R. Since R is commutative AB ab. Since ab M we have that the ideal ab M. This means AB M. Now consider: R = (M + A)(M + B) = M 2 + AM + MB + AB The second ideal property tells us that M 2, AM, and MB = BM are all subsets of M and we know that AB M so that tells us R M. This contradicts the definition of maximal ideal and thus we have proved by contradiction that at least one of a or b must be in M. It follows that M is a prime ideal. Proposition 5.28 If R is a commutative ring with one and M R is a maximal ideal then R/M is a field. Let F = R/M and let π M : R F be the canonical homomorphism. By Proposition 5.27 we know M is a prime ideal and so Proposition 5.26 tells us F is an integral domain. Let a R M. Then, since M is maximal, we see that {a} M = R. The image under π M of any element of {a} M is in π M (a) and so it follows that π(a) = F. This means that F = {π M (a) x : x F }. This means that for some b F, π M (a) b = 1. Since a was chosen to be any element of R that is not in M it follows that every nonzero element of F is a unit. This is sufficient to demonstrate that F is a field. Problems Problem 5.32 Prove Proposition Problem 5.33 Find an example of a ring R with 1 and a ring homomorphism f : R S so that f(1) is not the identity of S. Hint: Proposition 5.14 implies that f cannot be surjective. Problem 5.34 Prove Proposition Problem 5.35 Prove Proposition Problem 5.36 Define the map ι : R C by r r + 0i. Prove that this map is a ring homomorphism. Problem 5.37 Find an injective ring homomorphism from R into R n n.
13 5.3. POLYNOMIAL RINGS 135 Problem 5.38 Prove that the ring given in Problem 5.7 is not isomorphic, as a ring, to Z 4. Problem 5.39 Prove that the ring given in Problem 5.7 is isomorphic, as a ring, to the ring you found in Problem Problem 5.40 Prove that the group of units of Z 3 Z 3 is isomorphic to the Klein4 group. Problem 5.41 Suppose we take the matrix 0 1 M = 2 1 with entries from Z 3. Let R the ring of matrices {M} generated by closing the set {M} under both matrix addition and matrix multiplication. Prove that this ring is isomorphic to Z 3 Z 3, Problem 5.42 Suppose that F is a field. Prove that F has only two ideals. Problem 5.43 Examine S = {12, 18} Z. For I = S Z find the principle of I. Problem 5.44 Let R = Z n Z m be the direct product of the integers modulo n and m for n, m > 1. Find n and m so that there is an ideal of R that is not principle. Demonstrate this by giving the ideal. Problem 5.45 Prove that a field is a principle ideal domain. Problem 5.46 Prove that if π : Z R is a surjective ring homomorphism then R is one of Z or Z n. Problem 5.47 List all the ideals of Z 12 and prove your list is complete. Problem 5.48 Find, up to isomorphism, all homomorphic images of Z 12. Problem 5.49 Suppose that F is a field and f : F R is a surjective ring homomorphism. Prove that R is the trivial ring or R = F. Problem 5.50 Let C R be the set of continuous functions from the real numbers to the real numbers. (i) Prove that C R is a subring of the ring given in Example 5.1. (ii) Prove that the evaluation map r defined is Example 5.16, when restricted to C R, is a surjective homomorphism. (iii) Describe ker( 2 ). Problem 5.51 Generalize Examples 5.1 and 5.16 in the following manner. Given the usual definitions (f + g)(x) = f(x) + g(x) and (fg)(x) = f(x) g(x) and assuming R is a ring prove: (i) The set of all functions from R to itself form a ring. (ii) The evaluation map r for r R is a surjective group homomorphism. Problem 5.52 Let Q = R S be a direct product of rings. Show that the sets I R = {(r, 0) : r R} and I S = {(0, s) : s S} are both ideals of Q. Problem 5.53 Suppose that R is a commutative ring. Prove that the product of principle ideals is a principle ideal. Problem 5.54 Prove that all prime ideals of the integers are those generated by prime numbers. Problem 5.55 Find a maximal ideal in Z 5 Z 5. Problem 5.56 List all the maximal ideals of Z 24 and prove your list is correct. 5.3 Polynomial Rings We are already familiar with polynomials, over the real numbers, from our earlier studies in elementary algebra and calculus. In this section we will look at polynomials over arbitrary rings and derive some of their properties. Definition 5.28 If R is a ring then a polynomial is a finite sum of the form p(x) = r 0 + r 1 x + r 2 x 2 + r n x n where r i R, n is a natural number, and x is an unknown. We call the values r i the coefficients of
14 136 CHAPTER 5. RINGS the polynomial. The unknown x is called the variable and the number n is called the degree of the polynomial. We use the notation deg(p) = n for the degree of p(x). Strictly speaking in the above definition of degree, we require that the coefficient of x n be nonzero, so that the degree of a polynomial is the largest k such that the coefficient of x k is nonzero, (in the case of p(x) = 0 we say that deg(p) = 0). Definition 5.29 A polynomial of degree zero is called a constant polynomial. A polynomial of degree one or more is called a nonconstant polynomial. Definition 5.30 If R is a ring and x is an unknown then we denote the set of all polynomials with coefficients in R by R[x]. This set is called the polynomials over R. Example 5.15 f(x) = 1 2 x3 2 3 x2 + 4x 1 7 is an example of a polynomial over the rational numbers: f(x) Q[x]. g(x) = πx 2 + e R[x] is a polynomial over the real numbers. If we have two polynomials p(x) and q(x) of degree n and m with n < m then we may pad p(x) with terms like 0x k for n < k m to enable us to act as if they are of the same degree. This zero padding is used in many of the remaining definitions and proofs in this section to permit clarity of exposition. Definition 5.31 We define addition and multiplication of polynomials in the usual way: and n n r i x i + s i x i = i=0 n r i x i i=0 i=0 n s i x i = i=0 2n i=0 n (r i + s i )x i i=0 j+k=i r j s k x i Example 5.16 Suppose that f(x) = x 2 + x + 1 and g(x) = x 3 5x + 2 are polynomials with integer coefficients. Then f(x) + g(x) = x 3 + x 2 4x + 3 f(x) g(x) = x 5 + x 4 4x 3 3x 2 3x + 2 f(x) 2 = x 4 + 2x 3 + 3x 2 + 2x + 1 Proposition 5.29 If R is a ring then so is R[x] with the addition and multiplication given in Definition this proof is left as an exercise. Proposition 5.30 If R is a ring with 1, then so is R[x]. The polynomial 1 x 0 is a multiplicative identity using the multiplication given in Definition Proposition 5.31 If R is a commutative ring, then so is R[x]. then: Let p(x) = n m r i x i and q(x) = s j x j, i=0 p(x) q(x) = = = = j=0 n m r i x i s j x j i=0 n+m k=0 n+m k=0 i+j=k i+j=k j=0 r i s j x k s j r i x k m n s i x i r j x j i=0 = q(x) p(x) j=0 Notice that the key to the above calculations is that r i s j = s j r i, permitting R[x] to inherit the commutativity of R.
15 5.3. POLYNOMIAL RINGS 137 Proposition 5.32 If R is an integral domain then so is R[x]. Propositions 5.30 and 5.31 already demonstrate that R[x] is a commutative ring with one. It remains to show that R[x] contains no zero divisors. Suppose for p(x), q(x) R[x] we have that p(x) q(x) = 0. Let p i and q j be the nonzero coefficients of smallest degree in p(x) and q(x) respectively. Such coefficients exist by the well ordering principle unless the respective polynomials are zero. Then if r(x) = p(x) q(x) we see that p i q j x i+j is the nonzero coefficient in r(x) of least degree. Since p(x) q(x) = 0 it follows that p i q j = 0. Since R is an integral domain this means that either p i = 0 (and hence p(x) = 0) or q j = 0 (and hence q(x) = 0). This permits us to deduce that R[x] contains no zero divisors, completing the proof that R[x] is an integral domain. Example 5.17 Proposition 5.32 tells us that the polynomials over the integers Z[X] form an integral domain. Corollary 5.3 The ring of polynomials over a field is an integral domain. Proposition 5.9 tells us that fields are integral domains and so we may prove the corollary by citing Proposition There are some very small rings, e.g. Z 2, but polynomial rings (other than one special case) are always infinite. The only polynomial ring which is finite is of course the polynomials over the trivial ring. It is easy to see this, pick any nonzero element a of the ring R, then the polynomials of the form ax k where k N form a countably infinite set of polynomials. Definition 5.32 Let R be a ring. We define scalar multiplication of a polynomial p(x) R[x] by s R as follows. If n p(x) = r i x i then s p(x) = i=0 n (s r i )x i i=0 Definition 5.33 Suppose that R is a commutative ring with one and that p(x), q(x) R[x]. Then if there is some unit u R so that p(x) = u q(x) we say that p and q are associated by u or that p and q are associates. Definition 5.34 Suppose that R is a ring with 1. We say that a polynomial in R[x] is monic if the coefficient of the highest power of x is 1. Proposition 5.33 If F is a field then every nonzero polynomial is the associate of a unique monic polynomial. Let p(x) F [x] be a nonzero polynomial with p(x) = n a i x i i=0 Let b be the member of F so that b a n = 1. Such a unit exists because F is a field. Let q(x) = b p(x) then for n q(x) = (b a i )x i i=0 the coefficient of x n is 1. We see that q(x) is a monic associate of p(x). It remains to demonstrate uniqueness. Suppose that c p(x) is also monic. Then c = a 1 n but, because inverses in the group (F, ) are unique, we have c = b. This tells us that q(x) is the unique monic associate of p(x). There are a number of strong analogies between the integers and the polynomial ring over a field. At this point we begin to develop this analogy, starting with the idea of divisibility. Definition 5.35 We say that a polynomial a(x) divides a polynomial b(x) if there exists a polynomial c(x) so that b(x) = a(x)c(x) We write a(x) b(x) when a(x) divides b(x). Proposition 5.34 Let F be a field and let f(x), g(x) F [x] be nonzero polynomials. If f(x) g(x) and deg(f) = deg(g) then f(x) and g(x) are associates.
16 138 CHAPTER 5. RINGS the proof is left as an exercise. Proposition 5.35 Division Algorithm for Polynomials Let F be a field and let a(x), b(x) F [x] be monic polynomials with deg(b) deg(a). Then there exists a unique monic polynomial q(x) and a unique polynomial r(x) with deg(r) < deg(b) so that Let a(x) = b(x) q(x) + r(x) Q = {a(x) b(x)s(x) : s(x) F [x]} Then the set of degrees of elements of Q form a nonempty nonnegative set of integers. By the well ordering principle there is a least degree of elements in Q. Choose q(x) and r(x) so that r(x) = a(x) b(x)q(x) is a witness to this least degree. Then a(x) = b(x) q(x) + r(x). Since a and b are monic it follows that q is as well, to permit cancellation of the highest order term. To see that the degree of r is strictly less than the degree of b, notice that we can modify the coefficients of q to, in effect, subtract or add additional multiples of b until any coefficient in a(x) b(x) q(x) for a term of degree at least as great as the degree of b is zero. This adjustment of the coefficients of q is possible because every nonzero member F is a multiple of every other nonzero member of F. This gives us the proposition, except that it remains to demonstrate uniqueness. Suppose that a(x) = b(x)q 1 (x) + r 1 (x) = b(x)q 2 (x) + r 2 (x) both satisfy the proposition. Then: b(x)q 1 (x) + r 1 (x) = b(x)q 2 (x) + r 2 (x) b(x)q 1 (x) b(x)q 2 (x) = r 2 (x) r 1 (x) b(x)(q 1 (x) q 2 (x)) = r 2 (x) r 1 (x) Minimality tells us that deg(r 1 ) = deg(r 2 ); let this degree be m. The definition of polynomial addition tells us that deg(r 2 (x) r 1 (x)) m. Since m < deg(b), the fact that equal polynomials have equal degrees tells us that the last of the three equations above require q 1 (x) q 2 (x) = 0 and r 2 (x) r 1 (x) = 0. This follows from the fact that R[x] is an integral domain (Proposition 5.32) and hence lacks zero divisors. We thus see that q and r are unique. We now continue the analogy between the integers and the polynomials of a field by defining the GCD and showing that a version of Euclid s algorithm holds. Definition 5.36 If we have three polynomials a(x), b(x), and c(x) such that a(x) b(x) and a(x) c(x) then we say a(x) is a common divisor of b(x) and c(x). Definition 5.37 Suppose that R = F [X] is the ring of polynomials over a field. The greatest common divisor of two polynomials is a monic common divisor that is divisible by all other common divisors. Proposition 5.36 Let F be a field. The greatest common divisor of any two nonzero polynomials a(x) and b(x) in F [x] exists and is unique. Let I = {s(x)a(x) + t(x)b(x) : s, t F [X]} Then the set of degrees of members of I, which are not identically zero, is a nonempty set of nonnegative integers and so has a least value n by the well ordering principle. Let d(x) be a monic polynomial of degree n in I; Proposition 5.33 permits us to assume that d(x) is monic (since we have insisted that identically zero polynomials are not permitted). Then d(x) = s(x)a(x) + t(x)b(x) for some choice of s(x) and t(x). Note that deg(d) deg(a) and deg(d) deg(b), since a, b I. Set a(x) = q(x)d(x) + r(x) by applying the division algorithm. Compute: so that a(x) = q(x)d(x) + r(x) = q(x) (s(x)a(x) + t(x)b(x)) + r(x) r(x) = (1 q(x)s(x))a(x) t(x)b(x) which implies that r(x) I. This tells us that deg(r(x)) deg(d(x)), an impossibility because the
17 5.3. POLYNOMIAL RINGS 139 division algorithm forces the opposite, unless r(x) = 0. This then yields the result that a(x) = q(x)d(x) and so d(x) a(x). By repeating the above argument with a(x) and b(x) exchanged we see that it is also the case that d(x) b(x). If follows that d(x) is a common divisor of a(x) and b(x). Let c(x) be any common divisor of a(x) and b(x). Then we can find w(x) and z(x) so that a(x) = c(x)w(x) and b(x) = c(x)z(x). This means that, for arbitrary f(x) and g(x) that f(x)a(x) + g(x)b(x) = f(x)(c(x)w(x)) + g(x)(c(x)z(x)) = c(x)(f(x)w(x) + g(x)z(x)) and we see that, since f(x) and g(x) were chosen arbitrarily, that c(x) divides every member of I, in particular c(x) d(x) and so we see that d(x) is a greatest common divisor of a(x) and b(x). Uniqueness follows from Propositions 5.34 and 5.33, given that any two GCD s of a(x) and b(x) must divide one another. The following pair of corollaries are extracted directly from the proof of Proposition 5.36 and so no explicit proof is given. Corollary 5.4 Suppose that F is a field and that a(x), b(x) F [X] are nonzero polynomials. Then GCD(a(x), b(x)) is the minimal degree monic member of the set I = {s(x)a(x) + t(x)b(x) : s, t F [X]} Corollary 5.5 Suppose that F is a field and that a(x), b(x) F [X] are nonzero polynomials. Then if c(x) is a common divisor of a(x) and b(x) then it divides every member of I = {s(x)a(x) + t(x)b(x) : s, t F [X]} Continuing the analogy with the integers, we now produce a version of Euclid s algorithm for polynomials. Since the division algorithm for polynomials requires that a(x) and b(x) be monic, we will need to repeatedly correct the remainders in the Euclidean algorithm to be monic. We adopt the convention that if f(x) is a polynomial over a field then f (x) is the unique monic associate of f(x) guaranteed by Proposition Check carefully when using the algorithm that the associates do in fact come out in the laundry properly. Algorithm 5.1 Polynomial Euclid s Algorithm Let F be a field and assume a(x), b(x) F [x] are monic polynomials of positive degree. Apply the division algorithm repeatedly. The following series of equations: a(x) = b(x)q(x) 1 + r 1 (x) b(x) = r 1(x)q 2 (x) + r 2 (x) r 1 (x) = r 2 (x)q 3(x) + r 3 (x) r i(x) = r i+1(x)q i+1 (x) + r i+2 (x) r n 2(x) = r n 1(x)q n (x) + 0 eventually produces a remainder r n = 0. In this case r n 1(x) = GCD(a(x), b(x)). Notice that the remainders have the property that deg(r i (x)) > deg(r i+1 (x)) deg(0). If a remainder is not zero then the next remainder can be computed. The process thus stops when a remainder of zero is found. Since a(x) and b(x) are of finite degree, it follows that a finite number of remainders are computed. We see from this that there is a smallest positive remainder, r n 1 (x). If we set a(x) = r 1 (x) and b(x) = r 0 (x) then each remainder from the first one is a linear combination of the two remainders preceding it and hence all are linear combinations of a(x) and b(x). This means, by Corollary 3.2 that every common divisor of a(x) and b(x) divides each of the remainders. Notice that the last equation shows that r n 1 (x) divides r n 2 (x). Since r n 3 (x) is a linear combination of r n 1 (x) and r n 2 (x), we see r n 1 (x) also divides r n 3 (x). Following the equations back in like fashion we see that r n 1 (x) divides all the remainders and, as we reach the top of the stack of equations, a(x) and b(x) as well. This means that rn 1(x) is a common divisor of a(x) and b(x). The polynomial rn 1(x) thus satisfies the definition of the greatest common divisor of a(x) and b(x) and the algorithm works as claimed. By tracing back though the equations generated while running the Euclidean algorithm it is possible to actually find the polynomial coefficients s(x) and t(x)
18 140 CHAPTER 5. RINGS Example Find the GCD of x 3 +4x 2 +7x+6 and x 3 +3x 2 +3x+2 in Q[x] and find the polynomial coefficients that realize the GCD. Solution: x 3 + 4x 2 + 7x + 6 = (1)(x 3 + 3x 2 + 3x + 2) + (x 2 + 4x + 4) x 3 + 3x 2 + 3x + 2 = (x 1)(x 2 + 4x + 4) + (3x + 6) x 2 + 4x + 4 = 1 (x + 2)(3x + 6) and we have computed GCD(x 3 + 4x 2 + 7x + 6, x 3 + 3x 2 + 3x + 2) = 3x + 6 or x + 2 since we usually express the result as the unique monic associate. It remains to trace back: 3x + 6 = (x 3 + 3x 2 + 3x + 2) (x 1)(x 2 + 4x + 4) = (x 3 + 3x 2 + 3x + 2) (x 1) (x 3 + 4x 2 + 7x + 6) (x 3 + 3x 2 + 3x + 2) = [1 + (x 1)](x 3 + 3x 2 + 3x + 2) (x 1)(x 3 + 4x 2 + 7x + 6) = (x)(x 3 + 3x 2 + 3x + 2) (x 1)(x 3 + 4x 2 + 7x + 6) and so, passing to the associate, we have with s(x) = 1 3 x and t(x) = 1 3 (x 1) (x + 2) = 1 3 (x)(x3 + 3x 2 + 3x + 2) 1 3 (x 1)(x3 + 4x 2 + 7x + 6) Figure 5.2: An example of the polynomial Euclidean algorithm, including traceback. such that GCD(a(x), b(x)) = s(x)a(x) + t(x)b(x) An example of such a traceback appears in Figure 5.2. Our next step is to define the polynomial analog to a prime number. Definition 5.38 Let F be a field. A polynomial in F [x] is irreducible if all its divisors are units or associates. Notice that the units of F [x] are precisely the nonzero constant polynomials, i.e. all nonzero polynomials of degree zero. Example 5.18 The polynomials x, x+1, x 2 +x+1, x 3 +x+1, and x 3 +x 2 +1 are examples of irreducibles in Z 2 [x]. The polynomial x = (x + 1) 2 in Z 2 [x] and so is not irreducible. On the other hand x is irreducible in Q[x]. Proposition 5.37 Let F be a field. All polynomials of degree one in F [x] are irreducible. If a polynomial is of degree one its divisors are of degree 0 (and hence units) or 1 (and hence associates, by Proposition 5.34). Proposition 5.38 Suppose that F is a field and that p(x) F [x] is an irreducible polynomial. If f(x) F [x] is a polynomial such that p(x) f(x) then GCD(p(x), f(x)) = 1. The only divisors of p(x) are associates of p(x) and units. Since p(x) f(x) it follows that the only common divisors of p(x) and f(x) are units, making their greatest common divisor 1. Proposition 5.39 Suppose that F is a field. If p(x) F [x] is irreducible and a(x) and b(x) are poly
(Rgs) Rings Math 683L (Summer 2003)
(Rgs) Rings Math 683L (Summer 2003) We will first summarise the general results that we will need from the theory of rings. A unital ring, R, is a set equipped with two binary operations + and such that
More informationRings. Chapter 1. Definition 1.2. A commutative ring R is a ring in which multiplication is commutative. That is, ab = ba for all a, b R.
Chapter 1 Rings We have spent the term studying groups. A group is a set with a binary operation that satisfies certain properties. But many algebraic structures such as R, Z, and Z n come with two binary
More informationPolynomial Rings. i=0
Polynomial Rings 4152018 If R is a ring, the ring of polynomials in x with coefficients in R is denoted R[x]. It consists of all formal sums a i x i. Here a i = 0 for all but finitely many values of
More informationChapter 3. Rings. The basic commutative rings in mathematics are the integers Z, the. Examples
Chapter 3 Rings Rings are additive abelian groups with a second operation called multiplication. The connection between the two operations is provided by the distributive law. Assuming the results of Chapter
More informationRINGS: SUMMARY OF MATERIAL
RINGS: SUMMARY OF MATERIAL BRIAN OSSERMAN This is a summary of terms used and main results proved in the subject of rings, from Chapters 1113 of Artin. Definitions not included here may be considered
More informationAlgebraic structures I
MTH5100 Assignment 110 Algebraic structures I For handing in on various dates January March 2011 1 FUNCTIONS. Say which of the following rules successfully define functions, giving reasons. For each one
More informationCHAPTER I. Rings. Definition A ring R is a set with two binary operations, addition + and
CHAPTER I Rings 1.1 Definitions and Examples Definition 1.1.1. A ring R is a set with two binary operations, addition + and multiplication satisfying the following conditions for all a, b, c in R : (i)
More informationCourse 311: Michaelmas Term 2005 Part III: Topics in Commutative Algebra
Course 311: Michaelmas Term 2005 Part III: Topics in Commutative Algebra D. R. Wilkins Contents 3 Topics in Commutative Algebra 2 3.1 Rings and Fields......................... 2 3.2 Ideals...............................
More informationNOTES ON FINITE FIELDS
NOTES ON FINITE FIELDS AARON LANDESMAN CONTENTS 1. Introduction to finite fields 2 2. Definition and constructions of fields 3 2.1. The definition of a field 3 2.2. Constructing field extensions by adjoining
More informationPolynomial Rings. i=0. i=0. n+m. i=0. k=0
Polynomial Rings 1. Definitions and Basic Properties For convenience, the ring will always be a commutative ring with identity. Basic Properties The polynomial ring R[x] in the indeterminate x with coefficients
More informationFinite Fields: An introduction through exercises Jonathan Buss Spring 2014
Finite Fields: An introduction through exercises Jonathan Buss Spring 2014 A typical course in abstract algebra starts with groups, and then moves on to rings, vector spaces, fields, etc. This sequence
More informationRings and Fields Theorems
Rings and Fields Theorems Rajesh Kumar PMATH 334 Intro to Rings and Fields Fall 2009 October 25, 2009 12 Rings and Fields 12.1 Definition Groups and Abelian Groups Let R be a nonempty set. Let + and (multiplication)
More informationFactorization in Polynomial Rings
Factorization in Polynomial Rings Throughout these notes, F denotes a field. 1 Long division with remainder We begin with some basic definitions. Definition 1.1. Let f, g F [x]. We say that f divides g,
More informationDefinitions. Notations. Injective, Surjective and Bijective. Divides. Cartesian Product. Relations. Equivalence Relations
Page 1 Definitions Tuesday, May 8, 2018 12:23 AM Notations " " means "equals, by definition" the set of all real numbers the set of integers Denote a function from a set to a set by Denote the image of
More informationMATH 431 PART 2: POLYNOMIAL RINGS AND FACTORIZATION
MATH 431 PART 2: POLYNOMIAL RINGS AND FACTORIZATION 1. Polynomial rings (review) Definition 1. A polynomial f(x) with coefficients in a ring R is n f(x) = a i x i = a 0 + a 1 x + a 2 x 2 + + a n x n i=0
More informationOutline. MSRIUP 2009 Coding Theory Seminar, Week 2. The definition. Link to polynomials
Outline MSRIUP 2009 Coding Theory Seminar, Week 2 John B. Little Department of Mathematics and Computer Science College of the Holy Cross Cyclic Codes Polynomial Algebra More on cyclic codes Finite fields
More informationLecture 7: Polynomial rings
Lecture 7: Polynomial rings Rajat Mittal IIT Kanpur You have seen polynomials many a times till now. The purpose of this lecture is to give a formal treatment to constructing polynomials and the rules
More informationFactorization in Integral Domains II
Factorization in Integral Domains II 1 Statement of the main theorem Throughout these notes, unless otherwise specified, R is a UFD with field of quotients F. The main examples will be R = Z, F = Q, and
More informationExample: This theorem is the easiest way to test an ideal (or an element) is prime. Z[x] (x)
Math 4010/5530 Factorization Theory January 2016 Let R be an integral domain. Recall that s, t R are called associates if they differ by a unit (i.e. there is some c R such that s = ct). Let R be a commutative
More informationφ(xy) = (xy) n = x n y n = φ(x)φ(y)
Groups 1. (Algebra Comp S03) Let A, B and C be normal subgroups of a group G with A B. If A C = B C and AC = BC then prove that A = B. Let b B. Since b = b1 BC = AC, there are a A and c C such that b =
More informationAlgebra Homework, Edition 2 9 September 2010
Algebra Homework, Edition 2 9 September 2010 Problem 6. (1) Let I and J be ideals of a commutative ring R with I + J = R. Prove that IJ = I J. (2) Let I, J, and K be ideals of a principal ideal domain.
More informationGroups, Rings, and Finite Fields. Andreas Klappenecker. September 12, 2002
Background on Groups, Rings, and Finite Fields Andreas Klappenecker September 12, 2002 A thorough understanding of the Agrawal, Kayal, and Saxena primality test requires some tools from algebra and elementary
More information1. Algebra 1.5. Polynomial Rings
1. ALGEBRA 19 1. Algebra 1.5. Polynomial Rings Lemma 1.5.1 Let R and S be rings with identity element. If R > 1 and S > 1, then R S contains zero divisors. Proof. The two elements (1, 0) and (0, 1) are
More informationComputations/Applications
Computations/Applications 1. Find the inverse of x + 1 in the ring F 5 [x]/(x 3 1). Solution: We use the Euclidean Algorithm: x 3 1 (x + 1)(x + 4x + 1) + 3 (x + 1) 3(x + ) + 0. Thus 3 (x 3 1) + (x + 1)(4x
More information2a 2 4ac), provided there is an element r in our
MTH 310002 Test II Review Spring 2012 Absractions versus examples The purpose of abstraction is to reduce ideas to their essentials, uncluttered by the details of a specific situation Our lectures built
More informationTheorem 5.3. Let E/F, E = F (u), be a simple field extension. Then u is algebraic if and only if E/F is finite. In this case, [E : F ] = deg f u.
5. Fields 5.1. Field extensions. Let F E be a subfield of the field E. We also describe this situation by saying that E is an extension field of F, and we write E/F to express this fact. If E/F is a field
More informationA field F is a set of numbers that includes the two numbers 0 and 1 and satisfies the properties:
Byte multiplication 1 Field arithmetic A field F is a set of numbers that includes the two numbers 0 and 1 and satisfies the properties: F is an abelian group under addition, meaning  F is closed under
More informationALGEBRA II: RINGS AND MODULES OVER LITTLE RINGS.
ALGEBRA II: RINGS AND MODULES OVER LITTLE RINGS. KEVIN MCGERTY. 1. RINGS The central characters of this course are algebraic objects known as rings. A ring is any mathematical structure where you can add
More informationGEOMETRIC CONSTRUCTIONS AND ALGEBRAIC FIELD EXTENSIONS
GEOMETRIC CONSTRUCTIONS AND ALGEBRAIC FIELD EXTENSIONS JENNY WANG Abstract. In this paper, we study field extensions obtained by polynomial rings and maximal ideals in order to determine whether solutions
More informationMATH 326: RINGS AND MODULES STEFAN GILLE
MATH 326: RINGS AND MODULES STEFAN GILLE 1 2 STEFAN GILLE 1. Rings We recall first the definition of a group. 1.1. Definition. Let G be a non empty set. The set G is called a group if there is a map called
More informationMath 120 HW 9 Solutions
Math 120 HW 9 Solutions June 8, 2018 Question 1 Write down a ring homomorphism (no proof required) f from R = Z[ 11] = {a + b 11 a, b Z} to S = Z/35Z. The main difficulty is to find an element x Z/35Z
More information8 Appendix: Polynomial Rings
8 Appendix: Polynomial Rings Throughout we suppose, unless otherwise specified, that R is a commutative ring. 8.1 (Largely) a reminder about polynomials A polynomial in the indeterminate X with coefficients
More informationSection III.6. Factorization in Polynomial Rings
III.6. Factorization in Polynomial Rings 1 Section III.6. Factorization in Polynomial Rings Note. We push several of the results in Section III.3 (such as divisibility, irreducibility, and unique factorization)
More information(a + b)c = ac + bc and a(b + c) = ab + ac.
2. R I N G S A N D P O LY N O M I A L S The study of vector spaces and linear maps between them naturally leads us to the study of rings, in particular the ring of polynomials F[x] and the ring of (n n)matrices
More informationRings. EE 387, Notes 7, Handout #10
Rings EE 387, Notes 7, Handout #10 Definition: A ring is a set R with binary operations, + and, that satisfy the following axioms: 1. (R, +) is a commutative group (five axioms) 2. Associative law for
More informationMoreover this binary operation satisfies the following properties
Contents 1 Algebraic structures 1 1.1 Group........................................... 1 1.1.1 Definitions and examples............................. 1 1.1.2 Subgroup.....................................
More informationRings If R is a commutative ring, a zero divisor is a nonzero element x such that xy = 0 for some nonzero element y R.
Rings 10262008 A ring is an abelian group R with binary operation + ( addition ), together with a second binary operation ( multiplication ). Multiplication must be associative, and must distribute over
More informationClass Notes; Week 7, 2/26/2016
Class Notes; Week 7, 2/26/2016 Day 18 This Time Section 3.3 Isomorphism and Homomorphism [0], [2], [4] in Z 6 + 0 4 2 0 0 4 2 4 4 2 0 2 2 0 4 * 0 4 2 0 0 0 0 4 0 4 2 2 0 2 4 So {[0], [2], [4]} is a subring.
More informationCOMMUTATIVE RINGS. Definition 3: A domain is a commutative ring R that satisfies the cancellation law for multiplication:
COMMUTATIVE RINGS Definition 1: A commutative ring R is a set with two operations, addition and multiplication, such that: (i) R is an abelian group under addition; (ii) ab = ba for all a, b R (commutative
More informationPh.D. Qualifying Examination in Algebra Department of Mathematics University of Louisville January 2018
Ph.D. Qualifying Examination in Algebra Department of Mathematics University of Louisville January 2018 Do 6 problems with at least 2 in each section. Group theory problems: (1) Suppose G is a group. The
More informationHonors Algebra 4, MATH 371 Winter 2010 Assignment 3 Due Friday, February 5 at 08:35
Honors Algebra 4, MATH 371 Winter 2010 Assignment 3 Due Friday, February 5 at 08:35 1. Let R 0 be a commutative ring with 1 and let S R be the subset of nonzero elements which are not zero divisors. (a)
More informationMTH310 EXAM 2 REVIEW
MTH310 EXAM 2 REVIEW SA LI 4.1 Polynomial Arithmetic and the Division Algorithm A. Polynomial Arithmetic *Polynomial Rings If R is a ring, then there exists a ring T containing an element x that is not
More informationMath Introduction to Modern Algebra
Math 343  Introduction to Modern Algebra Notes Rings and Special Kinds of Rings Let R be a (nonempty) set. R is a ring if there are two binary operations + and such that (A) (R, +) is an abelian group.
More information1. Group Theory Permutations.
1.1. Permutations. 1. Group Theory Problem 1.1. Let G be a subgroup of S n of index 2. Show that G = A n. Problem 1.2. Find two elements of S 7 that have the same order but are not conjugate. Let π S 7
More informationPolynomial Rings : Linear Algebra Notes
Polynomial Rings : Linear Algebra Notes Satya Mandal September 27, 2005 1 Section 1: Basics Definition 1.1 A nonempty set R is said to be a ring if the following are satisfied: 1. R has two binary operations,
More informationMath 2070BC Term 2 Weeks 1 13 Lecture Notes
Math 2070BC 2017 18 Term 2 Weeks 1 13 Lecture Notes Keywords: group operation multiplication associative identity element inverse commutative abelian group Special Linear Group order infinite order cyclic
More informationChapter 14: Divisibility and factorization
Chapter 14: Divisibility and factorization Matthew Macauley Department of Mathematical Sciences Clemson University http://www.math.clemson.edu/~macaule/ Math 4120, Summer I 2014 M. Macauley (Clemson) Chapter
More informationHandout  Algebra Review
Algebraic Geometry Instructor: Mohamed Omar Handout  Algebra Review Sept 9 Math 176 Today will be a thorough review of the algebra prerequisites we will need throughout this course. Get through as much
More information0 Sets and Induction. Sets
0 Sets and Induction Sets A set is an unordered collection of objects, called elements or members of the set. A set is said to contain its elements. We write a A to denote that a is an element of the set
More informationALGEBRAIC GEOMETRY COURSE NOTES, LECTURE 2: HILBERT S NULLSTELLENSATZ.
ALGEBRAIC GEOMETRY COURSE NOTES, LECTURE 2: HILBERT S NULLSTELLENSATZ. ANDREW SALCH 1. Hilbert s Nullstellensatz. The last lecture left off with the claim that, if J k[x 1,..., x n ] is an ideal, then
More informationMath 4310 Solutions to homework 7 Due 10/27/16
Math 4310 Solutions to homework 7 Due 10/27/16 1. Find the gcd of x 3 + x 2 + x + 1 and x 5 + 2x 3 + x 2 + x + 1 in Rx. Use the Euclidean algorithm: x 5 + 2x 3 + x 2 + x + 1 = (x 3 + x 2 + x + 1)(x 2 x
More information2 Lecture 2: Logical statements and proof by contradiction Lecture 10: More on Permutations, Group Homomorphisms 31
Contents 1 Lecture 1: Introduction 2 2 Lecture 2: Logical statements and proof by contradiction 7 3 Lecture 3: Induction and WellOrdering Principle 11 4 Lecture 4: Definition of a Group and examples 15
More informationFactorization in Polynomial Rings
Factorization in Polynomial Rings These notes are a summary of some of the important points on divisibility in polynomial rings from 17 and 18. PIDs Definition 1 A principal ideal domain (PID) is an integral
More informationModern Computer Algebra
Modern Computer Algebra Exercises to Chapter 25: Fundamental concepts 11 May 1999 JOACHIM VON ZUR GATHEN and JÜRGEN GERHARD Universität Paderborn 25.1 Show that any subgroup of a group G contains the neutral
More informationALGEBRA. 1. Some elementary number theory 1.1. Primes and divisibility. We denote the collection of integers
ALGEBRA CHRISTIAN REMLING 1. Some elementary number theory 1.1. Primes and divisibility. We denote the collection of integers by Z = {..., 2, 1, 0, 1,...}. Given a, b Z, we write a b if b = ac for some
More informationRings. Chapter Homomorphisms and ideals
Chapter 2 Rings This chapter should be at least in part a review of stuff you ve seen before. Roughly it is covered in Rotman chapter 3 and sections 6.1 and 6.2. You should *know* well all the material
More informationIntroduction to finite fields
Chapter 7 Introduction to finite fields This chapter provides an introduction to several kinds of abstract algebraic structures, particularly groups, fields, and polynomials. Our primary interest is in
More information2 ALGEBRA II. Contents
ALGEBRA II 1 2 ALGEBRA II Contents 1. Results from elementary number theory 3 2. Groups 4 2.1. Denition, Subgroup, Order of an element 4 2.2. Equivalence relation, Lagrange's theorem, Cyclic group 9 2.3.
More informationHonors Algebra 4, MATH 371 Winter 2010 Assignment 4 Due Wednesday, February 17 at 08:35
Honors Algebra 4, MATH 371 Winter 2010 Assignment 4 Due Wednesday, February 17 at 08:35 1. Let R be a commutative ring with 1 0. (a) Prove that the nilradical of R is equal to the intersection of the prime
More informationFinite Fields. Sophie Huczynska. Semester 2, Academic Year
Finite Fields Sophie Huczynska Semester 2, Academic Year 200506 2 Chapter 1. Introduction Finite fields is a branch of mathematics which has come to the fore in the last 50 years due to its numerous applications,
More informationA MODELTHEORETIC PROOF OF HILBERT S NULLSTELLENSATZ
A MODELTHEORETIC PROOF OF HILBERT S NULLSTELLENSATZ NICOLAS FORD Abstract. The goal of this paper is to present a proof of the Nullstellensatz using tools from a branch of logic called model theory. In
More information2. Prime and Maximal Ideals
18 Andreas Gathmann 2. Prime and Maximal Ideals There are two special kinds of ideals that are of particular importance, both algebraically and geometrically: the socalled prime and maximal ideals. Let
More informationLECTURE NOTES IN CRYPTOGRAPHY
1 LECTURE NOTES IN CRYPTOGRAPHY Thomas Johansson 2005/2006 c Thomas Johansson 2006 2 Chapter 1 Abstract algebra and Number theory Before we start the treatment of cryptography we need to review some basic
More information1 First Theme: Sums of Squares
I will try to organize the work of this semester around several classical questions. The first is, When is a prime p the sum of two squares? The question was raised by Fermat who gave the correct answer
More informationIntegral Extensions. Chapter Integral Elements Definitions and Comments Lemma
Chapter 2 Integral Extensions 2.1 Integral Elements 2.1.1 Definitions and Comments Let R be a subring of the ring S, and let α S. We say that α is integral over R if α isarootofamonic polynomial with coefficients
More informationCHAPTER 10: POLYNOMIALS (DRAFT)
CHAPTER 10: POLYNOMIALS (DRAFT) LECTURE NOTES FOR MATH 378 (CSUSM, SPRING 2009). WAYNE AITKEN The material in this chapter is fairly informal. Unlike earlier chapters, no attempt is made to rigorously
More informationMath 547, Exam 2 Information.
Math 547, Exam 2 Information. 3/19/10, LC 303B, 10:1011:00. Exam 2 will be based on: Homework and textbook sections covered by lectures 2/33/5. (see http://www.math.sc.edu/ boylan/sccourses/547sp10/547.html)
More informationMath 121 Homework 5: Notes on Selected Problems
Math 121 Homework 5: Notes on Selected Problems 12.1.2. Let M be a module over the integral domain R. (a) Assume that M has rank n and that x 1,..., x n is any maximal set of linearly independent elements
More informationALGEBRA PH.D. QUALIFYING EXAM September 27, 2008
ALGEBRA PH.D. QUALIFYING EXAM September 27, 2008 A passing paper consists of four problems solved completely plus significant progress on two other problems; moreover, the set of problems solved completely
More informationHomework 8 Solutions to Selected Problems
Homework 8 Solutions to Selected Problems June 7, 01 1 Chapter 17, Problem Let f(x D[x] and suppose f(x is reducible in D[x]. That is, there exist polynomials g(x and h(x in D[x] such that g(x and h(x
More informationDiscrete Mathematics and Probability Theory Spring 2016 Rao and Walrand Discussion 6B Solution
CS 70 Discrete Mathematics and Probability Theory Spring 016 Rao and Walrand Discussion 6B Solution 1. GCD of Polynomials Let A(x) and B(x) be polynomials (with coefficients in R or GF(m)). We say that
More information(1) A frac = b : a, b A, b 0. We can define addition and multiplication of fractions as we normally would. a b + c d
The Algebraic Method 0.1. Integral Domains. Emmy Noether and others quickly realized that the classical algebraic number theory of Dedekind could be abstracted completely. In particular, rings of integers
More informationCommutative Rings and Fields
Commutative Rings and Fields 1222017 Different algebraic systems are used in linear algebra. The most important are commutative rings with identity and fields. Definition. A ring is a set R with two
More informationFurther linear algebra. Chapter II. Polynomials.
Further linear algebra. Chapter II. Polynomials. Andrei Yafaev 1 Definitions. In this chapter we consider a field k. Recall that examples of felds include Q, R, C, F p where p is prime. A polynomial is
More informationAlgebra Ph.D. Entrance Exam Fall 2009 September 3, 2009
Algebra Ph.D. Entrance Exam Fall 2009 September 3, 2009 Directions: Solve 10 of the following problems. Mark which of the problems are to be graded. Without clear indication which problems are to be graded
More informationMathematical Foundations of Cryptography
Mathematical Foundations of Cryptography Cryptography is based on mathematics In this chapter we study finite fields, the basis of the Advanced Encryption Standard (AES) and elliptical curve cryptography
More informationPolynomials. Chapter 4
Chapter 4 Polynomials In this Chapter we shall see that everything we did with integers in the last Chapter we can also do with polynomials. Fix a field F (e.g. F = Q, R, C or Z/(p) for a prime p). Notation
More informationChapter 4 Finite Fields
Chapter 4 Finite Fields Introduction will now introduce finite fields of increasing importance in cryptography AES, Elliptic Curve, IDEA, Public Key concern operations on numbers what constitutes a number
More informationTROPICAL SCHEME THEORY
TROPICAL SCHEME THEORY 5. Commutative algebra over idempotent semirings II Quotients of semirings When we work with rings, a quotient object is specified by an ideal. When dealing with semirings (and lattices),
More information32 Divisibility Theory in Integral Domains
3 Divisibility Theory in Integral Domains As we have already mentioned, the ring of integers is the prototype of integral domains. There is a divisibility relation on * : an integer b is said to be divisible
More informationMATH 403 MIDTERM ANSWERS WINTER 2007
MAH 403 MIDERM ANSWERS WINER 2007 COMMON ERRORS (1) A subset S of a ring R is a subring provided that x±y and xy belong to S whenever x and y do. A lot of people only said that x + y and xy must belong
More informationLecture 7.3: Ring homomorphisms
Lecture 7.3: Ring homomorphisms Matthew Macauley Department of Mathematical Sciences Clemson University http://www.math.clemson.edu/~macaule/ Math 4120, Modern Algebra M. Macauley (Clemson) Lecture 7.3:
More informationFinite Fields. Mike Reiter
1 Finite Fields Mike Reiter reiter@cs.unc.edu Based on Chapter 4 of: W. Stallings. Cryptography and Network Security, Principles and Practices. 3 rd Edition, 2003. Groups 2 A group G, is a set G of elements
More informationALGEBRA II: RINGS AND MODULES. LECTURE NOTES, HILARY 2016.
ALGEBRA II: RINGS AND MODULES. LECTURE NOTES, HILARY 2016. KEVIN MCGERTY. 1. INTRODUCTION. These notes accompany the lecture course Algebra II: Rings and modules as lectured in Hilary term of 2016. They
More informationMODEL ANSWERS TO HWK #10
MODEL ANSWERS TO HWK #10 1. (i) As x + 4 has degree one, either it divides x 3 6x + 7 or these two polynomials are coprime. But if x + 4 divides x 3 6x + 7 then x = 4 is a root of x 3 6x + 7, which it
More informationSome practice problems for midterm 2
Some practice problems for midterm 2 Kiumars Kaveh November 14, 2011 Problem: Let Z = {a G ax = xa, x G} be the center of a group G. Prove that Z is a normal subgroup of G. Solution: First we prove Z is
More informationMATH 361: NUMBER THEORY FOURTH LECTURE
MATH 361: NUMBER THEORY FOURTH LECTURE 1. Introduction Everybody knows that three hours after 10:00, the time is 1:00. That is, everybody is familiar with modular arithmetic, the usual arithmetic of the
More informationSUMMARY ALGEBRA I LOUISPHILIPPE THIBAULT
SUMMARY ALGEBRA I LOUISPHILIPPE THIBAULT Contents 1. Group Theory 1 1.1. Basic Notions 1 1.2. Isomorphism Theorems 2 1.3. Jordan Holder Theorem 2 1.4. Symmetric Group 3 1.5. Group action on Sets 3 1.6.
More informationBasic Algebra. Final Version, August, 2006 For Publication by Birkhäuser Boston Along with a Companion Volume Advanced Algebra In the Series
Basic Algebra Final Version, August, 2006 For Publication by Birkhäuser Boston Along with a Companion Volume Advanced Algebra In the Series Cornerstones Selected Pages from Chapter I: pp. 1 15 Anthony
More informationMath Introduction to Modern Algebra
Math 343  Introduction to Modern Algebra Notes Field Theory Basics Let R be a ring. M is called a maximal ideal of R if M is a proper ideal of R and there is no proper ideal of R that properly contains
More informationPolynomials, Ideals, and Gröbner Bases
Polynomials, Ideals, and Gröbner Bases Notes by Bernd Sturmfels for the lecture on April 10, 2018, in the IMPRS Ringvorlesung Introduction to Nonlinear Algebra We fix a field K. Some examples of fields
More informationPrime Rational Functions and Integral Polynomials. Jesse Larone, Bachelor of Science. Mathematics and Statistics
Prime Rational Functions and Integral Polynomials Jesse Larone, Bachelor of Science Mathematics and Statistics Submitted in partial fulfillment of the requirements for the degree of Master of Science Faculty
More informationMT5836 Galois Theory MRQ
MT5836 Galois Theory MRQ May 3, 2017 Contents Introduction 3 Structure of the lecture course............................... 4 Recommended texts..................................... 4 1 Rings, Fields and
More informationHomework 6 Solution. Math 113 Summer 2016.
Homework 6 Solution. Math 113 Summer 2016. 1. For each of the following ideals, say whether they are prime, maximal (hence also prime), or neither (a) (x 4 + 2x 2 + 1) C[x] (b) (x 5 + 24x 3 54x 2 + 6x
More informationϕ : Z F : ϕ(t) = t 1 =
1. Finite Fields The first examples of finite fields are quotient fields of the ring of integers Z: let t > 1 and define Z /t = Z/(tZ) to be the ring of congruence classes of integers modulo t: in practical
More information1/30: Polynomials over Z/n.
1/30: Polynomials over Z/n. Last time to establish the existence of primitive roots we rely on the following key lemma: Lemma 6.1. Let s > 0 be an integer with s p 1, then we have #{α Z/pZ α s = 1} = s.
More informationg(x) = 1 1 x = 1 + x + x2 + x 3 + is not a polynomial, since it doesn t have finite degree. g(x) is an example of a power series.
6 Polynomial Rings We introduce a class of rings called the polynomial rings, describing computation, factorization and divisibility in such rings For the case where the coefficients come from an integral
More information1 Rings 1 RINGS 1. Theorem 1.1 (Substitution Principle). Let ϕ : R R be a ring homomorphism
1 RINGS 1 1 Rings Theorem 1.1 (Substitution Principle). Let ϕ : R R be a ring homomorphism (a) Given an element α R there is a unique homomorphism Φ : R[x] R which agrees with the map ϕ on constant polynomials
More informationbe any ring homomorphism and let s S be any element of S. Then there is a unique ring homomorphism
21. Polynomial rings Let us now turn out attention to determining the prime elements of a polynomial ring, where the coefficient ring is a field. We already know that such a polynomial ring is a UFD. Therefore
More informationAbstract Algebra: Chapters 16 and 17
Study polynomials, their factorization, and the construction of fields. Chapter 16 Polynomial Rings Notation Let R be a commutative ring. The ring of polynomials over R in the indeterminate x is the set
More information