Rings. Chapter Definitions and Examples

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 3.11. 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

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, +, ). + 0 1 2 0 0 1 2 1 1 2 0 2 2 0 1 0 1 2 0 0 0 0 1 0 1 2 2 0 2 1 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 right-zero 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 non-commutative 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.

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 non-empty 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.

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 5.16. 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 non-zero 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.

5.1. 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 component-wise. 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.

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}: + 0 1 q v 0 0 1 q v 1 1 0 v q q q v 0 1 v v q 1 0 0 1 q v 0 0 0 0 0 1 0 1 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 non-zero 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.

5.1. DEFINITIONS AND EXAMPLES 129 0 1 2 a 1+a 2+a 2a 1+2a 2+2a 0 0 0 0 0 0 0 0 0 0 1 0 1 2 a 1+a 2+a 2a 1+2a 2+2a 2 0 2 1 2a 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 2 2 2 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 2 2 3 of 2 2 matrices over Z 3 : (i) a commutative ring with one? (ii) an integral domain? (iii) a field? Prove your answers.

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 non-commutative 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 3. 5.2 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.

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 5.14. 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 3.11. 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 non-empty 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.

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 + 0 = 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 5.22. 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.

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 2 + + 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 2 + + i n j n IJ. Then r(i 1 j 1 + i 2 j 2 + + i n j n ) = (ri 1 )j 1 + (ri 2 )j 2 + + (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 2 + + 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 2 + + ( 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.

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 non-zero element of F is a unit. This is sufficient to demonstrate that F is a field. Problems Problem 5.32 Prove Proposition 5.15. 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 5.19. Problem 5.35 Prove Proposition 5.22. 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.

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 5.12. Problem 5.40 Prove that the group of units of Z 3 Z 3 is isomorphic to the Klein-4 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

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 non-zero, (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 non-constant 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 5.31. 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 5.31. 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.

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 non-zero 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 non-zero 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 5.32. 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 non-zero 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 non-zero 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 non-zero polynomials. If f(x) g(x) and deg(f) = deg(g) then f(x) and g(x) are associates.

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 non-negative 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 non-zero 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 non-negative 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

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 non-zero 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 non-zero 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 5.33. 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)

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) + 0 3 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 non-zero 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 2 + 1 = (x + 1) 2 in Z 2 [x] and so is not irreducible. On the other hand x 2 + 1 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-