# Math Introduction to Modern Algebra

Size: px
Start display at page:

Transcription

1 Math 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. (B) is an associative operation. (C) For all x, y, z R we have the left- and right-distributive laws, x (y +z) = x y +x z and (y + z) x = y x + z x. A commutative ring is a ring where the operation is commutative. Let s go ahead and immediately verify an important detail for a ring R: Let 0 be the additive identity in R. Then = 0. Let r R. Then 0 r = (0 + 0) r = 0 r + 0 r. Then it follows that 0 r = 0. Similarly, r 0 = 0. ELFY: Show that if R is a ring then for all x, y R we have that x( y) = (xy) = ( x)y and ( x)( y) = xy. Hint: Use the distributive laws to simplify the sums xy + x( y), ( x)y + xy, and then conclude that ( x)( y) = (x( y)) = ( (xy)) = xy since (xy) + xy = 0 and that inverses are unique. Examples: (1) R = {0} forms the trivial ring with = 0 and 0 0 = 0. (2) Z, Q, R, and C all are commutative rings with the usual addition and multiplication. (3) Let n N. The set nz is a commutative ring with the usual addition and multiplication.

2 (4) Let n N. The set Z n = {0, 1,..., n 1} is a commutative ring with addition modulo n and multiplication modulo n. (5) We know that (Z n, +) = (Z/nZ, +) (isomorphic as additive groups) and that Z n is a ring (see above). Naturally one can ask whether Z/nZ forms a ring. Let s define as a binary operation on Z/nZ by (a + nz) (b + nz) = (ab) + nz. We must show that this is well-defined. Suppose (a+nz) = (c+nz) and (b+nz) = (d+nz) where a, b, c, d Z. Then (a b), (c d) nz. So a c = nk and b d = nj for some k, j Z. Therefore, (a + nz) (b + nz) = (ab) + nz = ((c + nk)(d + nj) + nz) = (cd) + nz since cd (c + nk)(d + nj) = n( cj dk nkj) nz. Associativity, commutativity, and the distribution laws are inherited from Z. Hence, Z/nZ forms a commutative ring (that we will see is isomorphic to Z n as a ring). (6) Let n N. The set M nn forms a noncommutative ring with the usual matrix addition and matrix multiplication as (M nn, +) is an abelian group, matrix multiplication is associative (but not commutative) and distributes over matrix addition. (7) Let R 1, R 2,..., R n be rings. The direct product R 1 R 2 R n forms a ring with addition and multiplication done componentwise.

3 Special kinds of rings: A ring R with a multiplicative identity (we will write 1 for the multiplicative identity in general) is called a ring with unity and the identity 1 is called unity. In a ring with unity an element u R with a multiplicative inverse u 1 R is called a unit. Consider the set of units in a ring with unity: R X = {r R r 1 R, rr 1 = r 1 r = 1}. ELFY: Show that this forms a group under multiplication (it s known as unit group of the ring). Consider Z 10, which is a ring with unity. ELFY: 1, 3, 7, 9 are units while 2, 4, 5, 6, 8 are not. Let R be a ring. Let r R. r is a zero-divisor if r 0 and there exists s R such that s 0 and r s = 0 (and/or s r = 0). For example, 2 and 3 are zero-divisors in Z 6 as 2 3 = 0. A commutative ring with unity is an integral domain if it has no zero-divisors. For example, Z is an integral domain. A ring R is a field if it is a commutative ring and R = {x R x 0} forms a group under multiplication. For example, Q, R, and C are fields. So is Z p where p N is prime (we shall prove this later). Let R be a ring [or field]. S R is a subring [or subfield] of R if it is a ring [or field] itself (using the same operations). For instance, 2Z is a subring of Z and Q are a subfield of R. Notation: in both cases we use (and < if proper). So we write 2Z < Z and Q < R. A ring is always a subring of its polynomial ring: Let R be a ring and R[X] be its polynomial ring (to be formally defined later; it s the set of polynomials with coefficients from R). Then since R R[X] we have that R < R[X]. Proposition 9.1: Let n N. In the ring Z n the zero divisors are {m Z n gcd(m, n) 1}. Proof : Let m Z n where m 0. Let gcd(m, n) = d. Suppose d > 1. zero-divisor. Then mn d = m ( n d ) = ( m d ) n = 0. But m 0 and n d 0. So m is a Suppose d = 1. Suppose s Z n and ms = 0. Then n divides ms. Then n divides s showing that s = 0. Hence m is not a zero-divisor

4 Corollary 9.2: For any prime p N, Z p forms an integral domain. Proof : Just apply the previous proposition. For all m Z p, if m 0 then gcd(m, p) = 1. Hence Z p has no zero-divisors, making it an integral domain ELFY: Suppose R is a commutative ring. x, y, z R, xy = xz implies y = z. Show that R is an integral domain iff for all Proposition 9.3: Every finite integral domain is a field. Proof : Formally we regard the trivial integral domain {0} as the trivial field (where 0 = 1). Clearly, D = {0, 1} = Z 2 is a field. Suppose D is a finite integral domain with at least 3 elements. Let 0, 1, a 1, a 2,..., a n be all the elements of D (where the order of D is n + 2 {3,...}). Let a D where a 0. Consider the elements a 1, a a 1, a a 2,..., a a n. Since a b = a c implies b = c for all b, c D we have that the n + 1 products above are all distinct and non-zero. So, they must be 1, a 1, a 2,..., a n in some new order. Hence either a 1 = 1 or a a i = 1 for some i {1, 2,..., n}. In the former case, a = 1 is a unit. In the latter case a is a unit with a 1 = a i. So every non-zero element of D is invertible and hence forms a multiplicative group. Hence D is a field Corollary 9.4: For any prime p N, Z p forms a field Let R be a nontrivial ring. Suppose every element in R is of finite additive order. Then R is said to be of characteristic n if n is the least positive integer such that r+r+ +r = n r = 0 for all r R. Suppose it is not the case that every element in R is of finite additive order. Then R is said to be of characteristic 0. Obviously Z is of characteristic 0. Consider Z n where n N. Clearly n is the least integer such that nm = 0 for all m Z n. Hence Z n is of characteristic 0. Here is a cute example: Let R = Z 2 Z where the addition in the first component is modulo 2 and in the second component is the usual addition on Z. Clearly R has elements of finite additive order, such as (1, 0). But R also has elements that are not of finite additive order, like (0, 1). Consequently, R is of characteristic 0.

5 In a ring with unity, one can check for the characteristic of the ring by just examining what happens to 1: Let R be a non-trivial ring with unity. Suppose n 1 0 for all n N. Then clearly R is of characteristic 0 as not all elements are of finite additive order. Now assume 1 is of finite additive order. Let n be the least positive integer such that n 1 = 0. Let r R. Then adding r to itself n times gives n r = (1+1+ 1) r = (n 1) r = 0 r = 0. Since n r = 0 for all r R and n is the least positive integer such that n 1 = 0 it follows that n is the characteristic of R. ( ) n ELFY: Show that the binomial theorem (x + y) n n! = k!(n k)! xn k y k k=0 x, y in a commutative ring and any positive integer n. Hint: Use induction. holds for

6 Fermat s and Euler s Theorems Let p N be prime. Here is a neat consequence of the fact that Z p is a field: Z p forms a multiplicative group of order p 1. Since the order of any element divides the order of the group, we have that a p 1 = 1 for all a Z p. Using that Z p and Z/pZ are isomorphic as rings (to be proven later) we have that for all a Z, if a pz then (a + pz) p 1 = a p 1 + pz = 1 + pz. So for a Z, if a pz then a p 1 1 pz which implies a p 1 1 (mod p). We have just proved the following: Proposition 10.1 (Fermat s Little Theorem): If a Z and p N is a prime not dividing a then a p 1 1 (mod p) Corollary 10.2: If a Z then a p a (mod p) for any prime p N. Proof : Let a Z. Let p N be a prime. We consider two cases. If p does not divide a then a p a (mod p) holds by using the previous proposition. If p divides a then a p a (mod p) holds as both a and a p are equivalent to 0 modulo p Let s compute the remainder of when divided by 17: By Fermat s Little Theorem, for any positive integer k, (14 k ) 16 1 (mod 17). Then we can use that = (14 10 ) 16 to conclude that (mod 17). So modulo 17 we get ( 3) So the remainder of when divided by 17 is 7. Proposition 10.3: Let n 2 be an integer. Let G n be the set of elements in Z n that are not zero-divisors. G n forms a group under multiplication modulo n. Proof : Let s just use ab to denote the multiplication of a Z and b Z modulo n (which is associative). First we show G n is closed under multiplication modulo n. Let a, b G n. Suppose ab G n. Then there exists c Z n such that c 0 and (ab)c = 0. That implies a(bc) = 0. Since b G n and c 0, bc 0 (by the definition of G n ). But then a G n and bc 0 implies a(bc) 0 (by the definition of G n ). That contradiction tells us G n is closed under multiplication modulo n. Of course 1 G n as for all c Z n, 1c = c 0.

7 Let a G n. Let 1, a 1, a 2,..., a r be the r + 1 elements of G n. Then Consider the elements a1, aa 1, aa 2,..., aa r. Since ab = ac implies a(b c) = 0 and since a G n we have that ab = ac implies b = c for all b, c G n. So the r + 1 products above are all distinct and in G n. So, they must be 1, a 1, a 2,..., a r in some new order. Hence either a 1 = 1 or a a i = 1 for some i {1, 2,..., r}. In the former case, a = 1 is a unit and a 1 = 1 G n. In the latter case a is a unit with a 1 = a i G n. Either way a 1 G n. Thus G n is a group under multiplication modulo n Let s solve 7x 3 (mod 15): 7 is in G 15 as the gcd(7, 15) = 1. In G n, we have that 7 1 = 13 (ELFY: Verify this is correct). So the equation 7x = 3 in G 15 has the unique solution x = 7 1 (3) = (13)(3) = 9 in G 15. Let n N. The Euler phi-function, φ(n), is defined as the number of positive integers k < n that are relatively prime to n (meaning that the gcd(k, n) = 1). By proposition 9.1 we have that φ(n) is the order of the group G n (elements of Z n that are not zero-divisors). Proposition 10.4 (Euler s Theorem): If n N and a Z is relatively prime to n then a φ(n) 1 (mod n). Proof : Let n N and a Z be relatively prime to n. Then a + nz = b + nz for some positive integer b < n. Let d = gcd(b, n) N. Then b = kd and n = jd for some positive integers k, j. Since a b nz we have that a b = ln for some l Z. But then a kd = ljd implies that a = d(k + lj). Thus d = 1 as d is a common divisor of a and n. b is in G n and hence is of an order that divides φ(n). Thus b φ(n) = 1 in G n. Thus a φ(n) + nz = (a + nz) φ(n) = (b + nz) φ(n) = b φ(n) + nz = 1 + nz. Therefore a φ(n) 1 nz which yields a φ(n) 1 (mod n) Note: Euler s Theorem is a generalization of Fermat s Little Theorem. ELFY: Show that G 12 is a non-cyclic group of order 4 (thus isomorphic to the Klein 4-group).

8 Ring Homomorphisms Suppose R and R are rings and φ : R R satisfies φ(a + b) = φ(a) + φ(b) and φ(ab) = φ(a)φ(b) for all a, b R. Then φ is a ring homomorphism. If φ : R R is a bijection and a homomorphism then φ is an ring isomorphism and the rings R and R are called isomorphic and write R = R. Example: Let n N. Consider the rings Z n and Z/nZ. Recall that we know these are isomorphic as additive groups where an isomorphism φ : Z n Z/nZ is given by φ(m) = m + nz. To show that they are isomorphic as rings it suffices to show that φ(lm) = φ(l)φ(m) for all l, m Z n. Let l, m Z n. Suppose lm = j in Z n. Then in Z we have lm = kn + j where k Z. Then φ(l)φ(m) = (l + nz)(m + nz) = lm + nz = j + nz since lm j nz. Hence φ(l)φ(m) = φ(j) = φ(lm). Thus φ is a ring isomorphism showing Z n = Z/nZ (as rings). Suppose φ : R R is ring homomorphism. Since a ring homomorphism is a homomorphism of additive groups, forgetting about the other binary operation, we get that φ(0) = 0 and φ( a) = φ(a) for free from group theory! Caution: Unity need not go to unity. Consider the function φ : Z 6 Z 6 given by φ(m) = 3m (mod 6). Let m, n Z 6. Let m + n = j and mn = k in Z 6. Then φ(m + n) = φ(j) = 3j (mod 6) = 3(m + n) (mod 6) = 3m + 3n (mod 6) = φ(m) + φ(n). Furthermore φ(mn) = φ(k) = 3k (mod 6) = 3mn (mod 6) = 9mn (mod 6) = (3m)(3n) (mod 6) = φ(m)φ(n) So φ is a ring homomorphism from Z 6 to itself, but φ(1) = 3 1. Note: If φ is an onto ring homomorphism (aka a ring epimorphism) from a ring with unity to another ring, then that other ring has unity and φ(1) = 1. Let s see why: Let φ : R R be an onto ring homomorphism and 1 R (unity). Let y R. Since φ onto, there exists x R such that φ(x) = y. Therefore, consider that φ(1)y = φ(1)φ(x) = φ((1)x) = φ(x) = y and yφ(1) = φ(x)φ(1) = φ(x(1)) = φ(x) = y. So therefore R has unity with φ(1) = 1. Finally we just mention that the kernel of a ring homomorphism φ : R R is the kernel of φ viewed as an additive group homomorphism: Ker(φ) = {r R φ(r) = 0}.

9 Polynomial Rings Let R be a ring. For convenience, we say p(x) (or just p) is a polynomial over R if p(x) = k=0 a k x k = k a k x k where all but finitely many of the coefficients a k R for k Z 0 are equal to 0. By these definitions, polynomials over R are essentially power series with coefficients from R, but finite sums (of ring products) when evaluated at any r R. We define the degree of a polynomial p(x) over R, denoted deg(p), in two cases: The degree of the zero polynomial (all coefficients are 0) is. The degree of a non-zero polynomial p(x) = a k x k over R is n where n is the largest positive integer such that a n 0. Polynomial notation: Instead of always writing infinite sums, we write p(x) = a 0 + a 1 x + a n x n (or p(x) = 0) when p(x) is of degree n (or ). k=0 Define R[X] = {a 0 + a 1 x a n x n n Z 0, a i R}. By the discussion above, R[X] is the set of all polynomials over a ring R. We have the following rule for polynomial multiplication in R[X]: If p(x) = k a k x k and q(x) = k b k x k are in R[X] then the product is in R[X] since p(x)q(x) = d k x k where d k = k=0 k a i b k i R. i=0 It s pretty clear that (R[X], +) is an abelian group. Associativity and the distributive laws are straightforward to check, but are fairly tedious, so they are left as an ELFY. So R[X] forms a ring with polynomial addition and polynomial multiplication. We call R[X] the polynomial ring over R.

10 Consider x + 1 as polynomial in Z 2 [X]. Then (x + 1) 2 = x (since 2x = 0 as 2 = 0). Who said you can t distribute exponents!? valid!] [Don t get the wrong idea, generally it s not Let R[X] be a polynomial ring over a commutative ring R. Then for any r R, the function φ r : R[X] R given by φ r (p) = p(r) is a homomorphism called the evaluation homomorphism. Let s check that it is indeed a ring homomorphism: Let r R and φ r be given as above. Let p = a 0 + a 1 x + + a n x n and q = b 0 + b 1 x + + b n x n in R[X] (we aren t assuming they are of equal degree; for instance a n could be 0 with b n not 0). So φ r (p+q) = φ r ((a 0 +b 0 )+(a 1 +b 1 )x+ +(a n +b n )x n ) = (a 0 +b 0 )+(a 1 +b 1 )r+ +(a n +b n )r n = = (a 0 + a 1 r + + a n r n ) + (b 0 + b 1 r + + b n r n ) = φ r (p) + φ r (q). Thus φ r is an additive group homomorphism, which means that it suffices to check that φ r satisfies the multiplicative property φ r (fg) = φ r (f)φ r (g) where f, g are monomials in R[X]. So let f = ax i and g = bx j in R[X]. Then φ r (fg) = φ(abx i+j ) = abr i+j = abr i r j = (ar i )(br j ) = φ r (f)φ r (g). (ELFY): Let R be a commutative ring. Show that the evaluation homomorphism is onto. φ 0 : R[X] R Proposition 12.1 (Division Algorithm): Let F be a field (commutative ring where the non-zero elements form a multiplicative group). Let f = a 0 + a 1 x + + a n x n and g = b 0 + b 1 x + + b m x m be elements in F [X] where a n 0, b m 0, and m > 0. Then there exist unique polynomials q, r in F [X] such that f = gq + r and deg(r) < m. Proof : Consider the set S = {f gs s F [X]} which at least contains f. If 0 S then f = gs for some s F [x] which shows there exists q, r described above by taking q = s and r = 0 (which has degree which we regard as less than m). Otherwise, let r S be of minimal degree. Then r = f gs for some s F [X] and thus f = gs + r. We need to show that the degree of r is less than m. Well, suppose r = c 0 + c 1 x + c t x t in F [X] with c t 0 and t m. Then f g(s + c t b 1 m x t m ) = f gs g(c t b 1 m x t m ) = r (c t x t + lower degree terms) must be in S and have degree less than r, contradicting how r was chosen. Hence deg(r) < m.

11 Now suppose f = gq 1 + r 1 and f = gq 2 + r 2 where q 1, q 2, r 1, r 2 F [X] and the degrees of r 1 and r 2 are less than m. Then 0 = g(q 1 q 2 ) + (r 1 r 2 ) which implies r 2 r 1 = g(q 1 q 2 ) and the degree of r 2 r 1 is less than m, the degree of g. That implies that q 1 q 2 = 0. Hence q 1 = q 2. But then r 2 r 1 = 0. Hence r 1 = r 2 finishing the proof To actually do this in any field we can use the classic procedure: Let s work in Z 3 [X]. Let f = x 3 + x + 2 and g = 2x + 1. Let s find q, r such that f = qg + r and r is of degree or 0. 2x 2 + 2x + 1 2x + 1)x 3 + 0x 2 + x + 2 x 3 + 2x 2 x 2 + x x 2 + 2x 2x + 2 2x So we have that x 3 + x + 2 = (2x + 1)(2x 2 + 2x + 1) + 1 in Z 3 [X]. (ELFY) The Root-Factor Correspondence: Use the division algorithm to show that a F is a root (aka zero) of f F [X] iff f = q(x a) for some q F [X]. As a consequence, we get that a non-zero polynomial f F [X] of degree n can have at most n roots in F. A non-constant polynomial f F [X] is irreducible over F if it cannot be written as a product f = gh where both g, h F [X] have lower degree than f(x). Otherwise (meaning f can be written as such a product) the polynomial f is reducible over F. Of course the particular field being used matters! Consider that x 2 2 is irreducible over Q, but reducible over R or C as in those fields x 2 2 = (x 2)(x + 2).

12 Consider f = x 2 + x + 1 as a polynomial in Z 2 [X]. Notice that 0 and 1 are not roots of f as f(0) = 1 0 and f(1) = 1 0 (remember we are working in Z 2 here). Consequently, f is irreducible over Z 2 as it cannot be written as a product of two degree 1 polynomials in Z 2 [X]. However, if we consider f Z 3 [X] the story changes. f(1) = 0 which means x 1 = x + 2 is a factor of f in Z 3 [x]. However, f(0) = 1 0 and f(2) = 1 0, so thus x + 2 is the only factor of f, so f = (x + 2) 2. Indeed (x + 2) 2 = x 2 + x + 1 in Z 3 [X]. (ELFY) Let F be a field and f F [X] be of degree at least 2. Prove that if f is irreducible over F then f(r) 0 for all r F (that is, f does not have a root in F ). Hint: Use the division algorithm with x r as the divisor! Proposition 12.2: If R is an integral domain then so is R[X]. Proof : Since R is commutative and contains unity, it is clear that R[X] is commutative and contains unity. Let f, g R[X] be non-zero polynomials. Let a 0 and b 0 be their leading coefficients. Then the product ab is the leading coefficient of fg. Since R is an integral domain ab 0. Thus fg 0 showing R[X] to be an integral domain The following is an easy and useful result related to rational polynomials: Proposition 12.3 (Rational Root Theorem): Let f(x) = a n x n +a n 1 x n 1 + +a 1 x+a 0, where a n 0, be in Z[X] and nonconstant. Assume f(x) has a root a/b in Q such that (WLOG) a, b are relatively prime. Then a is an integer factor of a 0 and b is an integer factor of a n. Proof : Consider that 0 = b n f(a/b) = b n (a n (a/b) n + a n 1 (a/b) n a 1 (a/b) + a 0 ) = a n a n + a n 1 a n 1 b + + a 1 ab n 1 + a 0 b n. Then we get that 0 = a(a n a n 1 + a n 1 a n 2 b + + a 1 b n 1 ) + a 0 b n. This implies a is an integer divisor of a 0 b n and hence a must be an integer divisor of a 0 as a and b n are relatively prime. Similarly 0 = a n a n + b(a n 1 a n a 1 b n 2 a + a 0 b n 1 ) implies that b must be an integer divisor of a n

14 (ELFY) Suppose R is a ring with unity and I R is an ideal of R. Show that R/I is a ring with unity with 1 + I as the unity. We have ALREADY seen an example of a quotient ring, namely Z/nZ. We have also shown that given any ring homomorphism φ : R S, Ker(φ) is an ideal, therefore R/Ker(φ) is a quotient ring. One thing to be aware of is that structure can be lost or gained between a ring R and a quotient ring of R. Consider that Z is an integral domain, but Z 6 = Z/6Z is not. Consider that Z6 is not even an integral domain, while (ELFY) Z 6 /{0, 3} = Z 3 is a field. Proposition 13.3 (First Isomorphism Theorem): Let φ : R S is a ring homomorphism then φ(r) is a subring of S and R/Ker(φ) and φ(r) are isomorphic (as rings). Proof : (ELFY) It s easy to show that φ(r) is a subring, since one just needs to justify that ab φ(r) for all a, b φ(r), as it s already known to be an additive subgroup from group theory. For convenience, let I = Ker(φ). Let µ : R/I φ(r) be given by µ(r + I) = φ(r). Recall that we ve already shown this is well-defined and an isomorphism of R/I and φ(r) as additive groups (Prop. 8.4). Let s show that we have the multiplicative property. Let r + I, s + I R/I. Then µ((r + I)(s + I)) = µ(rs + I) = φ(rs) = φ(r)φ(s) = µ(r + I)µ(s + I). Thus R/Ker(φ) = φ(r) as rings Let R be a commutative ring with unity. (ELFY) It s very easy to show that for any r R, the additive subgroup (r) = {ra a R} is an ideal, called a principal ideal generated by r. Proposition 13.4: Let F be a field. Then every ideal in F [X] is a principal ideal generated by some f F [X]. Proof : F [X] is a commutative ring with unity so the concept of principal ideal makes sense. Clearly the zero ideal {0} is principal as (0) = {0}. Suppose I is a non-zero ideal in F [X]. Choose a non-zero polynomial f I with minimal degree. Without loss of generality assume the leading coefficient is 1 because otherwise we can multiply by the inverse of the leading coefficient (a constant polynomial in F [X]) and thereby remain in the ideal I. Clearly (f) I since I is an ideal. Let s prove that (f) = I. Suppose the degree of f is 0. Then f = 1 by our assumption that the leading coefficient is 1. Then for any g F [X], we have that g = g 1 so g (f). Hence (f) = F [X] since F [X] (f). In particular we have (f) = I = F [X].

15 So assume the degree of f is at least 1. By the division algorithm, for any g I, it follows that there exists polynomials q, r F [X] such that g = fq + r and deg(r) < deg(f). Since g, fq I it follows that r = g fq I. f was picked to the be a non-zero polynomial of minimum degree in I. So r = 0 and thus g = fq (f) showing that I (f) and thus I = (f) is a principal ideal We know that F [X] is an integral domain because F is (any field is an integral domain). Consequently F [X] is called a principal ideal domain (aka PID) because it is an integral domain and all ideals are principal. Consider the ideal (x 2 + 1) in R[X] (a PID). R[X]/(x 2 + 1): Let s play around with the quotient ring For convenience, let s set I = (x 2 + 1). Let f R[X]. If the degree of f is at least two, we can use the division algorithm to find polynomials q, r in R[X] such that f = (x 2 + 1)q + r and deg(r) < 2. Then f r (x 2 + 1) implies f + (x 2 + 1) = r + (x 2 + 1). So polynomials of degree two or larger aren t needed as coset representatives. Therefore R[X]/I = {a + bx + I a, b R}. Let s investigate the multiplication in the quotient ring: Let a + bx + I, c + dx + I R[X]/I. Then (a + bx + I)(c + dx + I) = (a + bx)(c + dx) + I = ac + adx + bcx + bdx 2 + I. By applying the division algorithm to x 2 divided by x we get x 2 = (x 2 + 1)(1) (1) which implies that x 2 + I = 1 + I. Using the above we have that ac + adx + bcx + bdx 2 + I = ac + adx + bcx bd + I since (ac + adx + bcx + bdx 2 ) (ac + adx + bcx bd) = bd(x 2 + 1) I. So here is how multiplication works in R[X]/I: Look familiar? (a + bx + I)(c + dx + I) = (ac bd) + (ad + bc)x + I. (ELFY) Show that φ : C R[X]/I given by φ(a + bi) = a + bx + I is a ring isomorphism. Let s look at a finite example. Consider the PID Z 2 [X] and the ideal I = (x 2 + x + 1). (ELFY) Show that Z 2 [X]/I = {0 + I, 1 + I, x + I, x I}. Observe that (x + I) 2 = x 2 + I = x I, (x + I) 3 = x 2 + x + I = 1 + I. So the non-zero elements of the commutative quotient ring Z 2 [X]/I is a multiplicative group (the cyclic group of order 3). Thus we have a field F 4 = Z 2 [X]/(x 2 + x + 1) with 4 elements! It turns out there is a finite field F p n of any prime power order.

### Math 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

### Rings. 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

### Chapter 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

### RINGS: 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 11-13 of Artin. Definitions not included here may be considered

### CHAPTER 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)

### Supplement. Dr. Bob s Modern Algebra Glossary Based on Fraleigh s A First Course on Abstract Algebra, 7th Edition, Sections 0 through IV.

Glossary 1 Supplement. Dr. Bob s Modern Algebra Glossary Based on Fraleigh s A First Course on Abstract Algebra, 7th Edition, Sections 0 through IV.23 Abelian Group. A group G, (or just G for short) is

### φ(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 =

### Course 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...............................

### NOTES 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

### Finite Fields. Sophie Huczynska. Semester 2, Academic Year

Finite Fields Sophie Huczynska Semester 2, Academic Year 2005-06 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,

### Eighth Homework Solutions

Math 4124 Wednesday, April 20 Eighth Homework Solutions 1. Exercise 5.2.1(e). Determine the number of nonisomorphic abelian groups of order 2704. First we write 2704 as a product of prime powers, namely

### Math 547, Exam 1 Information.

Math 547, Exam 1 Information. 2/10/10, LC 303B, 10:10-11:00. Exam 1 will be based on: Sections 5.1, 5.2, 5.3, 9.1; The corresponding assigned homework problems (see http://www.math.sc.edu/ boylan/sccourses/547sp10/547.html)

### Rings 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 non-empty set. Let + and (multiplication)

### Algebraic structures I

MTH5100 Assignment 1-10 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

### Finite Fields. Sophie Huczynska (with changes by Max Neunhöffer) Semester 2, Academic Year 2012/13

Finite Fields Sophie Huczynska (with changes by Max Neunhöffer) Semester 2, Academic Year 2012/13 Contents 1 Introduction 3 1 Group theory: a brief summary............................ 3 2 Rings and fields....................................

### Section 18 Rings and fields

Section 18 Rings and fields Instructor: Yifan Yang Spring 2007 Motivation Many sets in mathematics have two binary operations (and thus two algebraic structures) For example, the sets Z, Q, R, M n (R)

### Ideals, congruence modulo ideal, factor rings

Ideals, congruence modulo ideal, factor rings Sergei Silvestrov Spring term 2011, Lecture 6 Contents of the lecture Homomorphisms of rings Ideals Factor rings Typeset by FoilTEX Congruence in F[x] and

### Math 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

### CHAPTER 14. Ideals and Factor Rings

CHAPTER 14 Ideals and Factor Rings Ideals Definition (Ideal). A subring A of a ring R is called a (two-sided) ideal of R if for every r 2 R and every a 2 A, ra 2 A and ar 2 A. Note. (1) A absorbs elements

### Foundations of Cryptography

Foundations of Cryptography Ville Junnila viljun@utu.fi Department of Mathematics and Statistics University of Turku 2015 Ville Junnila viljun@utu.fi Lecture 7 1 of 18 Cosets Definition 2.12 Let G be a

### Math 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

### Abstract Algebra II. Randall R. Holmes Auburn University

Abstract Algebra II Randall R. Holmes Auburn University Copyright c 2008 by Randall R. Holmes Last revision: November 30, 2009 Contents 0 Introduction 2 1 Definition of ring and examples 3 1.1 Definition.............................

### Groups, 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

### ABSTRACT ALGEBRA MODULUS SPRING 2006 by Jutta Hausen, University of Houston

ABSTRACT ALGEBRA MODULUS SPRING 2006 by Jutta Hausen, University of Houston Undergraduate abstract algebra is usually focused on three topics: Group Theory, Ring Theory, and Field Theory. Of the myriad

### Moreover 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.....................................

### Algebra Review. Instructor: Laszlo Babai Notes by Vincent Lucarelli and the instructor. June 15, 2001

Algebra Review Instructor: Laszlo Babai Notes by Vincent Lucarelli and the instructor June 15, 2001 1 Groups Definition 1.1 A semigroup (G, ) is a set G with a binary operation such that: Axiom 1 ( a,

### Homework 10 M 373K by Mark Lindberg (mal4549)

Homework 10 M 373K by Mark Lindberg (mal4549) 1. Artin, Chapter 11, Exercise 1.1. Prove that 7 + 3 2 and 3 + 5 are algebraic numbers. To do this, we must provide a polynomial with integer coefficients

### Quizzes for Math 401

Quizzes for Math 401 QUIZ 1. a) Let a,b be integers such that λa+µb = 1 for some inetegrs λ,µ. Prove that gcd(a,b) = 1. b) Use Euclid s algorithm to compute gcd(803, 154) and find integers λ,µ such that

### Polynomial 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

### a b (mod m) : m b a with a,b,c,d real and ad bc 0 forms a group, again under the composition as operation.

Homework for UTK M351 Algebra I Fall 2013, Jochen Denzler, MWF 10:10 11:00 Each part separately graded on a [0/1/2] scale. Problem 1: Recalling the field axioms from class, prove for any field F (i.e.,

### Factorization 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,

### MATH 3030, Abstract Algebra FALL 2012 Toby Kenney Midyear Examination Friday 7th December: 7:00-10:00 PM

MATH 3030, Abstract Algebra FALL 2012 Toby Kenney Midyear Examination Friday 7th December: 7:00-10:00 PM Basic Questions 1. Compute the factor group Z 3 Z 9 / (1, 6). The subgroup generated by (1, 6) is

### MATH 420 FINAL EXAM J. Beachy, 5/7/97

MATH 420 FINAL EXAM J. Beachy, 5/7/97 1. (a) For positive integers a and b, define gcd(a, b). (b) Compute gcd(1776, 1492). (c) Show that if a, b, c are positive integers, then gcd(a, bc) = 1 if and only

### 0 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

### Polynomial Rings. (Last Updated: December 8, 2017)

Polynomial Rings (Last Updated: December 8, 2017) These notes are derived primarily from Abstract Algebra, Theory and Applications by Thomas Judson (16ed). Most of this material is drawn from Chapters

### Prime 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

### Total 100

Math 542 Midterm Exam, Spring 2016 Prof: Paul Terwilliger Your Name (please print) SOLUTIONS NO CALCULATORS/ELECTRONIC DEVICES ALLOWED. MAKE SURE YOUR CELL PHONE IS OFF. Problem Value 1 10 2 10 3 10 4

Quadratic Congruences, the Quadratic Formula, and Euler s Criterion R. C. Trinity University Number Theory Introduction Let R be a (commutative) ring in which 2 = 1 R + 1 R R. Consider a quadratic equation

### CSIR - Algebra Problems

CSIR - Algebra Problems N. Annamalai DST - INSPIRE Fellow (SRF) Department of Mathematics Bharathidasan University Tiruchirappalli -620024 E-mail: algebra.annamalai@gmail.com Website: https://annamalaimaths.wordpress.com

### ALGEBRA AND NUMBER THEORY II: Solutions 3 (Michaelmas term 2008)

ALGEBRA AND NUMBER THEORY II: Solutions 3 Michaelmas term 28 A A C B B D 61 i If ϕ : R R is the indicated map, then ϕf + g = f + ga = fa + ga = ϕf + ϕg, and ϕfg = f ga = faga = ϕfϕg. ii Clearly g lies

### 1 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

### Ph.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

### Name: Solutions Final Exam

Instructions. Answer each of the questions on your own paper. Be sure to show your work so that partial credit can be adequately assessed. Put your name on each page of your paper. 1. [10 Points] All of

### 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

### First Semester Abstract Algebra for Undergraduates

First Semester Abstract Algebra for Undergraduates Lecture notes by: Khim R Shrestha, Ph. D. Assistant Professor of Mathematics University of Great Falls Great Falls, Montana Contents 1 Introduction to

### Congruences and Residue Class Rings

Congruences and Residue Class Rings (Chapter 2 of J. A. Buchmann, Introduction to Cryptography, 2nd Ed., 2004) Shoichi Hirose Faculty of Engineering, University of Fukui S. Hirose (U. Fukui) Congruences

### Part IV. Rings and Fields

IV.18 Rings and Fields 1 Part IV. Rings and Fields Section IV.18. Rings and Fields Note. Roughly put, modern algebra deals with three types of structures: groups, rings, and fields. In this section we

### Some 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

### 1 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

### Math 581: Skeleton Notes

Math 581: Skeleton Notes Bart Snapp June 7, 2010 Chapter 1 Rings Definition 1 A ring is a set R with two operations: + called addition and called multiplication such that: (i) (R,+) is an abelian group

### Math 4400, Spring 08, Sample problems Final Exam.

Math 4400, Spring 08, Sample problems Final Exam. 1. Groups (1) (a) Let a be an element of a group G. Define the notions of exponent of a and period of a. (b) Suppose a has a finite period. Prove that

### SUMMARY ALGEBRA I LOUIS-PHILIPPE THIBAULT

SUMMARY ALGEBRA I LOUIS-PHILIPPE 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.

### A few exercises. 1. Show that f(x) = x 4 x 2 +1 is irreducible in Q[x]. Find its irreducible factorization in

A few exercises 1. Show that f(x) = x 4 x 2 +1 is irreducible in Q[x]. Find its irreducible factorization in F 2 [x]. solution. Since f(x) is a primitive polynomial in Z[x], by Gauss lemma it is enough

### (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

### I216e Discrete Math (for Review)

I216e Discrete Math (for Review) Nov 22nd, 2017 To check your understanding. Proofs of do not appear in the exam. 1 Monoid Let (G, ) be a monoid. Proposition 1 Uniquness of Identity An idenity e is unique,

### Computations/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

### INTRODUCTION TO THE GROUP THEORY

Lecture Notes on Structure of Algebra INTRODUCTION TO THE GROUP THEORY By : Drs. Antonius Cahya Prihandoko, M.App.Sc e-mail: antoniuscp.fkip@unej.ac.id Mathematics Education Study Program Faculty of Teacher

### (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

### AN INTRODUCTION TO THE THEORY OF FIELD EXTENSIONS

AN INTRODUCTION TO THE THEORY OF FIELD EXTENSIONS SAMUEL MOY Abstract. Assuming some basic knowledge of groups, rings, and fields, the following investigation will introduce the reader to the theory of

### D-MATH Algebra II FS18 Prof. Marc Burger. Solution 26. Cyclotomic extensions.

D-MAH Algebra II FS18 Prof. Marc Burger Solution 26 Cyclotomic extensions. In the following, ϕ : Z 1 Z 0 is the Euler function ϕ(n = card ((Z/nZ. For each integer n 1, we consider the n-th cyclotomic polynomial

### Factorization 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

### Abstract 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

### Section 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)

### Algebraic Structures Exam File Fall 2013 Exam #1

Algebraic Structures Exam File Fall 2013 Exam #1 1.) Find all four solutions to the equation x 4 + 16 = 0. Give your answers as complex numbers in standard form, a + bi. 2.) Do the following. a.) Write

### 2. THE EUCLIDEAN ALGORITHM More ring essentials

2. THE EUCLIDEAN ALGORITHM More ring essentials In this chapter: rings R commutative with 1. An element b R divides a R, or b is a divisor of a, or a is divisible by b, or a is a multiple of b, if there

### Lecture Notes Math 371: Algebra (Fall 2006) by Nathanael Leedom Ackerman

Lecture Notes Math 371: Algebra (Fall 2006) by Nathanael Leedom Ackerman October 17, 2006 TALK SLOWLY AND WRITE NEATLY!! 1 0.1 Integral Domains and Fraction Fields 0.1.1 Theorems Now what we are going

### Theorems and Definitions in Group Theory

Theorems and Definitions in Group Theory Shunan Zhao Contents 1 Basics of a group 3 1.1 Basic Properties of Groups.......................... 3 1.2 Properties of Inverses............................. 3

### Module MA3411: Abstract Algebra Galois Theory Michaelmas Term 2013

Module MA3411: Abstract Algebra Galois Theory Michaelmas Term 2013 D. R. Wilkins Copyright c David R. Wilkins 1997 2013 Contents 1 Basic Principles of Group Theory 1 1.1 Groups...............................

### 1 2 3 style total. Circle the correct answer; no explanation is required. Each problem in this section counts 5 points.

1 2 3 style total Math 415 Examination 3 Please print your name: Answer Key 1 True/false Circle the correct answer; no explanation is required. Each problem in this section counts 5 points. 1. The rings

### Handout - 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

### MATH 4107 (Prof. Heil) PRACTICE PROBLEMS WITH SOLUTIONS Spring 2018

MATH 4107 (Prof. Heil) PRACTICE PROBLEMS WITH SOLUTIONS Spring 2018 Here are a few practice problems on groups. You should first work through these WITHOUT LOOKING at the solutions! After you write your

### Solutions to Some Review Problems for Exam 3. by properties of determinants and exponents. Therefore, ϕ is a group homomorphism.

Solutions to Some Review Problems for Exam 3 Recall that R, the set of nonzero real numbers, is a group under multiplication, as is the set R + of all positive real numbers. 1. Prove that the set N of

### Definitions, Theorems and Exercises. Abstract Algebra Math 332. Ethan D. Bloch

Definitions, Theorems and Exercises Abstract Algebra Math 332 Ethan D. Bloch December 26, 2013 ii Contents 1 Binary Operations 3 1.1 Binary Operations............................... 4 1.2 Isomorphic Binary

### ALGEBRA 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

### MATH 581 FIRST MIDTERM EXAM

NAME: Solutions MATH 581 FIRST MIDTERM EXAM April 21, 2006 1. Do not open this exam until you are told to begin. 2. This exam has 9 pages including this cover. There are 10 problems. 3. Do not separate

### Algebra Exam Fall Alexander J. Wertheim Last Updated: October 26, Groups Problem Problem Problem 3...

Algebra Exam Fall 2006 Alexander J. Wertheim Last Updated: October 26, 2017 Contents 1 Groups 2 1.1 Problem 1..................................... 2 1.2 Problem 2..................................... 2

### Modern Algebra I. Circle the correct answer; no explanation is required. Each problem in this section counts 5 points.

1 2 3 style total Math 415 Please print your name: Answer Key 1 True/false Circle the correct answer; no explanation is required. Each problem in this section counts 5 points. 1. Every group of order 6

### Algebra Exam Syllabus

Algebra Exam Syllabus The Algebra comprehensive exam covers four broad areas of algebra: (1) Groups; (2) Rings; (3) Modules; and (4) Linear Algebra. These topics are all covered in the first semester graduate

### 6]. (10) (i) Determine the units in the rings Z[i] and Z[ 10]. If n is a squarefree

Quadratic extensions Definition: Let R, S be commutative rings, R S. An extension of rings R S is said to be quadratic there is α S \R and monic polynomial f(x) R[x] of degree such that f(α) = 0 and S

### Factorization 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

### Discrete Mathematics with Applications MATH236

Discrete Mathematics with Applications MATH236 Dr. Hung P. Tong-Viet School of Mathematics, Statistics and Computer Science University of KwaZulu-Natal Pietermaritzburg Campus Semester 1, 2013 Tong-Viet

### M2P4. Rings and Fields. Mathematics Imperial College London

M2P4 Rings and Fields Mathematics Imperial College London ii As lectured by Professor Alexei Skorobogatov and humbly typed by as1005@ic.ac.uk. CONTENTS iii Contents 1 Basic Properties Of Rings 1 2 Factorizing

### φ(a + b) = φ(a) + φ(b) φ(a b) = φ(a) φ(b),

16. Ring Homomorphisms and Ideals efinition 16.1. Let φ: R S be a function between two rings. We say that φ is a ring homomorphism if for every a and b R, and in addition φ(1) = 1. φ(a + b) = φ(a) + φ(b)

### MATH 3030, Abstract Algebra Winter 2012 Toby Kenney Sample Midterm Examination Model Solutions

MATH 3030, Abstract Algebra Winter 2012 Toby Kenney Sample Midterm Examination Model Solutions Basic Questions 1. Give an example of a prime ideal which is not maximal. In the ring Z Z, the ideal {(0,

### Finite Fields. Saravanan Vijayakumaran Department of Electrical Engineering Indian Institute of Technology Bombay

1 / 25 Finite Fields Saravanan Vijayakumaran sarva@ee.iitb.ac.in Department of Electrical Engineering Indian Institute of Technology Bombay September 25, 2014 2 / 25 Fields Definition A set F together

### Math 581 Problem Set 3 Solutions

Math 581 Problem Set 3 Solutions 1. Prove that complex conjugation is a isomorphism from C to C. Proof: First we prove that it is a homomorphism. Define : C C by (z) = z. Note that (1) = 1. The other properties

### 36 Rings of fractions

36 Rings of fractions Recall. If R is a PID then R is a UFD. In particular Z is a UFD if F is a field then F[x] is a UFD. Goal. If R is a UFD then so is R[x]. Idea of proof. 1) Find an embedding R F where

### MATH 113 FINAL EXAM December 14, 2012

p.1 MATH 113 FINAL EXAM December 14, 2012 This exam has 9 problems on 18 pages, including this cover sheet. The only thing you may have out during the exam is one or more writing utensils. You have 180

### Algebra SEP Solutions

Algebra SEP Solutions 17 July 2017 1. (January 2017 problem 1) For example: (a) G = Z/4Z, N = Z/2Z. More generally, G = Z/p n Z, N = Z/pZ, p any prime number, n 2. Also G = Z, N = nz for any n 2, since

### Note that a unit is unique: 1 = 11 = 1. Examples: Nonnegative integers under addition; all integers under multiplication.

Algebra fact sheet An algebraic structure (such as group, ring, field, etc.) is a set with some operations and distinguished elements (such as 0, 1) satisfying some axioms. This is a fact sheet with definitions

### = 1 2x. x 2 a ) 0 (mod p n ), (x 2 + 2a + a2. x a ) 2

8. p-adic numbers 8.1. Motivation: Solving x 2 a (mod p n ). Take an odd prime p, and ( an) integer a coprime to p. Then, as we know, x 2 a (mod p) has a solution x Z iff = 1. In this case we can suppose

### Modern Algebra Math 542 Spring 2007 R. Pollack Solutions for HW #5. 1. Which of the following are examples of ring homomorphisms? Explain!

Modern Algebra Math 542 Spring 2007 R. Pollack Solutions for HW #5 1. Which of the following are examples of ring homomorphisms? Explain! (a) φ : R R defined by φ(x) = 2x. This is not a ring homomorphism.

### Rings, Modules, and Linear Algebra. Sean Sather-Wagstaff

Rings, Modules, and Linear Algebra Sean Sather-Wagstaff Department of Mathematics, NDSU Dept # 2750, PO Box 6050, Fargo, ND 58108-6050 USA E-mail address: sean.sather-wagstaff@ndsu.edu URL: http://www.ndsu.edu/pubweb/~ssatherw/

### ALGEBRA QUALIFYING EXAM SPRING 2012

ALGEBRA QUALIFYING EXAM SPRING 2012 Work all of the problems. Justify the statements in your solutions by reference to specific results, as appropriate. Partial credit is awarded for partial solutions.

### 2a 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

### MATH 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

### Informal Notes on Algebra

Informal Notes on Algebra R. Boyer Contents 1 Rings 2 1.1 Examples and Definitions................................. 2 1.2 Integral Domains...................................... 3 1.3 Fields............................................