Selected exercises from Abstract Algebra by Dummit and Foote (3rd edition). Bryan F´elix Abril 12, 2017 Section 7.1 Exercise 5. Decide which of the following are subrings of Q: (a) the set of all rational numbers with odd denominators (when written in lowest terms) Proof. Indeed, the set is a subring. First we show that it is a subgroup of Q by means of the subgroup criterion; i.e. a c Let b ; d be in the set. Then a c ad − bc − = b d bd is a rational number with odd denominator (since both b and d are odd). Furthermore a c ac · = b d bd is in the set by the same argument. Hence, the set is closed under multiplication, proving itself a subring. (b) the set of all rational numbers with even denominators (when written in lowest terms) 1 1 Proof. The set is not a subring as it ain't a group. Observe that 2 and 6 are in the set, but 1 1 1 − = 2 6 3 is not. (c) The set of nonnegative rational numbers Proof. Again, the set is not a subring since it is not a subgroup. Note that 1 and 3 are in the set, but 1 − 3 = −2 is not. (d) the set of squares of rational numbers 1 Proof. Again, the set is not a subring since it ain't a subgroup. Note that 4 is in the set, 1 1 1 1 but 4 + 4 = 2 is not since 2 is not the square of a rational number. (e) the set of all rational numbers with odd numerators (when written in lowest terms) 1 1 Proof. Again, the set is not a subring because it is not a group. Note that 3 is in the set, 1 1 2 but 3 + 3 = 3 is not. (f) the set of all rational numbers with even numerators (when written in lowest terms) 2 Proof. The set is a subring. Note that the set is not empty ( 3 is an element). We show that the set is a subgroup by means of the subgroup criterion. 2a 2c Let b and d be elements of the group. Note that both b and d are odd, otherwise the fraction is not written in lowest terms. Furthermore 2a 2c 2(ad − bc) − = b d bd is an element of the set, since the product bd is still odd and therefore, the factor 2 in 2a 2c the numerator survives. Then, the set is closed under multiplication, note that b · d = 2(2ac) bd : Exercise 12. Prove that any subring of a field which contains the identity is an integral domain. Proof. By construction, 1 is already in the set. Now we assume that the set is not an integral domain, and therefore, there exist two non-zero elements a and b, such that ab = 0. Since a and b are element of a field, they have multiplicative inverses and therefore the following holds a−1ab = a−10: Equivalently b = 0; a contradiction to the assumption that both a and b were non-zero elements. Exercise 14. Let x be a nilpotent element of a commutative ring R. (a) Prove that x is either zero or a zero divisor. Proof. Assume that xn = 0, where n is the least positive integer that satisfies the equation. Then x · xn−1 = 0, and therefore x is a zero divisor. (b) Prove that rx is nilpotent for all r 2 R. Proof. Let n be defined as in part (a). Then, since the ring is commutative, (rx)n = rnxn = 0 as desired. (c) Prove that 1 + x is a unit in R. Proof. Note that (1 + x)(1 − x + x2 − · · · ± xn−1) = (1 ± xn) = (1 + 0) = 1: Therefore 1 + x has a multiplicative inverse, and hence, it is a unit. (d) Deduce that the sum of a nilpotent element and a unit is a unit. Proof. Let a be a unit and x be a nilpotent element with xn = 0. Then, a+x = a(1+a−1x) where a−1x is nilpotent by part (b). Then, by part (c), (1 + a−1x) is a unit, and therefore a(1 + a−1x has a n inverse, namely, (1 + a−1x)−1a−1 . 2 Exercise 26. Let K be a field. A discrete valuation on K is a function v : K× ! Z satisfying (i) v(ab) = v(a) + v(b) (ii) v is surjective, and (iii) v(x + y) ≥ minfv(x); v(y)g for all x; y 2 K× with x + y 6= 0. The set R = fx 2 K×jv(x) ≥ 0g [ f0g is called the valuation ring of v: (a) Prove that R is a subring of K which contains the identity. Proof. Note that the homomorphism characterization of v guarantees that 1 is in R as v(1) = 0. Now we will show that R is a subgroup of the field by means of the subgroup criterion. Let a and b be elements of R. We want to prove that a − b is also an element of R. Assume that a − b 6= 0 (otherwise a − b is in R by construction) and observe that v(a−b) = v(a+(−b)) ≥ minfv(a); v(−b)g = minfv(a); v(−1·b)g = minfv(a); v(−1)+v(b)g: Note that 0 = v(1) = v(−1 · −1) = v(−1) + v(−1) = 2v(−1), therefore v(−1) = 0 and v(a − b) ≥ minfv(a); v(b)g ≥ 0 as both a and b are in R. It follows that a − b is in R and therefore R is an subgroup of the field. It is left to show that the group is closed under multiplication, to that end, note that for arbitrary a and b in R v(ab) = v(a) + v(b) ≥ 0 + 0 ≥ 0: We conclude that R is a subring that contains the multiplicative identity. (b) Prove that for each nonzero element x 2 K either x or x−1 is in R. Proof. Observe that v(x · x−1) = v(x) + v(x−1) = v(1) = 0 and assume that x is not in R. Then v(x) < 0 and the previous identity forces v(x) + v(x−1) = 0. It follows that v(x−1) = −v(x) > 0: Hence x−1 is in R: (c) Prove that an element x is a unit of R if and only if v(x) = 0: Proof. Assume x is a unit of R. Then, both v(x) ≥ 0 and v(x−1) ≥ 0. Furthermore v(x) + v(x−1) = v(xx−1) = v(1) = 0: The previous equation is satisfied only if v(x) = v(x−1) = 0. Now assume that v(x) = 0, then v(x−1) = −v(x) (by part (b)) and it follows that v(x−1) = 0: Therefore x−1 is in R and x is a unit in R. 3 Section 7.2 Exercise 5. Let F be a field and define the ring F ((x)) of formal Laurent series with coefficients from F by ( 1 ) X n F ((x)) = anx ; an 2 F and N 2 Z : n≥N (a) Prove that F ((x)) is a field. Proof. We prove the properties of a Field i. F ((x)) is an abelian group under addition. Note that 1 1 1 X n X n X n anx + bnx = (an + bn)x n≥N n≥M n≥minfN;Mg where we extend the series by using 0 coefficients if necessary. Likewise 1 1 1 X n X n X n bnx + anx = (bn + an)x : n≥M n≥N n≥minfN;Mg Since ai and bi are elements of a field, the previous resulting sums are equal. Associa- tivity and closure of + follows from the properties of polynomials. Lastly, the additive P n P n inverse of the element anx for n ≥ N is the sum (−an)x . ii. Multiplication distributes over addition and is commutative. Note that 1 1 1 ! 1 0 1 1 X n X n X n X n X n anx bnx + cnx = anx @ (bn + cn)x A n≥N n≥M n≥L n≥N n≥minfM;Lg To inspect the coefficients of the product we let di = (bi + ci) and K = minfM; Lg and we inspect the following multiplication table aN aN+1 aN+2 aN+3 ··· dK dK aN dK aN+1 dK aN+2 dK aN+3 ··· dK+1 dK+1aN dK+1aN+1 dK+1aN+2 dK+1aN+3 ··· dK+2 dK+2aN dK+2aN+1 dK+2aN+2 dK+2aN+3 ··· dK+3 dK+3aN dK+3aN+1 dK+3aN+2 dK+3aN+3 ··· . .. Note that the coefficient of the term xi are represented as diagonal sums of the entries of the table. Also, the order of multiplication is not important as all the entries come from a field. Furthermore, every coefficient is of the form aj · (bi + ci)ai and since the distributive property holds in F then every coefficient in our table can be rewritten as ajbi + ajci which proves the associative property. Commutativity follows in the same manner by rewriting diaj as ajdi. iii. There is a multiplicative identity. Our good ol' friend 1F (the unit in F ) satisfies. 4 iv. We have multiplicative inverses. We claim that the multiplicative inverse of 1 X n anx n≥N exist and is of the form 1 X n bnx : n≥−N This is clear from our multiplication table. The first term (in the upper left corner) is 0 the coefficient aN b−N of x .