MATH 205 HOMEWORK #1 OFFICIAL SOLUTION Problem 2: Show that if there exists an injective field homomorphism F ! F 0 then char F = char F 0. Solution: Let ' be the homomorphism, and suppose that char F = p. Note that '(1) = 1, since ' restricts to a homomorphism of abelian groups F × ! (F 0)×. Then 0 = '(1 + ··· + 1) = '(1) + ··· + '(1) = 1 + ··· + 1 : | {z } | {z } | {z } p p p Thus p ≥ char F 0. Let r be the remainder when p is divided by char F 0. Then we know that in F 0, 0 = 1 + ··· + 1 = 1 + ··· + 1 + ··· + 1 + ··· + 1 + 1 + ··· + 1 = 1 + ··· + 1 : | {z } | {z } | {z } | {z } | {z } p char F 0 char F 0 r r By the minimality of char F 0 we know that r = 0, so char F 0jp. Since p is prime we must have char F 0 = p, as desired. Problem 3: Suppose that F is a finite field. Show that jF j = pn, where p = char F . Solution: Let ` be any prime dividing jF j. By the Sylow theorems, we know that there exists an element x of order ` in (F; +). But then by the same argument as in the previous problem we know that x + ··· + x = x(1 + ··· + 1) = x(1 + ··· + 1) | {z } | {z } | {z } ` ` ` mod p so by the minimality of p we know that pj`, and thus that ` = p. Therefore jF j = pn, as desired. Problem 4: Prove that if F is a field of characteristic 0 then F contains a copy of Q. Solution: Let ': Z ! F be a ring homomorphism defined by sending 1 to 1. Since Z = h1i as an abelian group under addition, this defines a perfectly good homomorphism of abelian groups (Z; +) ! (F; +); now we just need to check that it extends to a ring homomorphism. We check that '(mn) = '(1 + ··· + 1) = '(1 + ··· + 1) + ··· + '(1 + ··· + 1) | {z } | {z } | {z } mn m m | {z } n = '(m)(1 + ··· + 1) = '(m)('(1) + ··· '(1)) = '(m)'(n): | {z } | {z } n n We now extend ' to a field homomorphism : Q ! F by defining (m=n) = (m) (n)−1: We need to check that this is well-defined. First, note that if m=n = p=q then mq = np so that (m=n) = '(m)'(n)−1 = '(m)'(n)−1'(mq)−1'(np) = '(m)'(n)−1'(m)−1'(q)−1'(n)'(p) = '(p)'(q)−1: 1 2 MATH 205 HOMEWORK #1 OFFICIAL SOLUTION We also need to check that this commutes with addition and multiplication. We have (m=n + p=q) = ((mq + np)=nq) = '(mq + np)'(nq)−1 = '(mq)'(nq)−1 + '(np)'(nq)−1 = '(m=n) + '(p=q); and ((m=n)(p=q)) = (mp=nq) = '(mp)'(nq)−1 = '(m)'(n)−1'(p)'(q)−1 = (m=n) (p=q): Problem 5: (a) Define F [x] to be the ring of polynomials with coefficients in F . More formally, we define an element a 2 F [x] to be a formal sum of the form 1 X n a = anx n=0 where there exists some N such that for all n ≥ N, an = 0. Multiplication and addition are defined as usual for polynomials. Prove that F [x] is a ring but not a field. Find F [x]×. (b) Let F (x) be the quotient of the set F [x]×(F [x]nf0g) by the equivalence relation (p(x); q(x)) ∼ (r(x); s(x)) if p(x)s(x) = q(x)r(x). We define (p(x); q(x)) + (r(x); s(x)) = (p(x)s(x) + q(x)r(x); q(x)s(x)) and (p(x); q(x)) · (r(x); s(x)) = (p(x)r(x); q(x)s(x)): Show that these operations are well-defined and make F (x) into a field. Why is F (x) often called the field of fractions of F [x]? (c) Let F x be the ring of formal Taylor series with coefficients in F . More formally, we define an elmentJ K a 2 F [x] to be a formal sum of the form 1 X n a = anx : n=0 Multiplication and addition are defined as usual for power series. Prove that F x is a ring but not a field. Find F x ×. J K (d) Let F ((x)) be the ringJ ofK formal Laurent series with coefficients in F . More formally, we define an element a 2 F ((x)) to be a formal sum of the form 1 X n a = anx n=−∞ such that for some N 2 Z, if n < N then an = 0. Show that F ((x)) is a field. (e) Show that there exist ring homomorphisms F ! F [x] ! F (x) ! F ((x)) and F ! F [x] ! F x ! F ((x)): J K Solution: L1 (a) As a group, F [x] is isomorphic to n=0 Z, so it is abelian. Let p(x); q(x); r(x) 2 F [x]; write 1 1 1 X n X n X n p(x) = anx q(x) = bnx r(x) = cnx : n=0 n=0 n=0 MATH 205 HOMEWORK #1 OFFICIAL SOLUTION 3 Then 1 1 X X n X X n p(x) q(x)r(x) = p(x) bjck x = aibjck x n=0 j+k=n n=0 i+j+k=n j;k≥0 i;j;k≥0 0 1 1 B X X n nC = B aibj x x C r(x) = p(x)q(x) r(x): @ n=0 i+j=n A i;j≥0 Thus multiplication is associative. Note that multiplication is also commutative, since multiplication in F is commutative. Thus it suffices to check one distributive law. X X n p(x) q(x) + r(x) = ai(bj + cj) x n≥0 i+j=n i;j≥0 X X n X X n = aibj x + aicj x = p(x)q(x) + p(x)r(x): n≥0 i+j=n n≥0 i+j=n i;j geq0 i;j≥0 Thus F [x] is a ring. However, it is not a field because x has no inverse: for any p(x), xp(x) has the first coefficient equal to 0, not 1. We define the degree of p(x) to be the greatest n such that an 6= 0. Note that the degree of p(x)q(x) is equal to the sum of the degrees of p(x) and q(x), and that the degree of 1 (the identity) is 0. Thus the degrees of all units in F [x] must be 0. On the other hand, if the −1 degree of p(x) is 0 and p(x) is not equal to 0 then it is just equal to a0; then a0 is clearly its multiplicative inverse. Thus F [x]× = F ×. (b) We need to check that F (x) has well-defined operations. Supopse that (p(x); q(x)) ∼ (p0(x); q0(x)) and (r(x); s(x)) ∼ (r0(x); s0(x)); we need to check that (p(x); q(x)) + (r(x); s(x)) ∼ (p0(x); q0(x)) + (r0(x); s0(x)) and (p(x); q(x))(r(x); s(x)) ∼ (p0(x); q0(x))(r0(x); s0(x)): These follow from exactly the same reasoning we used in problem 4 to extend ' to . The additive identity is represented by (0; 1) and the multiplicative identity is represented by (1; 1). To see that F (x) is a field, note that the additive inverse of (p(x); q(x)) is (−p(x); q(x)). The pair (p(x); q(x)) 6∼ (0; 1) if and only if p(x) 6= 0. In this case, (p(x); q(x))−1 = (q(x); p(x)). F (x) is called the field of fractions of F [x] because its elements can be con- sidered to be the rational functions p(x)=q(x), the equivalence relation expresses when two such fractions represent the same function, and addition and multiplication is defined as it is for rational functions. (c) The formulas for addition and multiplication here as exactly the same as in part (1), so the same proof works to show that F x is a ring. Again, it is not a field because x does not have a multiplicative inverse. J K P1 n P1 n Suppose that p(x) = n=0 anx has a multiplicative inverse q(x) = n=0 bnx . We 0 will compute a formula for q(x) by induction. The coefficient of x in p(x)q(x) is a0b0, so −1 1 b0 = a0 ; thus we see that a0 6= 0. The coefficient of x in p(x)q(x) is a0b1 + a1b0, so −1 b1 = −a0 a1b0; this always exists when a0 6= 0. By induction, we see that bn can be defined in terms of the a0; a1; : : : ; an; b0; : : : ; bn−1 and will be well-defined exactly when a0 6= 0. Thus 4 MATH 205 HOMEWORK #1 OFFICIAL SOLUTION every p(x) with nonzero constant term is invertible in F x , and all invertible elements have nonzero constant term. We thus conclude that J K F x × = fp(x) 2 F x j p(0) 6= 0g: J K J K P1 n P1 n (d) The formula for the product of p(x) = n=−∞ anx and q(x) = n=−∞ bnx is 1 X X n p(x)q(x) = aibj x : n=−∞ i+j=n Note that this is well-defined, since there exist M and N such that ai = 0 for i < M and bj = 0 for j < N. The fact that the axioms hold for addition and multiplication here follow from the same principle as in part (a); we simply remove the conditions on the subscripts that say that the indices i; j; k are nonnegative and check that all stages of the computation are well-defined.