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 ϕ be the homomorphism, and suppose that char F = p. Note that ϕ(1) = 1, since ϕ restricts to a homomorphism of abelian groups F × → (F 0)×. Then 0 = ϕ(1 + ··· + 1) = ϕ(1) + ··· + ϕ(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 0|p. Since p is prime we must have char F 0 = p, as desired.

Problem 3: Suppose that F is a finite field. Show that |F | = pn, where p = char F . Solution: Let ` be any prime dividing |F |. 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 p|`, and thus that ` = p. Therefore |F | = pn, as desired.

Problem 4: Prove that if F is a field of characteristic 0 then F contains a copy of Q. Solution: Let ϕ: 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 ϕ(mn) = ϕ(1 + ··· + 1) = ϕ(1 + ··· + 1) + ··· + ϕ(1 + ··· + 1) | {z } | {z } | {z } mn m m | {z } n = ϕ(m)(1 + ··· + 1) = ϕ(m)(ϕ(1) + ··· ϕ(1)) = ϕ(m)ϕ(n). | {z } | {z } n n We now extend ϕ to a field homomorphism ψ : Q → F by defining ψ(m/n) = ψ(m)ψ(n)−1. We need to check that this is well-defined. First, note that if m/n = p/q then mq = np so that ψ(m/n) = ϕ(m)ϕ(n)−1 = ϕ(m)ϕ(n)−1ϕ(mq)−1ϕ(np) = ϕ(m)ϕ(n)−1ϕ(m)−1ϕ(q)−1ϕ(n)ϕ(p) = ϕ(p)ϕ(q)−1. 1 2 MATH 205 HOMEWORK #1 OFFICIAL SOLUTION

We also need to check that this commutes with addition and multiplication. We have ψ(m/n + p/q) = ψ((mq + np)/nq) = ϕ(mq + np)ϕ(nq)−1 = ϕ(mq)ϕ(nq)−1 + ϕ(np)ϕ(nq)−1 = ϕ(m/n) + ϕ(p/q), and ψ((m/n)(p/q)) = ψ(mp/nq) = ϕ(mp)ϕ(nq)−1 = ϕ(m)ϕ(n)−1ϕ(p)ϕ(q)−1 = ψ(m/n)ψ(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 ∈ F [x] to be a formal sum of the form ∞ 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]\{0}) 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 ∈ F [x] to be a formal sum of the form ∞ 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 ∈ F ((x)) to be a formal sum of the form ∞ X n a = anx n=−∞

such that for some N ∈ 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: L∞ (a) As a group, F [x] is isomorphic to n=0 Z, so it is abelian. Let p(x), q(x), r(x) ∈ F [x]; write ∞ ∞ ∞ 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 ∞   ∞  
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    ∞     X X n n
=  aibj x x  r(x) = p(x)q(x) r(x).  n=0 i+j=n  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 ϕ to ψ. 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 P∞ n P∞ 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 × = {p(x) ∈ F x | p(0) 6= 0}. J K J K P∞ n P∞ n (d) The formula for the product of p(x) = n=−∞ anx and q(x) = n=−∞ bnx is ∞   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. α P∞ n Note that we can write every p(x) ∈ F ((x)) uniquely as x n=0 anx , where a0 6= 0. Let P∞ n −α −1 r(x) = n=0 anx . Then r(x) is invertible in F x ; then x r(x) is the multiplicative inverse of p(x). Thus F ((x)) is a field, as desired. J K (e) The homomorphism F → F [x] is given by sending an element a to the polynomial a + 0x + 0x2 + ··· . The homomorphism F [x] → F (x) is given by sending p(x) to (p(x), 1). The homomorphism F [x] → F x is given by sending p(x) to itself. The homomorphism F x → F ((x)) is given by sendingJ K p(x) to itself. JItK remains to construct the homomorphism ψ : F (x) → F ((x)). Let ϕ: F [x] → F ((x)) be the composite of the morphisms constructed above. Consider an element represented by (p(x), q(x)) in F (x) and define ψ((p(x), q(x))) = ϕ(p(x))ϕ(q(x))−1. We need to check that this is well-defined; this follows exactly the same outline as the proof in problem 4.

Problem 6: Let ι: F [x] → F (x) be the natural inclusion. Suppose that ϕ: F [x] → F 0 is any ring homomorphism. Show that there exists a field homomorphism ψ : F (x) → F 0 such that ϕ = ψι.

Solution: As before in problem 4, we define ψ(p(x), q(x)) = ϕ(p(x))ϕ(q(x))−1. The same proof as problem 4 shows that this is well-defined. Since ι takes p(x) to (p(x), 1), ψι(p(x)) = ϕ(p(x)), as desired.

P∞ n Problem 7: Suppose that p(x) ∈ F [x]; we write p(x) = n=0 anx . (a) Show that if a ∈ F such that p(a) = 0 then there exists a polynomial q(x) such that p(x) = (x − a)q(x). Use this to show that if p(x) is of degree n (the degree is the largest n such that an 6= 0) then there are at most n distinct elements r1, . . . , rn such that p(ri) = 0 for i = 1, . . . , n. (b) We define the derivative of p(x) to be

∞ 0 X n−1 p (x) = annx . n=0 Find a polynomial p(x) such that p0(x) = 0 but the degree of p(x) is greater than 0.

Solution: (a) We will prove this by induction on the degree of p(x). If the degree of p(x) is equal to 1 then p(x) = a1x + a0 = a1(x − a) + (a0 + a1a). Since p(a) = 0, we must have a0 + a1a = 0, and thus we can set q(x) = a1. Now suppose that this is true for all polynomials of degree MATH 205 HOMEWORK #1 OFFICIAL SOLUTION 5

Pn m n − 1 and consider p(x) = m=0 amx . We can write n−1 n−1 n−1 X m p(x) = anx (x − a) + aanx + amx . m=0 | {z } r(x) Since r(x) has degree less than n nad r(a) = 0 by the inductive hypothesis we can write r(x) = (x − a)s(x). Then n−1 n−1 p(x) = anx (x − a) + (x − a)s(x) = (x − a)(anx + s(x)), as desired. Suppose that we had n distinct elements r1, . . . , rn such that p(ri) = 0 for a p(x) of degree n. Using the above algorithm we can write p(x) = (x − r1)p1(x). Since r2 6= r1 and p(r2) = 0 we must have p1(r2) = 0, so we can apply the above algorithm again to p1(x) to get that p(x) = (x − r1)(x − r2)p2(x). Doing this inductively, we eventually conclude that p(x) = (x − r1) ··· (x − rn). But then for any r different from all of the ri’s,

p(r) = (r − r1) ··· (r − rn).

Since all of the ri were distinct, each term in the product is nonzero; thus since F is a field, their product cannot be zero. Thus there are no other elements r in F such that p(r) = 0, as desired. (b) If the characteristic of F is zero this is not possible. However, if char F = p then xp works.