D-MATH Algebra I HS17 Prof. Emmanuel Kowalski Solution 1 Algebraic closure, splitting field 1. Let K be a field of characteristic and L/K a field extension of degree. Show that there exists α L such that α K and L = K(α). What is irr(α, K)? Solution: First notice that char(l) as well, since otherwise 0 = 1 L = 1 K, contradiction. Hence 1 K L. Since [L : K] =, the extension is not trivial and there exists β L K. Then [L : K(β)][K(β) : K] = [L : K] = forces L = K(β) and in particular deg(irr(β, K)) = [K(β) : K] =. Write irr(β, K) = X + ax + b. Then ( 0 = β + aβ + b = β + a ) ( ) a 4 b, so that for α = β + a we see that α = a 4 b K and that K(α) = K(β) = L.. Let L = K(α)/K be a field extension such that [L : K] is odd. Prove that L = K(α ). Solution: Clearly, K(α ) K(α) = L. Notice that the element α L is a root of X α K(α )[X]. This implies that [L : K(α )] = deg(irr(α, K)). On the other hand, [L : K(α )][K(α ) : K] = [L : K] is odd, so that [L : K(α )] must be odd, too. Hence [L : K(α )] = 1, meaning that L = K(α ). 3. Let α = + 3 C (a) Show that α is algebraic over Q. (b) Compute [Q(α) : Q]. [Hint: α + α 1 Q(α)] (c) Determine irr(α, Q). Solution: (a) Since [Q( ) : Q] = and 3 is a root of X 3 Q( )[X], so that [Q(, 3) : Q( )], the extension Q(, 3)/Q is finite and hence algebraic. In particular, α = + 3 Q(, 3) is algebraic over Q. (b) Since Q(α) α+α 1 = + 3 + 3 = 3, we know that 3 Q(α) and = α 3 Q(α). This implies that Q(α) = Q(, 3). Notice that X 3 Q( )[X] is irreducible, because otherwise it would have a root in Q( ), which is not the case [indeed, writing a general element of Q( ) as s + t for s, t Q, we see that 3 = (s + t ) = s + t + st implies 1
that st = 0, so that 3 = s or 3 = t, which are impossible equalities]. Hence [Q(, 3) : Q( )] = and we can conclude that [Q(α) : Q] = [Q(, 3) : Q( )][Q( ) : Q] = = 4. (c) Squaring both sides of 3 = α we obtain 3 = α +, that is, = α 1. Squaring this equality, we get 8 = α 4 α + 1. Hence α is a root of the polynomial f := X 4 10X + 1, so that irr(α, Q) f. But deg(irr(α, Q)) = [Q(α) : Q] = 4 = deg(f), so that irr(α, Q) = f = X 4 10X + 1. 4. Let p be a prime number. For d 1, let N d be the number of monic irreducible polynomials in F p [X] of degree d. (a) Compute N 1 and N. (b) Let n Z >1. Using the description of finite fields stated in class, prove that X pn X is the product of all irreducible monic polynomials over F p whose degree divides n. (c) Deduce that dn d = p n, d n where d runs over divisors d 1 of n. (d) Prove that Solution: lim n N n (p n /n) = 1. [Hint: p n d n,d<n dn d = nn n p n. Notice that N d p d and d n/ for d < n. Use this to estimate 1 n d n,d<n dn d and conclude] (a) A monic polynomial of degree 1 in F p can be written as X + a for a F p. This is an irreducible polynomial for each a F p, so that N 1 = p. In degree, we see that there are p monic polynomials (corresponding to the choices of coefficients a 1, a in the expression X + a 1 X + a ). Among those, there are the non-irreducible polynomials, which can all written as (X b 1 )(X b ) in a unique way up to switching b 1 and b, so that there are p + ( ) p = p +p non-irreducible monic polynomials of degree. Hence N = p p + p = p p.
(b) Fix an algebraic closure F p of F p. As seen in class, for each m Z >1 there exists a unique subfield of F p containing p m elements, that is, F p m = {y F p : y pm = y}. In particular, F p n is the set of roots of X pn X and since its cardinality is equal to the degree of the polynomial and the polynomials X α F p n are coprime, we know that X pn X = (X α). α F p n For every α F p n we know that deg(irr(α, F p )) = [F p (α) : F p ] [F p n : F p ] = n, so that X α divides the product of all monic irreducible polynomials whose degree divides n. As the polynomials X α are pairwise coprime in F p n, we obtain that X pn X = (X α) f. α F p n f F p[x] irr. monic deg(f) n Conversely, let f F p [X] be a monic irreducible polynomial of degree d n. Let x F p be a root of f, so that f = irr(x, F p ) and [F p (x) : F p ] = d, which implies that Card(F p (x)) = p d. Hence F p (x) = F p d and in particular x pd = x. Write n = dl for l N. Since p n = (p d ) l, we see that x pn is obtained by repeatedly raising x to the p d -th power for l times, so that x pn = x. Hence x is a root of X pn X, which implies that f = irr(x, F p ) divides X pn X. By arbitrarity of f and since two distinct irreducible monic polynomials in F p [X] must be coprime, we obtain that f F p[x] irr. monic deg(f) n f X pn X. Since the two polynomials are associated in F p [X] and both monic, they must coincide. (c) This follows immediately from part (b), by comparing the degrees. Indeed, p n = deg(x pn X) = deg f F p[x] irr. monic deg(f) n f = ( d n f F p[x] irr. monic deg(f)=d = d n ) deg(f) dn d. 3
(d) The number of monic polynomials in F p of degree d is p d, so that N d p d for all d. If d is a proper divisor of n, then dl = n for l, so that d n/. In particular, the number of proper divisors of n is less than n/. Hence 1 n d n,d<n dn d 1 n n n pn = n 4 p n. (1) By the initial observation and by part (c), we know that p n d n,d<n dn d = nn n p n, which divided by p n gives By (1), we notice that 1 n p n 1 n d n,d<n dn d = N n p n /n 1 0 n p n 1 n d n,d<n dn d n p n n 4 p n = n 4p n n 0 and we can conclude that Nn p n /n 5. Prove that Q C is countable. 1 for n. Solution: First notice that Q is infinite because it contains N. If α is an algebraic number, there exists a unique irreducible polynomial f α Z[X] (in particular, f is primitive) with positive leading coefficient it is obtained by multiplying irr(α, Q) by the greatest common divisor of its coefficients. Thus for each α Q there exist unique n α N and a α 0,..., a α n α Z >0 such that a α n α > 0 and f α = a α n α X nα +... + a α 1 X + a α 0. Define, for N N, By construction, X N := {α Q : n α N, i = 1,..., n α a i N}. N N X N = Q. Moreover, each X N has a finite cardinality. Indeed, for a fixed N, there are N + 1 possible values of n α, for each of which there are no more than (N + 1) nα values for the coefficients a i and for each of the finitely many admissible tuples (n α, a α 0,..., a α n α ) which actually gives an irreducible polynomial there are at most n α roots that can be chosen as initial α Q. This implies that Q is a countable union of finite sets and as such it is countable. 4
6. Let K be a field and f K[X] a non-constant polynomial. Let L be a splitting field of f. Show that [L : K] deg(f)!. Solution: We recall the procedure used to prove existence of the splitting field. Let g be an irreducible factor of f. Then K can be seen as a subfield of K 1 := K[X]/(g) [X] =: α. In L 1, the image of f is divisible by X α. By uniqueness of the splitting field, the splitting field L of f over K is isomorphic as a K-extension to the splitting field L of h = f over X α K. Since deg(h) = deg(f) 1, we can work by induction on deg(f), and get [L : K] = [L : K] = [L : K 1 ][K 1 : K] = deg(g)[l : K 1 ] deg(f)[l : K 1 ] deg(f)(deg(f) 1)! = deg(f)! where in the last inequality we supposed that our result works for degree deg(f) 1. 7. (Trace and norm for finite field extensions) Let L/K be a finite field extension. (a) For x L, show that the following is a K-linear map: m x : L L y xy (b) Show that the map r L/K : L End K (L) sending x m x is a ring homomorphism. (c) Consider the maps Prove: Tr L/K : L K x Tr(m x ), N L/K : L K x det(m x ). (trace map) (norm map) Tr L/K is K-linear. N L/K (xy) = N L/K (x)n L/K (y) for all x, y L and N L/K (x) = 0 if and only if x = 0. (d) Given a tower of finite extensions L 1 /L /K, show that Tr L1 /K = Tr L /K Tr L1 /L. [Hint: Write down a K-basis of L 1 starting from a K-basis of L and an L -basis of L 1, then evaluate the right-hand side on α L 1 ]. (e) Prove that if x L is such that L = K(x), and irr(x, K)(X) = X d + a d 1 X d 1 + + a 1 X + a 0 K[X], then Tr L/K (x) = a d 1 and N L/K (x) = ( 1) d a 0. [Hint: (1, x,..., x d 1 ) is a K-basis of L.] 5
(f) Let p be an odd prime number, ζ = e πi p and K = Q(ζ) (see Assignment 11, Exercise 4). Compute: Tr K/Q (ζ), N K/Q (ζ) and N K/Q (ζ 1). Solution: (a) It is immediate to check K-linearity of each map m x. Indeed, m x is additive by distributivity of the multiplication with respect to addition, and it respect scalar multiplication by commutativity of the multiplication in L. (b) We immediately notice that m 0 = 0 and m 1 = id L. For x, y, z L, we have m x+y (z) = (x + y)z = xz + yz = m x (z) + m y (z) and m xy (z) = (xy)z = x(yz) = m x (m y (z)) = (m x m y )(z). This means that r L/K respects both sum and multiplication, and we can conclude that it is a ring homomorphism. As r L/K is not the zero map (since it sends 1 id L 0) and L is a field, the kernel is equal to (0), so that r L/K is injective. (c) First, we prove linearity of Tr L/K. Let n = [L : K] and fix a K-basis B for L. Then by basic linear algebra we have a K-linear ring isomorphism ϕ : End K (L) M n (K). Also, the trace map tr : M n (K) K is easily seen to be K-linear. Then by construction we have that Tr L/K = tr ϕ r L/K, which is K-linear as it is a composition of K-linear maps. As concerns norm, we have N L/K = det ϕ r L/K. Since all the composed maps respect multiplication, so does N L/K. Moreover, we have N L/K (x) = 0 if and only if det(m x ) = 0, which is equivalent to saying that m x is not an invertible endomorphism, and this happens precisely when x = 0 (since for x 0, me have m x 1 = m 1 x ). (d) Let B 1 = (e 1,..., e k ) be an L -basis for L 1, and B = (f 1,..., f l ) be an K-basis for L. As seen in class, B := (e 1 f 1, e 1 f,..., e 1 f l, e f 1,..., e f l,..., e k f 1,..., e k f l ) is a K-basis for L 1. For α L 1, we can find coefficients λ ij L, with 1 i, j k, so that for each i one has k α e i = λ ij e j. j=1 Then for each i, j as above and 1 s, t l we can find coefficients µ ijst L such that for each i, j and s one has l λ ij f s = µ ijst f t. Putting those two equalities together we get, for each i and t as above, k l α e i f s = µ ijst e j f t 6 j=1 t=1 t=1
Then the matrix correspondent to m α as a L -linear map of L 1, with respect to the basis B 1, is [m α ] L1 /L = T (λ ij ) i,j, so that Tr L1 /L (α) = k i=1 λ ii. Moreover, the matrix correspondent to m α as a K-linear map of L 1, with respect to the basis B, is [m α ] L1 /K = T (µ ijst ) (i,s),(j,t), where the row index is the couple (i, s) and the column index is the couple (j, t), and row (column) indexes are ordered with lexicographical order, so that Tr L1 /K(α) = k l i=1 s=1 µ iiss. Furthermore, for each i, j as before, the matrix correspondent to m λi,j as a K-linear map of L, with respect to the basis B, is so that Tr L /K(λ ij ) = l s=1 µ ijss. In conclusion, we have [m λi,j ] L /K = T (µ ijst ) s,t, Tr L /K(Tr L1 /L (α)) = Tr L /K( = k i=1 k λ ii ) = i=1 k Tr L /K(λ ii ) i=1 l µ iiss = Tr L1 /K(α). (e) Since L = K[X]/(irr(x, Q)) as field extensions of K and (1, x,..., x d 1 ) is a K-basis of L, we are interested in the matrix M x = (λ ij ) 0 i,j d 1 associated to m x. For j = 0,..., d, we have x x j = x j+1 so that we have { 1 if i = j + 1 λ ij =, for j = 0,..., d. 0 else. Moreover, x x d 1 = x d = a 0 a 1 x a d 1 x d 1, so that s=1 λ i,(d 1) = a i. What we have found is 0 0... 0 a 0 1 0... 0 a 1 M x = 0 1............. 0 ad 0... 0 1 a d 1 Then we get Tr L/K (x) = tr(m x ) = a d 1, and using Legendre form for the determinant on the first row we also obtain N L/K (x) = det(m x ) = ( 1) d a 0. 7
(f) By Assignment 11, Exercise 4, the minimal polynomial of ζ is Φ p := Xp 1 X 1. By part (e), Tr K/Q (ζ) = 1 and N K/Q (ζ) = 1, since p is odd. Notice that Q(ζ) = Q(ζ 1), so that Irr(ζ 1, Q) has degree p 1. Since ζ 1 satisfies G(X) := ϕ(x + 1) which is irreducible of degree p 1, we get Irr(ζ 1, Q) = (X + 1)p 1, X whose constant coefficient has been seen in Assignment 11, Exercise 4 to be equal to p. Then N L/K (ζ 1) = p. 8