Solutions of MATH 322 Homeworks

(1) Express the symmetric polynomial

over Z in terms of the elementary symmetric polynomials.

Solution. We observe that the leading monomial of f(x, y, z) = x³y² + x²y³ + x³z² + x²z³ + y³z² + y²z³ is x³y² = x³y²z⁰, so we subtract s₁^3-2 s₂^2-0s₃⁰ from f and find

The leading monomial of f₁ is -2x³yz = -2x³y¹z¹, so we subtract -2s₁^3-1 s₂^1-1s₃¹ from f₁ and find

The leading monomial of f₂ is -x²y²z = -x²y²z¹, so we subtract -s₁^2-2 s₂^2-1s₃¹ from f₂ and find

Hence

(2) Find a polynomial over Z whose roots are the squares of the roots of t³ + 5t² + 7t + 1 in Z[t].

Solution. Let's denote the roots of p(t) : = t³ + 5t² + 7t + 1 by a, b, c. The coefficients of p(t) are the elementary symmetric polynomials in a, b, c, with a suitable sign:

Let q(t) = t³ + q₁t² + q₂t + q₃ be the monic polynomial having a², b², c² as roots. The relationship between its coefficients and roots is as follows:

so we have to find s_i(x², y², z²) and substitute a, b, c for x, y, z.

(i) Newton's formula for s₂ yields s₁(x², y², z²) = x² + y² + z² = s₂(x, y, z) = [s₁(x, y, z)]² - 2s₂(x, y, z). So s₁(a², b², c²) = [s₁(a, b, c)]² - 2s₂(a, b, c) = (-5)² - 2·7 = 11 and q₁ = -s₁(a², b², c²) = - 11.

(ii) The leading monomial of s₂(x², y², z²) = x²y² + x²z² + y²z² is x²y²z⁰, so we subtract s₁^2-2s₂^2-0s₃⁰ from it and find s₂(x², y², z²) - s₂² = . . . = -2(x²yz + xy²z + xyz²) = -2(xyz)(x + y + z) = -2s₁s₃. So s₂(x², y², z²) = s₂² - 2s₁s₃, so s₂(a², b², c²) = s₂²(a, b, c) - 2s₁(a, b, c)s₃(a, b, c) = 7² - 2(-5)(-1) = 39 and q₂ = s₂(a², b², c²) = 39.

(iii) We have s₃(x², y², z²) = x²y²z² = (xyz)² = [s₃(x, y, z)]², so s₃(a², b², c²) = [s₃(a, b, c)]² = (-1)² = 1 and q₃ = -s₂(a², b², c²) = -1.

Thus the required polynomial is t³ - 11t² + 39t - 1.

(3) Determine whether Q × Q is a vector space if the operations are defined by

for all a, b, c, d in Q.

Solution. The given system Q × Q is not an abelian group under addition, so it cannot be a vector space over Q. Indeed, as (1, 1) + (e, f) = (1 + e, 0) ≠ (1, 1) whatever (e, f) maybe, there is no (e, f) in Q × Q for which (1, 1) + (e, f) = (1, 1), so Q × Q has no additive identity, so Q × Q is not a group.

(4) If q is a positive integer and K is a field with q elements, how many elements does Kⁿ have?

Solution. The first entry in an n-tuple in Kⁿ can take on one of the q values from K, likewise the second, third, … , nth entries can take on q values, independently of the other entries, so there are q ·q ·q . . . q = qⁿ elements in Kⁿ.

Quiz Question Construct a field having 125 elements in it.

Solution. From MATH 321 we know that Z₅ is a field, from question (4) we know that Z₅³ has 125 elements in it, from the book we know that Z₅³ is a vector space over Z₅, where the operations are defined as follows:

for all a, b, c, d in Z₅.

Thus Z₅³ is a vector space (over Z₅) and has 125 elements in it.

(5) Find all Z₂-bases of Z₂³.

Solution. Since Z₂³ is finite dimensional over Z₂, and since dim _Z2 Z₂³ = 3, the Z₂-bases of Z₂³ are exactly the Z₂-linearly independent subsets of Z₂³ having 3 elements. So we must find three-element subsets of Z₂³ that are linearly independent over Z₂.

There are just two elements of Z₂, namely 0 and 1, so two nonzero vectors in Z₂³ are linearly dependent over Z₂ if and only if one of them is the only nonzero scalar 1 times the other one, i.e., if and only if they are equal. Also, if u and v are linearly independent over Z₂, then their span is {0, u, v, u + v}, so {u, v, w} is linearly independent over Z₂ if and only if w is distinct from 0, u, v, u + v.

So we are to find subsets {u, v, w} of Z₂³ such that u ≠ 0, v ≠ 0, v ≠ u, w ≠ 0, w ≠ u, w ≠ v, w ≠ u + v. The simplest method of doing this seems writing all the sets {u, v, w} satisfying all these conditions except the last one, and deleting those sets not enjoying the last property. The elements of Z₂³ being (0, 0, 0), (0, 0, 1), (0, 1, 0), (0, 1, 1), (1, 0, 0), (1, 0, 1), (1, 1, 0), (1, 1, 1) we obtain the following 35 = 7 choose 3 sets, where {(a, b, c), {d, e, f), (g, h, j)} is abbreviated as abc def ghj:

Quiz Question Find the automorphism group Aut(Q(³√5)).

Solution. Put u := ³√5. Let f be any automorphism of K := O(u) = Q(³√5) subset R. Then f|Q is an automorphism of Q, and from the homework solution, we know that the only automorphism of Q is id_Q, therefore f|Q is the identity function on Q. Besides, from u³ - 5 = 0, we get (u³)f-5f = 0f, so (uf)³-5 = 0, so uf is a root of the polynomial x³-5. The roots of this polynomial are u, uw, uw², and uw, uw² are nonreal complex numbers, so they are not in our field K. Thus uf, being an element of K, is necessarily the unique root u of this polynomial that belongs to K. Then we get also u²f = (uf)² = u². Hence (a + bu + cu²)f = af + bf uf + cf (u²)f = a + bu + cu² for any a, b, c in Q, that is to say, f is the identity function on K.

Thus Aut(K) = {id_K}.

Quiz Question Find a field having 7³ elements, describe its elements, explain how they are added and multiplied.

Solution. If f(x) is an irreducible cubic polynomial in F₇[x], then F₇[x]/<f(x)> is a field with exactly 7³ elements. Thus all we have to do is find a cubic polynomial that is irreducible over F₇. Put f(x) := x³ - 2 in F₇[x]. Then f(0) = - 2 ≠ 0, f(1) = - 1 ≠ 0, f(3) = 6 ≠ 0, f(4) = -1 ≠ 0, f(5) = 4 ≠ 0, f(6) = 4 ≠ 0, so f(x) has no root in F₇, so f(x) has no linear factor in F₇[x], so f(x) is irreducible over F₇. It follows that E := F₇[x]/<f(x)> is a field and |E| = 7³.

For a in F₇, we write a in place of a + <f(x)> and u in place of x + <f(x)>. Then the elements of E have the form a + bu + cu², with a, b, c in F₇. Addition and multiplication are carried out in the usual way, and u³ is replaced by 2 wherever it occurs. Thus

(a + bu + cu²) + (d + eu + fu²) = (a + d) + (b + e)u + (c + f)u²,

and

for a, b, c, d, e, f in F₇.

(12) Let E/K be a field extension, and write G := Aut_KE for short. Prove that, if H is a normal subgroup of G, then H'' is also a normal subgroup of G.

Solution. We know that H'' is a subgroup of G. We need only prove that it is normal in G. This follows from the definitions:

H is normal in G,

For all f in G, for all h in H, we have f^-1hf in H,

For all b in H', for all f in G, for all h in H, we have b(f^-1hf) = b,

For all b in H', for all f in G, for all h in H, we have (bf^-1)h = bf^-1,

For all b in H', for all f in G, we have bf^-1 in H',

For all a in H'', for all b in H', for all f in G, we have bf^-1a = bf^-1,

For all a in H'', for all b in H', for all f in G, we have bf^-1af = b,

For all a in H'', for all f in G, we have f^-1af in H'',

H'' is normal in G.

(13) Let E/K be a field extension, G : = Aut_KE and L : = G¢. Prove that E/L is Galois.

Solution. We put H : = Aut_LE. For any intermediate field M of E/L, we set M^° : = {j Î H : aj = a for all a Î M}; for any subgroup J of H, we set J^° : = {a Î E: aj = a for all j Î J}. We are to prove that L^°° = L.

Since L Ê K, any automorphism of E that fixes every element of L fixes also every element of K, so H = Aut_LE Í Aut_KE = G. On the other hand, by the definition of L = G¢, any automorphism in G fixes every element of L, so any automorphism in G belongs to H and G Í H. Thus H = G.

We get now

(15) Let p, k be natural numbers, p prime. Let F_p(x) denote the p^th cyclotomic polynomial over Q. Prove that, if d divides F_p(k), then d is congruent to 0 or 1 mod p.

Solution. In this solution, let ≈ stand for the congruence sign.

Suppose first that d is a prime divisor of F_p(k). Then d divides (k^p-1)/(k-1), so d divides k^p-1, so k^p ≈ 1 mod d, so k^*p = 1 in F_d^×. this shows that the order of k^*, as an element in the group F_d^×, is a divisor of p, so it is 1 or p.

If the order of k^* is 1, then k is congruent to 1 mod d, so p ≈ 1 + 1 + … + 1 ≈ k^p-1 + k^p-2 + … + 1 ≈ F_p(k) ≈ 0 mod d, so d divides p, and since both p and d are prime numbers, d = p. Thus d ≈ 0 mod p.

If the order of k^* is p, we use the fact that the element of any finite group is a divisor of the order of that group. We obtain that p divides f(d) = d-1, so d ≈ 1 mod p.

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