MATH 202 - ALGEBRA IV - HW 4 SOLUTIONS

CHRIS LEBAILLY, MIGUEL-ANGEL MANRIQUE, WEI YUAN

1. Section 14.3

4. Construct the finite field of 16 elements and find a generator for the multiplicative group. How manygenerators are there?

To construct the finite field of 16 elements, we begin with F2 and adjoin a root with a minimum polynomialof degree four. One such degree four irreducible polynomial is x4 + x + 1 (0 and 1 are not roots, and theonly irreducible polynomial of degree 2, namely x2 + x+ 1, does not divide it). As such, the field

F24 = F2[x]/(x4 + x+ 1)

is a field with 16 elements, and has a basis over F2 given by 1, x, x2, x3.It follows by Lagranges Theorem and the observation #F22 = 15 that x generates the unit group. Indeed,

x3 6= 1 and x5 = x x4 = x (x+ 1) = x2 + x 6= 1.There are (15) = (3)(5) = 2 4 = 8 generators.

6. Suppose K = Q() = Q(D1,D2) with D1, D2 Z, is a biquadratic extension and that = a+b

D1+

cD2 + d

D1D2 where a, b, c, d Z are integers, and at least two of b, c, d are nonzero. Prove that the

minimum polynomial m(x) for over Q is irreducible of degree 4 over Q but is reducible modulo everyprime p. In particular, show that the polynomial x4 10x+ 1 is irreducible in Z[x] but is reducible moduloevery prime.

Proof. The problem as stated in the book is missing a condition that we have included above: we need thatat least two of b, c, and d are nonzero, so let us suppose that this is the case.

The minimal polynomial of is of course irreducible; we must show it has degree 4. If not, then since[K : Q] = 4, we would have Q or [Q() : Q] = 2. Since 1, D1,

D2,D1D2 form a basis of K over

Q, we have 6 Q as long as at least one of b, c, and d is nonzero. If [Q() : Q] = 2, then is fixed by anontrivial element of Gal(K/Q) = Z/2Z Z/2Z. The elements of Gal(K/Q) are the automorphisms givenby

D1 7 D1,

D2 7

D2

for all 4 possible choices of . For each of the nontrivial automorphisms , we find that () = impliesthat at least two of b, c, and d are zero. For example, if :

D1 7

D1,D2 7

D2, then () =

implies

a+ bD1 c

D2 d

D1D2 = a+ b

D1 + c

D2 + d

D1D2,

which implies that c = d = 0. The other two cases are similar. Therefore, we see K = Q() and hence thatm(x) has degree 4.

Suppose that m(x) is irreducible modulo a prime p. Let be a root of m(x) in Fp, so [Fp() : Fp] = 4.Yet clearly Fp(

D1,D2). Since Fp/(Fp)2 has size 2, it follows that at least one of D1, D2, or D1D2 is

a square (possibly zero). Thus [Fp(D1,D2) : Fp] 2, giving a contradiction.

The polynomial x4 10x2 + 1 is the special case of = 2 +3.

Date: 06/01/2011.

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8. Determine the splitting field of xp x + a over Fp where a 6= 0, a Fp. Show explicitly that the Galoisgroup is cyclic. Such an extension is called an Artin-Schreier extension.

Proof. Let K be the splitting field of xpx+a over Fp. We have seen already in problem 13.5.5 (assignment6) that xp x+ a is irreducible and separable over Fp. Notice that if is a root of xp x+ a and k Fp,then

(+ k)p (+ k) + a = p + kp k + a = p + a = 0,since kp = k. Therefore, the roots of xp x+ a in K are precisely + k for k Fp, and K = Fp() = Fpp .

Since the Galois group is transitive on the roots, we furthermore have a bijection

Gal(K/Fp) = Fp = Z/pZgiven by 7 k such that () = +k. It is easy to verify that this bijection is in fact a group homomorphism,hence a group isomorphism:

() = + k1, () = + k2

implies

() = (+ k2) = () + (k2) = + k1 + k2.

9. (a) If x Fq, then q(x) = xq = x, so q fixes Fq.(b) Let L be a finite extension of Fq of degree n. Then L has qn elements, so by Lagranges theorem

aqn1 = 1 for all a L, and hence aqn = a for all a L. The polynomial xqn x Fq[x] can have at

most qn roots in any extension field, and weve demonstrated this many in L already, namely the qn distinctelements of L. Therefore xq

nx splits completely into linear factors over L, and not over any subfield (sinceevery element of L is a root). Since splitting fields are unique up to isomorphism, we see that any two fieldextensions of Fq of degree n are isomorphic via an isomorphism fixing Fq.

(c) We saw in (a) that the automorphism q fixes Fq and hence is an element of Gal(Fqn/Fq). Furthermore,if mq is the identity, then a

qm a = 0 for all a Fqn . Since the polynomial xqm x can have at most qmroots, we must therefore have m n. Furthermore, every element a Fqn does satisfy aqn a = 0 as notedabove, so nq = 1 in Gal(Fqn/Fq). Therefore, the order of q in Gal(Fqn/Fq) is n; since this is the size of theGalois group, we see that Gal(Fqn/Fq) is the cyclic group of size n generated by q.

(d) If Fqd Fqn , then Fqn is vector space over Fqd . By counting sizes, we see that qn is a power of qd;in other words, d divides n. Conversely, if d divides n then xq

d x divides xqn x, and hence the splittingfield of xq

d x is contained in the splitting field of xqn x, i.e. Fqd Fqn

2. Section 14.4

1. The minimal polynomial of the element =

1 +

2 is f(x) = (x2 1)2 2, which has degree 4. Theroots of this polynomial are and , where =

12. So the Galois closure of K = Q() over Q is

Q(, ), which has degree 2 over K and degree 8 over Q.

3. By the Theorem of Primitive Element, F = Q() for some F . To prove that [F : Q] n, it sufficesto prove that the minimal polynomial of over Q has degree at most n. By the Cayley-Hamilton Theorem, satisfies its characteristic polynomial, which has degree n. Therefore the minimal polynomial of hasdegree at most n, as desired.

6. Let K = Fp(x, y) and F = Fp(xp, yp). For each c F , let Lc = Fp(x+cy). Since (x+cy)p = xp+cpyp F ,we have [Lc : F ] p. However since clearly K = Lc(y) and y has degree p over F , we have [K : Lc] p.Since [K : F ] = p2, we must therefore have [K : Lc] = [Lc : F ] = p.

Suppose now that Lc = Lc for distinct c, c F . Let L = Lc = Lc . Since x+ cy, x+ cy L, we obtain

by subtracting that (c c)y L, so y L since c, c F L. Then clearly x L as well, so L = K,a contradiction to the calculation above. Therefore the fields Lc are distinct, and there are infinitely manysince they are indexed by c F , and F is infinite.

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3. Section 14.5

4. Since n is a primitive nth root of unity, any primitive nth root of unity can be written = bn for some

integer b. Thena() = a(

bn) = a(n)

b = (an)b = abn = (

bn)a = a

as desired.

5. Recall that p(x) = xp1 + xp2 + + 1 = p1i=1 (x i). By comparing the coefficients of xp2, we

see thatp1i=1 i = 1. Suppose first that p - n. Let n denoted the element of Gal(Q(p)/Q) such that

n() = n for all p. Then n(

p1i=1 i) =

p1i=1

ni . But the left hand side is n(1) = 1, giving the

desired result. Finally, if p | n, then of course ni = 1, sop1i=1

ni = p 1.

7. In C, any nth root of unity has the form = e2piia/n, and so

= e2piia/n = e2piia/n = 1.

Therefore complex conjugation restricts to the automorphism 1 on Q(). The subfield of real elements ofQ(), denoted Q()+, is therefore the fixed field of 1. Since 1 has order 2, we see that [Q() : Q()+] = 2.It is clear that + 1 is fixed by 1, since 1 swaps and 1. Therefore Q( + 1) Q()+. On theother hand, the element satisfies the polynomial

(x )(x 1) = x2 ( + 1)x+ 1 Q( + 1)[x].Therefore [Q() : Q( + 1)] 2, so we must have equality and we must have Q()+ = Q( + 1).13. (a) The fact that a(paii

) = apaii

follows from #4. Since paiiis a paii th root of unity, clearly a(paii

) =

apaii

depends only on a modulo paii .

(b) The map Gal(Q(n)/Q) =i Gal(Q(paii )/Q) is just the restriction map a 7 (a (mod paii ))i discussed

in part (a). In the Chinese Remainder Theorem, the isomorphism (Z/nZ) = i(Z/paii Z) is simply givenby reduction modulo paii for each i, i.e. a 7 (a (mod paii ))i. The result follows.

References

[1] Dummit and Foote, Abstract Algebra, Third Algebra.

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