math202hw2sols

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MATH 202 - ALGEBRA IV - HW 6 SOLUTIONS

CHRIS LEBAILLY, MIGUEL-ANGEL MANRIQUE, JONATHAN SIEGEL, WEI YUAN

13.5 - Separable and Inseparable Extensions.(3) Prove thatd divides n if and only ifxd 1 divides xn 1.

Proof. Suppose that d | n. Then n = qd, soxn 1 =xqd 1 = (xd 1)(xdqd +xdq2d +xdq3d + +x2d +xd + 1)

Accordinglyxd 1 divides xn 1.Now suppose thatd n. Using the Euclidean Algorithm we can writen = qd + r where 0< r < d,

so

xn 1 = (xqd+r xr) + (xr 1) = xr(xqd 1) + (xr 1)

=x

r

(x

d

1)(xdqd

+x

dq2d

+ + 1) + (xr

1)It is clear that xd 1 divides the first term on the last line. However it does notdivide the secondsincexr 1 is non-zero and has degree less than xd 1 (since 0< r < d). Hencexd 1xn 1.

(4) Let a > 1 be an integer. Prove that for any positive integersn, d that d divides n if and only ifad 1 divides an 1 (cf. the previous exercise). Conclude in particular that Fpd Fpn if and onlyifp divides n.

Proof. The proof from the previous problem shows that d divides n if and only if ad 1 dividesan 1; the only change is that in the last step we argue by size rather than degree (if 0 < r < d,thenad 1 cannot divide ar 1 since ar 1 is nonzero, positive, and less than ad 1).

Applying this result with a = p, we see that pd 1 divides pn 1 if and only ifd divides n. Wecan think ofFpn as the union of

{0

}and Fpn =pn1

Fp, the group of (p

n

1)st roots of unity

in Fp. Therefore we have

d | n pd 1 | pn 1 pd1 pn1 Fpd Fpn.Note that the backwards part of the second equivalence requires a little argument here, using thefact that p (pd 1) and p (pn 1), so there exist primitive (pd 1)th and (pn 1)th roots ofunity. However, since the backwards implication (i.e. Fpd Fpn d|n) is clear from field degreeconsiderations, we dont delve into this further here.

(5) For any primep and any nonzeroa Fp prove that xp x +a is irreducible and separable over Fp.Proof. Let be a root off(x) = xp x+a in Fp. Accordingly, p +a = 0, and we see thata+k is another root for every k

Fp:

(+k)p (+k) +a= p +kp k+a= p +a= 0,since kp =k. This implies that f(x) is separable, since the p roots +k fork Fp are distinct. Itis also possible to prove separability using the derivative: sincef(x) = pxp1 1 =1, f(x) hasno roots and hence it is impossible for f(x) to have any multiple roots (by Proposition 33).

It remains to prove that f(x) is irreducible. Let m(x) be the minimal polynomial of over Fp.Write f(x) =m(x) g(x) for someg (x) Fp[x]. Ifg(x) = 1, then f(x) =m(x) is irreducible and

Date: 06/01/2011.

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were done. If not, g(x) has some roots in Fp; these roots must be roots off(x), and hence mustbe of the form +k for some k Fp. But it is easy to see that the minimal polynomial of+kis m(x k). Therefore,g(x) can be written as g(x) =m(x k)h(x) for some h(x)Fp[x]. Wethen obtainf(x) = m(x)m(x k)h(x).

Ifh(x) = 1, then we stop; if not we continue as before and see that h(x) is divisible bym(xj)for somej Fp. Continuing in this fashion, we eventually end afterd steps (for some d) and obtaina factorization of the form

f(x) = m(x k1)m(x k2) m(x kd)for some ki Fp with k1 = 0. Now the degrees of the polynomialsm(x ki) are all the same; letus call this degree n. By equating degrees we find p = nd. Since p is prime, we have either d = 1(in which case f(x) is irreducible as mentioned above) or n = 1, in which case Fp. But Fpimplies that 0 = f() =p +a = a, contrary to the assumption that a= 0. Thereforef(x) isirreducible as desired.

(7) Suppose K is a field of characteristic p which is not a perfect field K= Kp. Prove there existsirreducible inseparable polynomials over K. Conclude that there exist inseparable finite extensionsofK.

Proof. Pick a K such that a= cp

for any c F. We claim that xp

a is irreducible andinseparable.Let be a root ofxp a in a splitting field K. Then xp a= (x )p in K[x], so it is clearly

inseparable. We need to show that xpais irreducible inF[x]. Letm(x) be the minimal polynomialof overF[x], so m(x) is irreducible.

Since m(x) divides (x )p, we have m(x) = (x )d in K[x] for some positive integer d p.Lete be the maximal exponent such thatm(x)e divides xp a in F[x], say

(x )p =xp a= m(x)eg(x)Then g()= 0 since g() = 0 m(x)| g (x) m(x)e+1 | xp a, contrary to the definition ofe.Now sinceg (x) = (x )ped and is not a root ofg , we find that p = ed. This means thatd = 1ord = p. Ifd = 1 thenm(x) = xwhich means that F, which is a contradiction. Thus d = pwhich means that xp a= m(x), and hence is irreducible.

13.6 - Cyclotomic Polynomials and Extensions.(4) Prove that ifn = pkm wherep is a prime and m is relatively prime to p then there are precisely m

distinctnth roots of unity over a field of characteristic p.

Proof. To find the roots of unity we need to find the roots ofxn 1. Note that xn 1 = xpkm 1 =(xm1)pk . Sincem andp are relatively prime,xm1 and its derivativemxm1 are relatively prime.Thusxm1 does not have any multiple roots, meaning xm1 has preciselym distinct roots. Hencethere are precisely m distinct roots ofxn 1 as desired.

(8) Let l be a prime and let l(x) = xl1x1 = x

l1 +xl2 + +x + 1 Z[x] be the lth cyclotomicpolynomial, which is irreducible over Z by Theorem 41. This exercise determines the factorizationof l(x) modulo p for any prime p. Let denote any fixed primitive l

th root of unity.

(a) Show that ifp = l then l(x) = (x 1)l1 Fl[x].Proof. In Fp[x], we have (x 1)p =xp 1, and therefore xp1x1 = (x 1)p1 as desired.

(b) Suppose p =l and let fdenote the order ofp modulo l, i.e., fis the smallest power ofp withpf 1 (modl). Use the fact thatFpn is a cyclic group to show that n = f is the smallestpowerpn ofp with Fpn. Conclude that the minimal polynomial of overFp has degreef.Proof. First of all, note that a primitivelth root of unity exists in Fp by #4. Since l is a primeand= 1,has order dividing l, hence exactlyl. If Fpn for somen, thenl | pn1. Sincef

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is the smallest power ofp with that property, we see that Fpn n f. Now, since Fpf iscyclic, we can choose a generatorg . Then g is a primitive (pf 1)st root of unity, so g (pf1)/lis a primitivelth root of unity. Therefore Fpf.Weve therefore proved that Fp() = Fpf, so the minimal polynomial ofhas degree f.

(c) Show that Fp() = Fp(a

) for any integer a not divisible by l. [One inclusion is obvious. Forthe other, note that = (a)b where b is the multiplicative inverse ofa modulo l.] Concludeusing (b) that, in Fp[x], l(x) is the product of

l1f distinct irreducible polynomials of degree

f.Proof. It is obvious that Fp(

a) Fp(). For the reverse containment, letb be such that ab 1(mod l). Then (a)b = , whence Fp() Fp(a). Thus Fp() = Fp(a), so the minimalpolynomial ofa has degree f as well.

Since l(x) factors as the productl1

a=1(xa) over Fp[x], it follows that l(x) factors over Fp[x]as the product of the minimal polynomials of various as. Each of these minimal polynomialsis irreducible and has degree f; there are (l1)/fof them by comparing degrees. Finally, theseminimal polynomials are distinct since l(x) is separable. (l(x) is separable by #4.)

(d) In particular, prove that, viewed in Fp[x], 7(x) = x6 +x5 + +x+ 1 is (x1)6 for p = 7, a

product of distinct linear factors for p 1 (mod 7), a product of 3 irreducible quadratics forp 6 (mod 7), a product of 2 irreducible cubics forp 2, 4 (mod 7), and is irreducible forp 3, 5 (mod 7).Proof. Forp = 7, this is part (a). Ifp1 1 (mod 7), we havef= 1 and part (c) tells us 7(x)splits into 6 linear factors. If p 6 (mod 7), squaring both sides we see that p2 36 1(mod 7) sof= 2. Thus in this case 7(x) splits into 2 cubics. Now ifp 2, 4 (mod 7), we getf= 3 and 7(x) splits into 3 quadratics. Lastly, for p 3, 5 (mod 7), we getf= 6 and 7(x)is irreducible.

(10) Let denote the Frobenius mapx xp on the finite field Fpn. Prove that gives an isomorphismofFpn to itself (such an isomorphism is called an automorphism). Prove that

n

is the identity mapand that no lower power of is the identity.

Proof. Letx1, x2 Fpn , then(1) = 1p = 1,

(x1+x2) = (x1+x2)p =xp1+x

p2

and(x1 x2) = (x1 x2)p =xp1 xp2,

i.e. is a endomorphism. On the other hand, let x Fpn be the element satisfying (x) = 0. Then(x) = xp = 0.

Hence x = 0 and is injective. Since Fpn is a finite set, is then also surjective. Hence, isbijective. Therefore, is an automorphism ofFpn .

SinceF

pn

is a cyclic group of order p

n

1, then forx F

pn

,xp

n1 = 1.

Hencen(x) = xpn

=x,x Fpn, i.e. n =id.Suppose for somek Z with 1 k n satisfies k =id. Then forx Fpn

k(x) = xpk

=x,

i.e. xpkx= 0, x Fpn . But the equation above has at mostpk solutions overFpn and |Fpn | =pn.

Therefore,k = n, i.e. no lower power of is the identity.

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(11) Let denote the Frobenius map x xp on the finite field Fpn as in the previous exercise. De-termine the rational canonical form over Fp for considered as an Fp-linear transformation of then-dimensionalFp-vector space Fpn .

Solution. We saw in the previous exercise thatn = 1. Furthermore, the argument given there thatno lower power of is the identity generalizes to show that satisfies no polynomial of degree lowerthann overFp; indeed given such a polynomial an1

n1 +an2n2 +

+a0 = 0 with ai

Fp,

we would have that an1xpn1 +an2xp

n2 + +a0x= 0 for all x Fpn, which is impossible bydegree considerations.

This means that the minimal polynomial of is xn 1. Since Fpn is a dimension n-vector spaceover Fp, this means that the minimal polynomial is the same as the characteristic polynomial. Inparticular, we have one invariant factor, namely xn 1, and the rational canonical form of overFp is the companion matrix ofx

n 1:

0 0 0 0 11 0 0 0 00 1 0 0 0...

......

. . . ...

...0 0 0

0 0

0 0 0 1 0

.

(12) Let denote the Frobenius mapx xp on the finite fieldFpn as in the previous exercise. Determinethe Jordan canonical form (over a field containing all the eigenvalues) for considered as an Fp-lineartransformation of then-dimensionalFp-vector spaceFpn.

Solution. We saw in the previous exercise that the minimal polynomial and characteristic polynomialof are both equal to xn 1. In particular, the eigenvalues of overFp are thenth roots of unityin Fp. As in #4, let us writen = pkm with p m. Then the number ofnth roots of unity is m,and each occurs in the polynomial xn

1 with multiplicity pk. Since the minimal polynomial and

characteristic polynomial of are equal, we see that each eigenvalue occurs in exactly one Jordanblock. Therefore, the Jordan canonical form of contains m Jordan blocks of size pk, one for eachmth root of unity in Fp:

J0 0 0 0 00 J1 0 0 00 0 J2 0 0...

......

. . . ...

...0 0 0 Jm2 00 0 0 0 Jm1

,

whereJi is a pk pk Jordan block as follows:

i 1 0 0 00 i 1 0 00 0 i 0 0...

......

. . . ...

...0 0 0 i 10 0 0 0 i

,

witha primitive mth root of unity.

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14.1 - Basic Definitions.(1) (a) Show that if the fieldKis generated overFby the elements 1, . . . , n then an automorphism

of K fixing F is uniquely determined by (1), . . . , (n). In particular, show that anautomorphism fixesKif and only if it fixes a set of generators for K.Proof. Suppose Aut(K/F). Since is a homomorphism and fixes F, for any polynomialf(x1, . . . , xn)F[x], we find that (f(1, . . . , n)) =f((1), . . . , (n)). Furthermore givenanother such polynomial g , we find that

f(1, . . . , n)

g(1, . . . , n)

=

f((1), . . . , (n))

g((1), . . . , (n)). (1)

Since K is generated over F by1, , n, every element ofKcan be expressed as a rationalfunction of1, . . . , n, and hence equation (1) shows that the image under of every elementofK is determined by (1), . . . , (n).In particular, if fixes each i, then equation (1) implies that fixes every element ofK.

(b) Let G Gal(K/F) be a subgroup of the Galois group of the extension K/F and suppose1, . . . , k are generators for G. Show that the subfieldE/F is fixed by G if and only if it isfixed by the generators 1, . . . , k.Proof. It suffices to prove that the set of elements in Gal(K/F) fixing Eis a subgroup ofG.(Because then this subgroup contains G if and only if it contains the generators 1, . . . , k.)But this verification is easy:

id(x) = x. (x) = x and (x) = x implies ((x)) = x. (x) = x implies 1(x) = x.

(5) Determine the automorphisms of the extension Q( 4

2)/Q(

2) explicitly.

Solution. Since the minimal polynomial for 4

2 over Q(

2) isx22 and 42, 42 are its roots, thenAut(Q( 4

2)/Q(

2)) consists of two elements id and , where satisfies ( 4

2) = 42. Therefore,

Aut(Q( 4

2)/Q(

2)) =Z/2Z.

(7) This exercise determines Aut(R/Q).

(a) Prove that any Aut(R/Q) takes squares to squares and takes positive reals to positive reals.Conclude thata < b implies a < b for everya, b R.Proof. Let s R. Thens2 is a square in R. Since is a homomorphism, we have (s2) =(s)(s), a square in R. Now letp R be a positive real number. Note thatp R as well. Wehave(p) = (

p

p) = (

p)(

p), a square, which is in particular a positive real number.Nowa < bis equivalent to 0 < b a. By the above considerations, 0 < (b a). This implies(a)< (b).

(b) Prove that 1m < a b < 1m implies 1m < ab < 1m for every positive integer m. Concludethat is a continuous map on R.Proof. Suppose that 1m < ab < 1m . By the previous part we have( 1m)< (ab)< ( 1m).Since preserves Q, and since (a b) = (a) (b), we obtain

1m

< (a) (b)< 1m

. (2)

Now let x R and xn x. Let| | denote the standard Euclidean norm. For any > 0,choose m s.t. 1/m < ; we can find an N such that|xn x| < 1/m for n > N. By (2), wehave|(xn) (x)| < 1/m < for n > N, which implies that (xn) (x). Therefore, iscontinuous.

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(c) Prove that any continuous map on R which is the identity on Q is the identity map, henceAut(R/Q) = 1.Proof. SinceQ is dense in R, for any r R there exists a sequence{qn} Q that converges tor. Let be continuous on R and the identity on Q. Part (b) then gives

limn

(qn) = (r)

= limn qn= (r).Now since Q is Hausdorff, qn r implies r= r, i.e., is the identity on R.

(8) Prove that the automorphisms of the rational function fieldk(t) which fixkare precisely thefractionallinear transformationsdetermined by t at+bct+d fora, b, c, d k, ad bc = 0 (so f(t) k(t) maps tofat+bct+d

).

Proof. First we state the result from Section 13.2, Exercise 18. Let P, Qbe relatively prime polyno-mials in k [t], not both constant, with Q = 0. Ifu = P(t)/Q(t), thenk(u) k(t) and

[k(t) : k(u)] = max(deg P, deg Q). (3)

Lets grant this for now and solve the problem.

Given any element u k (t), u k , it is clear that we can define a field embedding k(t) k(t)such that (t) = u; namely, we let (f(t)) = f(u) for any rational function f(t)k(t). The pointof this problem is to determine for which u this map is an automorphism, i.e. when does it have aninverse.

Let Aut(k(t)/k). Let u = (t) = P(t)/Q(t) k(t), with P, Q as above. (Note P andQ are not both constant polynomials, or else u k, and fixes k so (t) k.) Similarly write1(t) = R(t)/S(t) for some polynomials R, S. Then

t= (1(t)) =

R(t)

S(t)

=

R((t))

S((t)) =

R(u)

S(u). (4)

In other words, we haveu k(t) andt k(u), sok(t) = k(u). By (3), this implies thatP andQ eachhave degree at most 1. If we writeP(t) = at+bandQ(t) = ct+d, then these polynomials are relativelyprime if and only ifad

bc

= 0. It is easy to solve the equation u = (at + b)/(ct + d) fort in terms of

u in order to show that the map (t) = P(t)/Q(t) has an inverse; we obtaint = (du b)/(cu + a).Therefore Aut(k(t)/k) is precisely the group of linear fractional transformations.

It remains to prove (3). Note thatt satisfies the polynomial f=P(x) u Q(x) k(u)[x]. It isclear thatfhas degree max(deg P, deg Q) as polynomial inx. We claim that fis irreducible; indeed,Gauss Lemma implies that we may check irreducibility over k [u][x] =k[u, x]. But as a polynomialin u, f is linear (since Q= 0) and hence clearly irreducible. Thereforef is irreducible, and is theminimal polynomial oft over k(u). Thus

[k(u) : k(t)] = deg f= max(deg P, deg Q),

proving (3) and finishing the problem.

(9) Determine the fixed field of the automorphism t t+ 1 ofk(t).Proof. Denote by the automorphism ofk defined by t t + 1. The answer to the problem willturn out to depend on the characteristic ofk .

LetP(t)/Q(t) k(t), with P, Q relatively prime polynomials in k [t] and Q = 0. Then P(t)/Q(t)is fixed by ifP(t)/Q(t) = P(t+ 1)/Q(t+ 1). In other words, we have

P(t)Q(t+ 1) =Q(t)P(t+ 1) in k [t]. (5)

Suppose thatP(t) is nonconstant, and let be any root ofP(t) in k. Then by (5), we haveQ() = 0or P(+ 1) = 0. But Q() = 0 is impossible, since then both P and Q would be divisible by theminimal polynomial of, and P and Q are relatively prime. ThereforeP(+ 1) = 0. Iterating

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this, we find that + j is a root of P(t) for every integer j. If k has characteristic zero, this isimpossible since P(t) can have only finitely many roots whereas there are infinitely many elements+j. Therefore, whenk has characteristic zero,P(t) is a constant; working similarly with purportedroots ofQ(t) shows that Q must be constant as well. Therefore, whenk has characteristic zero, thefixed field of onk(t) is just k .

When k has characteristicp, it is possible for the polynomial Pto satisfy the property that +j

is a root whenever is a root. In fact we saw such polynomials in 13.5, #5. Motivated by this, letu= P(t) = tp t. Thenu is fixed by , and hence the entire subfield k(u) k(t) is fixed as well. Weclaim that [k(t) : k(u)] = p. Indeed, the minimal polynomial oft overk(u) isxpxu= 0, which isirreducible by the argument of13.5, #5. Therefore, the fixed field of on k(t) in the characteristic

p case is exactly k(u) since this subfield is fixed, the whole field k(t) is not fixed (since (t)= t),and there are no intermediary subfields by degree considerations.

(10) LetKbe an extension of the fieldF. Let : K K be an isomorphism ofKwith a fieldK whichmaps F to the subfield F of K. Prove that the map 1 defines a group isomorphismAut(K/F) Aut(K/F).Proof. Denote the map 1 to be , where Aut(K/F). Clearly, is a map fromAut(K/F) to Aut(K/F).

For1, 2 Aut(K/F),(12) = 12

1 =112

1 = (1)(2),

and(id|K) = id|K 1 =1 =id|K.

Therefore, : Aut(K/F) Aut(K/F) is a homomorphism.Suppose () = id|K for some Aut(K/F), i.e. 1 =id|K . Then

= 1 id|K = id|K.Thus, is injective.

For Aut(K/F), let = 1, then() = 1 =(1)1 =id|K id|K = .

Thus, is surjective.Therefore, : Aut(K/F) Aut(K/F) is an isomorphism of groups.

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