MATH 202 - ALGEBRA IV - HW 1 SOLUTIONS

13.1 - Basic Theory of Field Extensions.

1. Let p(x) = x3 + 9x + 6. Then since all coefficients except the leading coefficient are divisible by3 and the constant term is not divisible by 9 we have that p is irreducible by the Eisenstein Criterion.Notice that (1)3 + 9(1) + 10 = 0 so that x + 1 divides x3 + 9x + 10. Now we factor and see thatx3 + 9x+ 10 = (x+ 1)(x2 x+ 10). Hence (1 + )(2 + 10) = 3 + 9+ 6 + 4 = 4. Hence and inverse to1 + is

2+104 .

4. Clearly this map preserves the additive structure since a + b

2 + c + d

2 = a + c + (b + d)

2 andab2+cd2 = a+c(b+d)2. Now we need to check that the multiplicative structure is also preserved.Now we see that (a+b

2)(c+d

2) = (ac+2bd)+(ad+bc)

2 and (ab2)(cd2) = (ac+2bd)(ad+bc)2.

Hence the map a+ b

2 a b2 is a field automorphism of Q[2].5. Assume that = a/b with a, b Z, b 6= 0. Then consider n + an1n1 + ...+ a0 = 0 where all of the

ai are integers. Now multiply this equation by bn. Then we get that an+b(an1an1 + ...+a0bn1) = 0. Letp be a prime integer dividing b. Then the previous equation implies that p divides an and thus that p dividesa. Hence if a/b is in lowest terms we must have that no primes divide b, hence b = 1 which completes theproof since this implies that a is an integer.

13.2 - Algebraic Extensions.

3. The roots of the minimal polynomial of 1 + i must be 1 + i and 1 i. Thus this polynomial is(x (1 + i))(x (1 i)) = x2 2x+ 2. Also, this polynomial has 1 + i as a root and is irreducible by theEisenstein test with p = 2.

15. Assume that we have a minimal counterexample f(x). Then it must be that deg(f) > 1 since if fis linear, then F () = F . Let deg(f) = 2k + 1. So we have that 1 can be written as a sum of squares inF () = F [X]/(f(x)). But then 1 + f(x)g(x) = f1(x)2 + + fn(x)2 with deg(fi) < 2k + 1 (since everyelement in F () can be written as a polynomial in with degree less than deg(f)). But then the degree ofthe right hand side is less than 4k + 2 and so deg(g) < 2k + 1.

We claim now that the degree of f1(x)2 + + fn(x)2 is even. Indeed, if d is the maximal degree amongthe fi, the the largest degree term that occurs in the sum is x2d. Furthermore, the coefficient of x2d is thesum of the squares of the leading coefficients of the fi(x) with degree d. Since F is formally real, 0 is notthe sum of nonzero squares in F (proof: if

ri=1 a

2i = 0, then

r1i=1 (ai/ar)

2 = 1.) Thus the degree off1(x)2 + + fn(x)2 is the even number 2d, as claimed.

The equation 1 + f(x)g(x) = f1(x)2 + + fn(x)2 now implies that deg(g) must be odd (since thedegree of f is odd and the degree of the right side is even). Now let p(x) be an irreducible factor of g of odddegree, which must exist since deg(g) is odd. Then deg(p) < deg(f) and

1 + p(x)(g(x)p(x)

f(x)) = f1(x)2 + + fn(x)2

so 1 is a square in F [X]/(p(x)), contradicting the minimality of f .19. a) Let K. By the axioms for a commutative ring we have that (a+b) = a+b and (cx) = c(x)

for all c K so in particular for all c F . Hence multiplication by is an F -linear transformation of K.b) Choose a basis for K/F , which will be of size n. This gives an isomorphism between Mnn(F ) and

HomF (K,K). Now we showed previously that for any K we have that m : K K, given by1

m(x) = x, is an F -linear transformation of K. Moreover, the map m : K HomF (K,K) given bym() = m is an injective ring homomorphism. This is easily verified since ( + )x = x + x and som+ = m +m and ()x = (x) so m = m m . Also we have that x = x for all x K implies = which proves the injectivity. Hence K = A where A = im(m) HomF (K,K) = Mnn(F ) and weare done.

22. Notice that K1 F K2 is a [K1 : F ][K2 : F ] dimensional vector space over F . Now consider the Fbilinear map K1 K2 K1K2 given by (a, b) ab. This induces an F -algebra map K1 F K2 K1K2.

If K1 F K2 is a field, this map must be injective or trivial. Since it clearly is not trivial as 1 1 7 1we have that [K1K2 : F ] [K1 : F ][K2 : F ]. In class we proved the reverse inequality, so we conclude that[K1K2 : F ] = [K1 : F ][K2 : F ] as desired.

For the reverse direction, we claim that the map K1 F K2 K1K2 constructed above is surjective.Indeed, the image R is an F algebra that contains K1 and K2. Since it is a subring of the finite extensionK1K2 of F , the inverse of every element R can be written as a polynomial in with coefficients in F ,and hence lies in R. Thus R is a field containing K1 and K2, and hence must equal K1K2 since K1K2 is bydefinition the smallest such field.

Therefore our map K1FK2 K1K2 is a surjective homomorphisms of F -algebras of the same dimensionover F ; hence it must be an isomorphism, and thus K1 F K2 is a field.13.3 - Classical Straightedge and Compass Constructions.

1. If one could contruct a regular 9-gon, then 9 = e2ipi/9 would be constructible. However, [Q[9] : Q] =(9) = 6 and 6 is not a power of 2. Hence a regular 9-gon is not constructible.

13.4 - Splitting Fields and Algebraic Closures.

5. Assume that the finite extension K/F is normal. Now let L be a maximal field between K and F , i.eF L K with L maximal. This L must exist since K/F is finite and F is a splitting field over itself.I claim that L = K. If not choose K, / L. Then the minimal polynomial f of over F splitscompletely and so by adjoining all of its roots to L we have a strictly larger splitting field between K andF , namely the splitting field of f(x)l(x) where L is the splitting field of l(x), contradicting the maximalityof L. Hence L = K and K is a splitting field.

Now assume that K is a splitting field of k(x). Let K be the root of an irreducible polynomial f overF . Let be another root. By theorem 8 there exists an F -algebra isomorphism between F () and F ().Also it is clear that K() is the splitting field of k(x) over F () and K() = K is the splitting field of k(x)over F (). Then by theorem 27 the F -algebra isomorphism between F () and F () lifts to an F -algebraisomorphism between K() and K. Hence K() = K as F algebras (in particular as F -vector spaces) andhence since K() K we have that K() = K by dimension considerations, and hence K is normal.

6. a) Let K1 and K2 be the splitting fields of k1(x) and k2(x) respectively over F . Then the compositumK1K2 is a splitting field for k1(x)k2(x) over F since clearly this polynomial splits in K1K2 and moreover ifk1(x)k2(x) splits in L, then L K1 and L K2 so L K1K2.

b) I will prove that K1 K2 is a normal extension of F which by the previous problem implies that K1 K2is a splitting field. So let K1 K2 be the root of an irreducible polynomial in F and let be a differentroot. Then since K1 and K2 are both normal extensions we have that K1 and K2 since K1 and K2. Hence K1 K2 and we are done.

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