## Math 503: Abstract Algebra

Homework 11 <me@tylerlogic.com>
May 7, 2014
http://coursework.tylerlogic.com/courses/upenn/math503/homework11

### 1

Given the Fundamental Theorem of Galois Theory, in order to find fields F K L such that K∕F and L∕K are finite Galois, but L∕F is not Galois, it suffices to find a finite Galois extension M∕F with Galois group G with subgroups H1,H2 such that H1 H2 and H2 G, but H1 G. In which case, we can put K = MH2 and L = MH1 in order to achieve the desired result.

In the last homework we saw that the splitting field of T4 - 3 over is finite Galois with Galois group D8. In the group D8 we have subgroups srsr,r2D8 where s is the mirror symmetry and r is the rotational symmetry.

The fact that r(sr)r3 = rsr4 = rs = sr3 demonstrates that sris not normal in D8. Since sr,r2= {1,r2,sr,sr3}, then

shows that srsr,r2, and
shows that sr,r2D8.

Thus for F = , K = Ωsr,r2 and L = Ωsr where Ω = (,i) is the splitting field of T4 - 3 over , we will have our desired scenario.

We can determine the fixed fields K and L as follows. Define s and r as the automorphisms

where θ = . This then sets

Because sris a subgroup of size two, then L∕ must be an extension of degree four. Since

and (θ(i- 1)) Ω is an extension of degree four, then we must have L = (θ(i- 1)). Similarly, sr,r2is a subgroup of size four, and (2) (θ(i - 1)) Ω has degree two. Then since

and

then we have K = (2).

### 3

Let L be a finite Galois extension of .

#### (a) Prove Gal(L∕ℝ) is a 2-group

Lemma 3.1. Any non-trivial finite extension of has even degree.

Proof. If K is a non-trivial finite extension of , then it is algebraic over . For any α K - , it’s minimal polynomial mα(x) [x] is nonlinear, irreducible. Since all odd degree polynomials in [x] have a real root, we conclude deg(mα(x)) is even. However, [K : ] = deg(α) = deg(mα(x)) and thus K has even degree. __

By the preceeding lemma, [L : ] is even. Then so is the size of Gal(L∕ ). So define H to be a 2-Sylow subgroup of Gal(L∕ ). Then the index of H in Gal(L∕ ) is not divisible by two and therefore E = LH is an extension of with odd degree. However, according to Lemma 3.1, the only such extension is itself. Thus E = . This furthermore implies that H = Gal(L∕ ), which, since H is a 2-Sylow subgroup, also means Gal(L∕ ) is a 2-group.

#### (b) Prove ℂ is algebraic

Lemma 3.2. There is no irreducible quadratic over

Proof. Any quadratic over has roots in provided by the quadratic formula. __

Let f(T) [T] and α be a root of f(T). Assume by way of contradiction that (α) is a nontrivial extension over . Then (α) is an extension of of even degree by Lemma 3.1. Thus, Gal((α) ) would be a nontrivial 2-group with size greater than or equal to 4, by part (a) of this problem. But then Gal((α) ) would have a subgroup of size 4 (Theorem 6.1 [DF04]: “p-groups have subgroups of all applicable sizes”), i.e. there would exist an extension K∕ with degree 2. However, this contradicts the Lemma 3.2, and hence (α) is trivial, implying α .

### References

[DF04]   D.S. Dummit and R.M. Foote. Abstract Algebra. John Wiley & Sons Canada, Limited, 2004.