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Chapter 6: Problem 15

Let \(K / k\) be a Galois extension, and let \(F\) be an intermediate field between \(k\) and \(K\). Let \(H\) be the subgroup of \(\mathrm{Gal}(K / k)\) mapping \(F\) into itself. Show that \(H\) is the normalizer of \(\mathrm{Gal}(K / F)\) in \(\mathrm{Gal}(K / k)\).

Short Answer

Expert verified
We have shown that the subgroup \(H\) of \(\mathrm{Gal}(K/k)\) mapping the intermediate field \(F\) into itself satisfies the properties required to be the normalizer of \(\mathrm{Gal}(K/F)\) in \(\mathrm{Gal}(K/k)\). Specifically, we proved that for every \(\sigma \in H\) and \(g \in \mathrm{Gal}(K/F)\), we have \(\sigma g \sigma^{-1} \in \mathrm{Gal}(K/F)\). Therefore, \(H\) is the normalizer of \(\mathrm{Gal}(K/F)\) in \(\mathrm{Gal}(K/k)\).

Step by step solution

01

Definition of a normalizer

A subgroup \(H\) of a group \(G\) is called the normalizer of another subgroup \(N\) in \(G\) if: 1. \(H \subseteq G\) 2. For every \(h \in H\) and \(n \in N\), we have \(hnh^{-1} \in N\) 3. \(H\) is the largest subgroup of \(G\) satisfying these properties
02

Proving that \(H\) maps \(F\) into itself

By definition, \(H\) is the subgroup of \(\mathrm{Gal}(K/k)\) that maps \(F\) into itself. Therefore, for any \(\sigma \in H\), and for any \(f \in F\), we have \(\sigma (f) \in F\).
03

Proving that \(H\) is the normalizer of \(\mathrm{Gal}(K/F)\)

We want to show that for every \(\sigma \in H\) and \(g \in \mathrm{Gal}(K/F)\), we have \(\sigma g \sigma^{-1} \in \mathrm{Gal}(K/F)\). Let \(\sigma \in H\) and \(g \in \mathrm{Gal}(K/F)\). We need to show that \(\sigma g \sigma^{-1}\) fixes \(F\). To do this, let \(f \in F\). We know that \(\sigma\) maps \(F\) to itself since \(\sigma \in H\). Therefore, \(\sigma^{-1}(f) \in F\). Now, we know that \(g \in \mathrm{Gal}(K/F)\), meaning that \(g\) fixes \(F\). Therefore, \(g\sigma^{-1}(f) = \sigma^{-1}(f)\). Applying \(\sigma\) on both sides, we get \(\sigma g \sigma^{-1}(f) = f\). This means that the element \(\sigma g \sigma^{-1}\) fixes \(F\), and therefore \(\sigma g \sigma^{-1} \in \mathrm{Gal}(K/F)\). This verifies condition 2 for being a normalizer. Since \(H\) was already defined to be a subgroup of \(\mathrm{Gal}(K/k)\) and it satisfies condition 2 for being the normalizer, we can conclude that \(H\) is the normalizer of \(\mathrm{Gal}(K/F)\) in \(\mathrm{Gal}(K/k)\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Intermediate Field
An 'intermediate field' F is a field that lies between two fields k and K in a field extension, such that k is a subfield of F and F is a subfield of K. Essentially, it satisfies the condition of k 鈯� F 鈯� K. The concept of an intermediate field is central to the study of field theory and helps to understand the structure and properties of field extensions.

Consider the analogy of a matryoshka doll: with the smallest doll representing the field k, the intermediate doll representing the field F, and the largest doll representing the field K. Each 'doll' perfectly fits within the next, representing how each field contains the one before it.
Normalizer of a Subgroup
In group theory, the 'normalizer' of a subgroup N in a group G is the largest subgroup H in which N is a normal subgroup. This means H consists of all elements in G that 'behave nicely' with respect to subgroup N. When an element `h` of H is combined with an element `n` of N and then followed by `h^-1`, the result is still an element in N. This is written in mathematical terms as hnh^(-1) 鈭� N for all h 鈭� H and n 鈭� N.

A simplified way to understand the normalizer is to think of N as a pattern within a larger picture represented by G. The normalizer H would then be all the parts of the picture that can be rotated or reflected without disturbing the pattern. This encapsulates the idea that elements of N combine with those of H to remain well-positioned within N. The exercise provided shows the application of this concept to Galois groups, which govern the symmetries of the roots of polynomials.
Galois Group
The 'Galois group' of a Galois extension K over k, denoted by Gal(K/k), consists of all field automorphisms of K that fix every element of k. These automorphisms are essentially symmetries of the field K that leave the smaller field k unchanged. The elements of the Galois group can be thought of as the 'moves' that rearrange the elements of the field K while preserving the structure of k.

In relation to the intermediate field F, the Galois group Gal(K/F) comprises those automorphisms that also fix every element of F, showing a more restricted set of symmetries. The interplay between these groups gives great insight into the inner workings of fields and their extensions, forming the cornerstone of Galois theory, which elegantly links field theory with group theory.

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Most popular questions from this chapter

Let \(k\) be a field of characteristic \(0 .\) Assume that for each finite extension \(E\) of \(k\), the index \(\left(E^{*}: E^{* \pi}\right)\) is finite for every positive integer \(n .\) Show that for each positive integer \(n\), there exists only a finite number of abelian extensions of \(k\) of degree \(n .\)

Let \(k\) be a field, \(k^{a}\) an algebraic closure, and \(\sigma\) an automorphism of \(k^{a}\) leaving \(k\) fixed. Let \(F\) be the fixed field of \(\sigma .\) Show that every finite extension of \(F\) is cyclic. (The above two problems are examples of Artin. showing how to dig holes in an algebraically closed field.)

Let \(K=C(x)\) where \(x\) is transcendental over \(\mathbf{C}\), and let \(\zeta\) be a primitive cube root of unity in C. Let \(\sigma\) be the automorphism of \(K\) over \(\mathrm{C}\) such that \(\sigma x=\zeta x\). Let \(\tau\) be the automorphism of \(K\) over \(\mathrm{C}\) such that \(\tau x=x^{-1}\). Show that $$ \sigma^{3}=1=\tau^{2} \text { and } \tau \sigma=\sigma^{-1} \tau . $$ Show that the group of automorphisms \(G\) generated by \(\sigma\) and \(\tau\) has order 6 and the subfield \(F\) of \(K\) fixed by \(G\) is the field \(C(y)\) where \(y=x^{3}+x^{-3}\)

Let \(Q^{n}\) be a fixed algebraic closure of Q. Let \(E\) be a maximal subfield of \(Q^{a}\) not containing \(\sqrt{2}\) (such a subfield exists by Zorn's lemma). Show that every finite extension of \(E\) is cyclic. (Your proof should work taking any algebraic irrational number instead of \(\sqrt{2}\).)

Let \(E\) be a finite separable extension of \(k\), of degree \(n\). Let \(W=\left(w_{1}, \ldots, w_{n}\right)\) be elements of \(E\). Let \(\sigma_{1}, \ldots, \sigma_{n}\) be the distinct embeddings of \(E\) in \(k^{a}\) over \(k\). Define the discriminant of \(W\) to be $$ D_{E / k}(W)=\operatorname{det}\left(\sigma_{i} w_{j}\right)^{2} $$ Prove: (a) If \(V=\left(v_{1}, \ldots, v_{n}\right)\) is another set of elements of \(E\) and \(C=\left(c_{i j}\right)\) is a matrix of elements of \(k\) such that \(w_{i}=\sum c_{i j} v_{j}\), then $$ D_{E k}(W)=\operatorname{det}(C)^{2} D_{E / k}(V) $$ (b) The discriminant is an element of \(k\) (c) Let \(E=k(\alpha)\) and let \(f(X)=\operatorname{Irr}(\alpha, k, X)\). Let \(\alpha_{1}, \ldots, \alpha_{n}\) be the roots of \(f\) and say \(\alpha=\alpha_{1}\). Then $$ f^{\prime}(\alpha)=\prod_{j=2}^{n}\left(\alpha-\alpha_{j}\right) $$ Show that $$ D_{E / \lambda}\left(1, \alpha, \ldots, \alpha^{n-1}\right)=(-1)^{m i n-1 W^{2}} N_{k}^{\xi}\left(f^{\prime}(\alpha)\right) $$ (d) Let the notation be as in (a). Show that det( \(\left(\operatorname{Tr}\left(w_{i} w_{j}\right)\right)=\left(\operatorname{det}\left(\sigma_{i} w_{j}\right)\right)^{2}\). [Hint: Let \(A\) be the matrix \(\left(\sigma_{i} w_{j}\right)\). Show that ' \(A A\) is the matrix \(\left(\operatorname{Tr}\left(w_{i} w_{j}\right)\right)\).]
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