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Chapter 7: Problem 2

Let \(K\) be a Galois extension of \(F\) whose Galois group is solvable, Let \(E\) be any extension of \(F\). Prove that \(K E / E\) has solvable Galois group.

Short Answer

Expert verified
In summary, to prove that the Galois group of the composite extension \(KE/E\) is solvable, we first showed that the Galois group of \(KE\) over \(K\) or \(E\) is solvable. We then applied the Galois correspondence and the properties of solvable groups to derive that the Galois group of \(KE/E\) is solvable.

Step by step solution

01

Show that the Galois group of \(KE\) over \(K\) or \(E\) is solvable

First, we need to show that the Galois group of \(KE\) over \(K\) or \(E\) is solvable. Let \(L = KE\). Since \(K\) is a Galois extension of \(F\) with the solvable Galois group \(G = \operatorname{Gal}(K/F)\), by the Galois correspondence, there exists a subgroup \(H\) of \(G\) such that \(L\) is the fixed field of \(H\), i.e. \(L = K^H\). Since \(H\) is a subgroup of the solvable group \(G\), \(H\) is also solvable. Let the extension \(K\) over \(L\) have degree \(m\) and let \(E\) over \(F\) have degree \(n\). Then, because \(K\) and \(E\) are linearly disjoint over \(F\), the degree of \(L\) over \(E\) is \(mn\). Now, consider the group \(\operatorname{Gal}(L/E)\), which has order dividing \(mn\). Since \(H\) is solvable, there exists a normal subgroup \(N \trianglelefteq H\) such that \(\frac{H}{N}\) is abelian. Now, let \(L_1 = K^N\). Then \(L_1\) is Galois over \(L\) and, by the Galois correspondence, \(\operatorname{Gal}(L_1/L)\) is isomorphic to \(\frac{H}{N}\). Therefore, \(\operatorname{Gal}(L_1/E)\) is solvable.
02

deriving the Galois group of \(KE/E\) being solvable

Now, let's consider the group \(\operatorname{Gal}(KE/E)\). To do this, consider the series of fields: \[E \subseteq L_1 \subseteq KE\] By step 1, we have that the group \(\operatorname{Gal}(L_1/E)\) is solvable. Next, we will show that \(\operatorname{Gal}(KE/L_1)\) is solvable. Notice that \(L_1 \subseteq KE = K L_1\) and \(K\) is a Galois extension of \(F\) with a solvable Galois group. Therefore, the restriction map: \[\operatorname{Res}: \operatorname{Gal}(KE/F) \rightarrow \operatorname{Gal}(K/F)\] is well-defined. Notice that \[\operatorname{ker}(\operatorname{Res}) = \operatorname{Gal}(KE/ K L_1) = \operatorname{Gal}(KE/ K)= \operatorname{Gal}(KE/L_1)\] Here, \(\operatorname{ker}(\operatorname{Res})\) is a normal subgroup of the solvable group \(\operatorname{Gal}(KE/F)\). Therefore, \(\operatorname{Gal}(KE/L_1)\) is a solvable group. Now, notice the series: \[\operatorname{Gal}(KE/E) \supseteq \operatorname{Gal}(KE/L_1) \supseteq \{1\}\] By the conclusion of Step 1 and the above result, this gives us a chain of normal subgroups with each quotient being solvable, thus making \(\operatorname{Gal}(KE/E)\) solvable. In summary, we have shown that for any extension \(E\) of \(F\), the Galois group of the composite extension \(KE/E\) is solvable.

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Key Concepts

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

Galois extension
A Galois extension is a type of field extension that features a perfect blend of algebraic and geometric properties. Imagine having a field (let's call it \( F \)) and creating a new, larger field from it (we鈥檒l call this \( K \)). If \( K \) has some nice characteristics, like being a normal and separable extension of \( F \), it is considered a Galois extension.

What makes a Galois extension special is its symmetry, akin to how a cube looks the same no matter how you spin it. This symmetry is represented by the Galois group, which we鈥檒l talk about in a bit. Galois extensions are crucial in solving polynomial equations, as they allow us to understand the roots through the structure of the field. Think of Galois extensions as tools that help unlock complex algebraic mysteries.
Solvable group
A solvable group is a concept borrowed from group theory. Imagine a puzzle that can be solved in steps. Similarly, a group is called solvable if it can be broken into simpler pieces, step by step, until it is fully decomposed. This means we can write it as a series of normal subgroups, each a solution to the earlier puzzle.

Why do we care about solvable groups here? Well, in the realm of Galois theory, if the Galois group of an extension is solvable, the roots of the polynomial related to that extension can be found using radicals (like square roots, cube roots, etc.). It ensures that certain equations aren't too wild and can, in fact, be broken down into more manageable pieces. Think of solvable groups as a beacon of hope for mathematicians trying to crack tough algebraic problems.
Galois group
The Galois group is like the security system of a field. Given a Galois extension \( K \) of \( F \), the Galois group, expressed as \( \operatorname{Gal}(K/F) \), consists of all automorphisms of \( K \) that keep \( F \) untouched in place.

In simpler terms, these are the different ways to shuffle the elements of \( K \) without messing up \( F \). The Galois group reveals the field's structural secrets, showing how complicated (or not) the roots of our polynomial are. This group plays a central role in the puzzle-solving nature of Galois theory, especially when determining if roots can be neatly expressed by radicals, tying directly into the concept of solvability.
Field extension
A field extension is like adding extra features to a toolkit. Start with a base field \( F \), and then expand it to include new elements that weren鈥檛 part of \( F \). This expansion forms a new field, \( K \), turning it into an extension \( K/F \).

Why is this done? Well, to solve equations or explore new algebraic territories that \( F \) alone cannot navigate. Field extensions are tools that help mathematicians solve polynomial equations by embedding more advanced numbers (like roots of polynomials) into the existing field. Understanding how degrees of extensions work and how they relate gives insight into the polynomial's character and solvability. It's like building a taller tower using larger base blocks for a more expansive view of mathematical landscapes.

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Most popular questions from this chapter

Let \(F\) be a field of chacteristic 0 , let \(a \in F, a \neq 0\). Let \(n\) be an odd integer \(\geq 3\). Assume that for all prime numbers \(p\) such that \(p \mid n\) we have \(a \notin F^{P}\) (where \(F^{\prime}\) is the set of \(p\) -th powers in \(F\). Show that \(t^{*}-a\) is irreducible in \(F[t]\). [Hint: Write \(n=p^{r} m\) with \(p X m\). Assume inductively that \(t^{m}-a\) is irreducible in \(F[t]\). Show that \(\alpha\) is not a \(p\) -th power in \(F(\alpha)\) and use induction. ] Remark. When \(n\) is even, the analogous result is not quite truc because of the factorization of \(t^{4}+4\). Fssentially this is the only exception, and the general result can be stated as follows. Theorem. Let \(F\) he a field and \(n\) an integer \(\geqq 2 .\) Let \(a \in F, a \neq 0\). Assume that for all prime numbers p such that \(p \mid n\) we hate \(a \notin F^{p}\), and if \(4 \mid n\) then \(a \notin-4 F^{4}\). Then \(t^{n}-a\) is irreducible in \(F[t]\) It is more tedious to handle this general case, but you can always have a try at it. The main point is that the prime 2 causes some trouble.

Give an example of extension fields \(F \subset E \subset K\) such that \(E / F\) is Galois, \(K / E\) is Galois, but \(K / F\) is not Galois.

\begin{array}{l} \text { Let } K \text { be Galois over } F \text { and let } E \text { be finite over } F \text { . Assume that } K \cap E=F \text { . }\\\ \text { Prove that }[K E: K]=[E: F] \end{array}

(a) Let \(p\) be an odd prime. Let \(F\) be a field of characteristic 0, and let \(a \in F\). \(a \neq 0\). Assume that \(a\) is not a \(p\) -th power in \(F\). Prove that \(t^{\prime \prime}-a\) is irreducible in \(F[t]\). [Hint: Suppose \(t^{\prime \prime}-a\) factors over \(F\). L.ook at the constant term of one of the factors, expressed as a product of some roots, and deduce that \(a\) is a \(p-t h\) power in \(F\).]

Let \(E_{1}, E_{2}\) be finite extensions of a field \(F_{,}\) and assume \(E_{1}, E_{2}\) contained in some field. If \(\left[E_{1}: F\right]\) and \(\left[E_{2}: F\right]\) are relatively prime, show that $$ \left[E_{1} E_{2}: F\right]=\left[E_{1}: F\right]\left[E_{2}: F\right] \quad \text { and } \quad\left[E_{1} E_{2} ; E_{2}\right]=\left[E_{1}: F\right] $$
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