Notes - Galois Theory HT25, Fields and field extensions

Created: October 17, 2024
Updated: October 28, 2024

Flashcards

Suppose:

  • $F$ is a subfield of $\mathbb C$
  • $y \in \mathbb C$

Give two equivalent @definitions for what it means for $y$ to be algebraic over $F$, and state the word used when $\alpha$ is not transcendental.
  • There exist $\alpha _ 0, \alpha _ 1, \ldots, \alpha _ d \in F$, where $\alpha _ d \ne 0$ and $\alpha _ d y^d + \cdots + \alpha _ 1 y + \alpha _ 0 = 0$
  • $\mathbb Q$ has a finite degree extension which contains $y$

If $y$ is not algebraic, then it is transcendental.

Suppose:

  • $F$ is a subfield of $\mathbb C$
  • $\alpha \in \mathbb C$ is algebraic over $F$

@Define what it means for $\alpha$ to be solvable by radicals over $F$.

There exists a chain of subfields

\[F = F_0 \subset F_1 \subset F_2 \subset \cdots \subset F_n\]

such that:

  • $\alpha \in F _ n$
  • For each $1 \le i \le n$, there exists $\alpha _ i \in F _ i$ and a positive integer $d _ i$ such that $F _ i = F _ {i-1}(\alpha _ i)$ and $\alpha _ i^{d _ i} \in F _ {i-1}$.

Intuitively, it means that $\alpha$ can be written as an expression containing elements of $F$ and using the operations $+$, $-$, $\times$, $\div$ and $\sqrt[k]{\cdot}$.

@Prove that if:

  • $K/F$ is a finite field extension
  • $\alpha \in K$

Then $\alpha$ is algebraic over $F$.

Let $n = [K : F]$, the degree of the field extension. Then $\{1, \alpha, \cdots, \alpha^n\}$ is linearly dependent over $F$. So then there exists $\lambda _ n, \ldots, \lambda _ 0 \in F$ such that

\[\lambda_n \alpha^n + \cdots \lambda_1 \alpha + \lambda_0 = 0\]

So $\alpha$ is algebraic over $F$.

@Define what it means for an extension $E / F$ to be simple.
\[E = F(\alpha)\]

for some $\alpha \in E$.

Suppose $K/F$ is a field extension where $\alpha \in K$. What is the difference between $F[\alpha]$ and $F(\alpha)$, and when are they equal?
  • $F[\alpha]$ is the subring of $K$ generated by $F$ and $\alpha$
  • $F(\alpha)$ is the smallest subfield of $K$ that contains both $F$ and $\alpha$

If $\alpha$ is algebraic over $F$, then $F(\alpha) = F[\alpha]$.

Suppose $K / F$ is a field extension and $\alpha \in K$. @State a lemma that connects $F[\alpha]$ and the minimal polynomial $m _ {F, \alpha}$.
\[F[\alpha] \cong F[t]/\langle m_{F, \alpha} \rangle\]

(why? the first isomorphism theorem for rings applied to the evaluation map $\text{ev} _ \alpha$).

Suppose:

  • $K/F$ is a finite field extension
  • $\alpha \in K$
  • $m _ {F, \alpha}$ is the corresponding minimal polynomial

@State and quickly @prove a lemma which connects the degree of the polynomial with the degree of the field extension.

Let $d := \deg m _ {F, \alpha}$. Then $[F[\alpha] : F] = d$.

By the division algorithm, for every $f \in F[t]$, there are unique $q, r \in F[t]$ with $\deg r < d$ such that $f = q m _ {F, \alpha} + r$. Since $\{1, t, \ldots, t^{d-1}\}$ is a basis of $F[t]/\langle m _ {F, \alpha} \rangle$, since $F[t]/\langle m _ {F, \alpha} \rangle \cong F[\alpha]$, it follows that $\{1, \alpha, \ldots, \alpha^{d-1}\}$ is a basis for $F[\alpha]$ over $F$ and so it has degree $d$.

Give an @example of a field extension that is not Galois.

Consider $\mathbb Q[\alpha] / \mathbb Q$ where $\alpha = \sqrt[3]{2}$. Then $m _ \mathbb Q(x) = x^3 - 2$, but $m(x)$ does not split completely, since it has two roots which are not in $\mathbb Q[\alpha]$.

Suppose:

  • $F \subset \mathbb C$
  • $K/F$ is a Galois extension

@Define the Galois group of $K/F$, written $\text{Gal}(K/F)$.

The group of all $F$-linear field automorphisms of $K$.

@Define an $F$-linear field automorphism of $k$.

A bijective map $\theta : K \to K$ such that for all $k _ 1, k _ 2 \in K$, and $f \in F$:

\[\begin{aligned} \theta(k_1 k_2) &= \theta(k_1) \theta(k_2) \\\\ \theta(k_1 + k_2) &= \theta(k_1) + \theta(k_2) \\\\ \theta(f k_1) &= f \theta(k_1) \end{aligned}\]

Suppose:

  • $K/F$ is a Galois extension
  • $H$ is a subgroup of $\text{Gal}(K/F)$

@Define the fixed field of $H$, written $K^H$.
\[K^H = \\{x \in K \mid \forall \sigma \in H,\text{ }\sigma(x) = x \\}\]

Suppose:

  • $K/F$ is a field extension
  • $\alpha _ 1, \ldots, \alpha _ n \in K$ are algebraic over $F$

@State a result on the size of the field $F(\alpha _ 1, \ldots, \alpha _ n)$..
\[[F(\alpha_1, \ldots, \alpha_n) : F] < \infty\]

@Prove that if:

  • $K/F$ is a field extension
  • $\alpha _ 1, \ldots, \alpha _ n \in K$ are algebraic

then:

\[[F(\alpha_1, \ldots, \alpha_n) : F] < \infty\]

Induct on $n$.

Base case: $n = 1$. This is equivalent to showing that $[F(\alpha _ 1) : F] < \infty$, which follows from the result that says for all $\alpha \in K$ that are algebraic over $F$, $[F(\alpha) : F] = \deg m _ \alpha$.

Inductive step: Suppose that $n > 1$. Then:

\[\begin{aligned} \,[F(\alpha_1, \ldots, \alpha_n) : F] &= [F(\alpha_1, \ldots, \alpha_{n-1})(\alpha_n) : F] \\\\ &=[F(\alpha_1, \ldots, \alpha_{n-1})(\alpha_n) : F(\alpha_1, \ldots, \alpha_{n-1})] \times [F(\alpha_1, \ldots, \alpha_{n-1}) :F ] \\\\ &< \infty \end{aligned}\]

where:

  • The second equality is justified by the tower law, and
  • The final inequality is justified by the inductive hypothesis: on the left for $n = 1$, and on the right for $n-1$.


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