The following Daily Questions were asked this week.

## Thursday, June 1

## First Question

Suppose $F\subseteq E$ is the splitting field extension of a separable polynomial $f(x)$ of degree $n$. If $f(x)$ is irreducible in $F[x]$, then $\operatorname{Gal}_F(E)$ is (isomorphic to) what type of subgroup of $S_n$?

## Reveal Answer

### Answer

A **transitive** subgroup. (The converse statement is also true.)

## Second Question

Suppose $f(x)\in {\bf Q}[x]$ is a monic irreducible quartic polynomial. Let ${\bf Q}\subseteq E$ be a splitting field extension of $f(x)$, and let $h(y)$ be the resolvent cubic of $f(x)$. If $h(y)$ splits completely in ${\bf Q}[y]$, then $\operatorname{Gal}_{\bf Q}(E)$ is isomorphic to …

## Reveal Answer

### Answer

… the Klein 4-group, i.e., $({\bf Z}/2{\bf Z})\times ({\bf Z}/2{\bf Z})$.

## Friday, June 2

For today’s questions, suppose $F$ is a field of characteristic not dividing $n$ that contains the $n$th roots of unity.

## First Question

For every $a\in F$, the extension $F\subseteq F(\sqrt[n]{a})$ is a Galois extension with what type of Galois group?

## Reveal Answer

### Answer

A **cyclic** Galois group, of ordering dividing $n$.

## Second Question

If $F\subseteq E$ is a cyclic extension of degree $n$, then what can we say about $E$?

## Reveal Answer

### Answer

It is a simple extension of the form $E=F(\sqrt[n]{a})$ for some $a\in F$. (We can even use Lagrange resolvents to help us find $\sqrt[n]{a}$.)

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