[[Simple extension]]
# Simplicity of an algebraic extension

Let $F:K$ be an [[Algebraic element|algebraic extension]].
Then $F:K$ is a [[simple extension]] iff the number of distinct [[Intermediate field extension|intermediate fields]] $K : E : F$ is finite. #m/thm/field 

> [!check]- Proof
> Suppose $F = K(\alpha)$ is simple and algebraic over $K$,
> so that $q_{K}(x) \in K[x]$ is the [[Algebraic element|minimal polynomial]] for $\alpha$.
> If $E$ is an intermediate field, then $F = E(\alpha)$ is also simple and algebraic over $E$,
> so that $q_{E}(x) \in E[x]$ is the minimal polynomial.
> Moreover, $q_{E}(x)$ is a factor of $q_{K}(x)$.
> 
> $q_{E}(x)$ will turn out to completely determine the intermediate field $E$, and since $q_{K}(x)$ only has finitely many factors in $\overline{K}[x]$, this proves the forward direction.
> In fact, $E$ will be generated over $K$ by the coëfficients of $q_{E}(x)$.
> 
> To show this, let $E'$ be the subfield of $E$ generated by the coëfficients and $K$.
> Then $q_{E}(x) \in E'[x]$, and since $q_{E}(x)$ is irreducible in the extension field $E'$,
> so too is it irreducible in $E'$.
> Since $E'(\alpha) = F = E(\alpha)$ and $F : E : E' : K$, it follows (see [[Intermediate field extension|tower of field extensions]])
> $$
> \begin{align*}
> \deg q_{E} = [F: E'] = [F : E][E : E'] = \deg q_{E} \, [E:E'],
> \end{align*}
> $$
> so $[E : E'] = 1$, i.e. $E = E'$, as required.
> 
> For the converse, assume that there are only finitely many intermediate fields $F : E : K$.
> The extension $F : K$ must be finitely generated, for otherwise the infinite sequence of subextensions
> $$
> \begin{align*}
> K \hookrightarrow K(\alpha_{1}) \hookrightarrow K(\alpha_{2}) \hookrightarrow \cdots \hookrightarrow F
> \end{align*}
> $$
> would give infinitely many intermediate fields.
> If $K$ is a Galois field so is $F$, and then by [[Finite extension of a Galois field]] $F:K$ is simple.
> 
> Let us consider the case $K$ is infinite, where we need to show that every finitely generated algebraic extension $F=K(\alpha_{1},\dots,\alpha_{r})$ is simple.
> 
> Arguing inductively, we may assume w.l.o.g. that $F=K(\alpha,\beta)$.
> For every $c \in K$, we have the intermediate field
> $$
> \begin{align*}
> K(\alpha,\beta) : K(c\alpha+\beta) : K.
> \end{align*}
> $$
> But there are only finitely many such intermediate extensions.
> Since $K$ is infinite, we must have
> $$
> \begin{align*}
> K(c'\alpha + \beta) = K(c\alpha + \beta)
> \end{align*}
> $$
> for some $c' \neq c$ in $K$, whence
> $$
> \begin{align*}
> \alpha &= \frac{(c'\alpha + \beta)- (c\alpha + \beta)}{c'-c} \in K(C\alpha+\beta), &
> \beta &= (c\alpha + \beta) - c\alpha \in K(c\alpha+\beta);
> \end{align*}
> $$
> so $K(\alpha,\beta) \sube K(c\alpha+ \beta)$.
> Thus $K(c\alpha+\beta)= K(\alpha,\beta)$, and we are done. <span class="QED"/>

  [^2009]: 2009\. [[Sources/@aluffiAlgebraChapter02009|Algebra: Chapter 0]], §VII.5.3, pp. 449–450.

## See also

- [[Primitive element theorem]]

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