[[Field extension]]
# Simple extension

A [[field extension]] $L:K$ is called **simple** iff $L$ is generated by the [[Adjunction of a ring|adjunction]] of a single element, #m/def/field
i.e. $L = K(\vartheta)$ for some $\vartheta \in L$.[^2009]
Such a $\vartheta$ is called a **primitive element**.

  [^2009]: 2009\. [[Sources/@aluffiAlgebraChapter02009|Algebra: Chapter 0]], §VII.1.2, pp. 387–388

## Classification

Let $K(\vartheta) : K$ be a simple extension, and consider the **evaluation map**
$$
\begin{align*}
\epsilon : K[t] &\to K[\vartheta] \sube K(\vartheta) \\
f(t) &\mapsto f(\vartheta )
\end{align*}
$$
Then

1. If $\epsilon$ is injective then $K(\vartheta) : K$ is an infinite extension, whence $K(\vartheta)$ is isomorphic to the [[field of rational functions]] $K(t)$;
2. If $\epsilon$ is not injective then $K(\vartheta) : K$ is an extension of finite degree $n$, and
  $$
  \begin{align*}
  K(\vartheta) \cong_{\cat{Fld}} \frac{K[t]}{\langle m_{\vartheta}(t) \rangle }
  \end{align*}
  $$
where $m_{\vartheta}(t) \in K[t]$ is the [[Algebraic element|minimal polynomial]] of $\vartheta$.

> [!check]- Proof
> By the [[Ring isomorphism theorems#First isomorphism theorem]], the image of $\epsilon$ is isomorphic to $K[t]/ \ker \epsilon$.
> Since $K(\vartheta)$ is an [[integral domain]], so is $K[\vartheta]$, and thus by [[Condition for a quotient commutative ring to be an integral domain]], $\ker \epsilon$ must be a [[prime ideal]] in $K[t]$.
> 
> First, consider $\ker \epsilon = 0$, i.e. $\epsilon$ is injective.
> By the universal property for the [[Field of fractions]], $\epsilon$ extends to a unique homomorphism
> $$
> \begin{align*}
> \bar{\epsilon} : K(t) \to K(\vartheta)
> \end{align*}
> $$
> where $K(t) \cong \bar{\epsilon} (K(t)) \leq K(\vartheta)$ is a field containing $K$ and $\vartheta$,
> whence by definition $\bar{\epsilon}(K(t)) = K(\vartheta)$.
> By injectivity, $\{ \vartheta^i \}_{i=0}^\infty$ are linearly independent so we have an infinite extension.
> 
> Now consider $\mathfrak{a} = \ker \epsilon \neq 0$.
> Since $K[t]$ is a [[Euclidean domain]] and thus a [[Principal ideal domain|PID]], it follows $\mathfrak{a} = \langle p(t) \rangle$ for a unique monic irreducible nonconstant polynomial $p(t) \in K[t]$.
> Since $\langle p(t) \rangle$ is maximal in $K[t]$, the image of $\epsilon$ is a subfield containing $\vartheta = \epsilon(t)$, and by the same token as above we have $\epsilon(K(t)) = K(\vartheta)$, giving the claimed isomorphism. <span class="QED"/>

  [^2009]: 2009\. [[Sources/@aluffiAlgebraChapter02009|Algebra: Chapter 0]], pp. 387–388

## Properties

1. If $K(\alpha)$ and $K(\beta)$ have the same minimal polynomial $f(x) \in K[x]$, then there exists a unique [[Morphism of field extensions|isomorphism of field extensions]] $\varphi : K(\alpha) \to K(\beta)$ such that $\varphi(\alpha) = \beta$. ^P1
2. More generally, an isomorphism of ground fields $\psi : K_{1} \to K_{2}$ such that $\psi(m_{\alpha}(x)) = m_{\beta}(x)$ lifts to an isomorphism of extensions $\bar{\psi} : K_{1}(\alpha) \to K_{2}(\beta)$. ^P2

> [!check]- Proof of 1–2
> Note that an isomorphism of simple field extensions is completely determined by the image of the primitive element.
> 
> Now using the isomorphism described in [[#Classification]],
> $$
> \begin{align*}
> K(\alpha) \cong \frac{K[x]}{\langle f(x) \rangle } \cong K(\beta)
> \end{align*}
> $$
> which necessarily fixes $K$ and maps $\alpha \mapsto \beta$, proving [[#^P1]], whereof [[#^P2]] is a straightforward generalization. <span class="QED"/>

### Results

- [[Bound on the automorphism group of a finite simple extension]].
- [[Simplicity of an algebraic extension]]
- By the [[Primitive element theorem]], every [[Finite field extension|finite]] [[separable extension]] is simple.

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