Why are these statements about separability equivalent?

Question

I am reading something about separable extensions and passed by the following definitions:

(Separable degree) Let $E$ be an algebraic extension of a field $F$. and let $\sigma :F\rightarrow L$ be an embedding of $F$ in an algebraically closed field $L$. Let $S_\sigma$ be the set of extensions of $\sigma$ to an embedding of $E$ in $L$. When we define $$card(S_\sigma):=[E:F]_S$$ and call it the separable degree of $E$ over $F$.

(Separable extension) Let $E$ be a finite extension of $k$. We shall say that $E$ is separable over $k$ if $$[E:k]_S=[E:k]$$

(Separable element) An element $\alpha$ algebraic over $k$ is said to be separable over $k$ if $k(\alpha)$ is separable over $k$. This is equivalent to say that the minimalpolynom of $\alpha$ has no multiple roots

I somehow don't see why in the third definition we have the equivalence, i.e. the statement written in bolt is equivalent to the one don't written in bolt.

Can someone maybe explain this to me?

Thanks for your help

Answer 1

Let $f(x)$ be the minimal polynomial of $\alpha$. An embedding $k(\alpha) \to L$ is completely determined by where it maps $\alpha$. $\alpha$ must be mapped to a root of $f$, and any root defines an embedding. Thus $[k(\alpha):k]_S$ is equal to the number of distinct roots of $f$. Whereas $[k(\alpha):k]$ is equal to the degree of $f$. If $f$ has a multiple root, then the number of distinct roots would be fewer.

Answer 2

We see by definition $\alpha$ is separable over $k$ if and only if $[k(\alpha):k]=[k(\alpha):k]_S$. Let us try to compute the latter quantity in a different way:

Recall $k(\alpha)\simeq k[x]/(m_\alpha(x))$ where $m_\alpha$ is the minimal polynomial of $\alpha$ over $k$. If $\beta$ is a root of $m_\alpha$ inside $L$, then there is an evaluation homomorphism $k[x]\to L$ sending $f\mapsto f(\beta)$ (where we are really using the embedding $\sigma$ to make sense of the expression $f(\beta)$ for $f\in k[x]$). This will send $m_\alpha\mapsto0$ by hypothesis so it induces an embedding $k(\alpha)\simeq k[x]/(m_\alpha(x))\to L$, and this embedding extends $\sigma$. Thus we have a function

$$(\text{roots of $m_\alpha$ in $L$})\to\,(\text{embeddings $k(\alpha)\hookrightarrow L$ extending $\sigma$}) $$

and you can check this is actually a bijection, the inverse being given by sending some $\tau:k(\alpha)\hookrightarrow L$ to the element $\tau(\alpha)\in L$.

From this bijection we see that $[k(\alpha):k]_S$ is exactly equal to the number of distinct roots of $m_\alpha$ in $L$, and since $[k(\alpha):k]=\deg(m_\alpha)$, you see

\begin{align*} \text{$\alpha$ is separable}&\iff [k(\alpha):k]=[k(\alpha):k]_S\\ &\iff\deg(m_\alpha)=\#\{\text{roots of $m_\alpha$ in $L$}\}\\ &\iff\text{$m_\alpha$ has no multiple roots}. \end{align*}