Definition of separable extension via $k$-algebra homomorphisms

Question

I came across with a definition of separable field extension $F/k$ as below :

Let $F/k$ be any finite extension. The extension is said to be separable if there exist $[F:k]$ distinct homomorphisms of $k-$algebras $F \to \bar{k}$. An arbitrary algebraic extension $E / k$ is said to be separable if every finite subextension of $E$ is separable.

I am not familiar with this definition. I know the definition of what a $k-$algebra and $k-$homomorphism is. So

i) Can you explain the definition in a 'basic' way?

ii) When can we say that a $k-$algebra $A$ is an extension of $k$?

Answer 1

(i): It's easier to think about this in the special case where $F$ is obtained from $k$ by adjoining a single algebraic element, say $F = k(\beta)$ for some $\beta \in \overline{k}$. Let $f \in k[T]$ be the minimal polynomial of $\beta$ over $f$. Let $\beta = \beta_1, ... , \beta_n \in \overline{k}$ be all the roots of $f$, counting multiplicity. Then we have

$$f(T) = (T - \beta_1) \cdots (T - \beta_n)$$

A $k$-algebra homomorphism of $F = k(\beta)$ into $\overline{k}$ is a homomorphism $\phi: F \rightarrow \overline{k}$ such that $\phi(x) = x$ for all $x \in k$. It is therefore completely determined by where it sends $\beta$. And since the coefficients of $f$ lie in $k$, we must have $\phi(\beta) = \beta_i$ for some $1 \leq i \leq n$. Therefore, the number of distinct $k$-algebra homomorphisms $F \rightarrow \overline{k}$ is the number of distinct roots of $f$.

In particular, there are at most $[F : k ] = \deg f$ distinct $k$-algebra homomorphisms, and $F$ is separable over $k$ if and only if $f$ has distinct roots.

What do things looks like when $F/k$ is not separable? In this case, we are necessarily in prime characteristic $p$, and there is multiplicity among the roots. Let's say $\beta_1, ... , \beta_r$ are the distinct roots, occurring with multiplicities $m_1, ... , m_r$. One can show that all these $m_i$ are equal, and are actually powers of $p$. Let's say $p^t = m = m_i$. Then the minimal polynomial looks like

$$f(T) = (T - \beta_1)^{p^t} \cdots (T -\beta_r)^{p^t} = (T^{p^t} - \beta_1^{p^t}) \cdots (T^{p^t} - \beta_r^{p^t})$$

there are $r$ distinct $k$-algebra homomorphisms $F \rightarrow \overline{k}$, and $\deg f = [F : k ] = rp^t$.

(ii): A $k$-algebra is a ring (I assume commutative with identity) containing $k$ as a subring. So a $k$-algebra is an extension of $k$ if and only if it is a field.

Answer 2

There are two things going on.

Every algebraic extension of $k$ is a subfield of $\overline{k}/k$. This is true, since algebraic closures are uniue, so whenever we have a finite $F/k$, we can consider $\overline{F}/k$, there exists an isomorhism $\overline{F}/k \to \overline{k}/k$, and hence a monomorphism (embedding) for $F/k$.

The second thing is that there is another criteria for separability:

It is generally true that for an extension $L/k$, with $[L:k]=n$, and $\sigma:k \to F$ some field embedding, the number of extensions of $\sigma$ to $L \hookrightarrow F$ is at most $n$, and equality is achieved when $L/k$ is separable, so that there exists some field $L^{\prime}$ with precisely $n$ embeddings (make the field large enough),

Putting these together proves the first claim in your question (aothough it's taken as a definition.) A reference for the second point is Theorem 3.8 in these notes

and I think that the first can be found in any algebra text.