Is it possible to extend a homomorphism from an etale algebra into the separable closure? I am currently learning etale algebras and they seem to be generalizations of separable algebraic field extensions, so I wanted to try to explore which properties hold and which don't. My question is as follows:
Let $L$ be an etale algebra over a field $K$,$L'$ a subalgebra of $L$ and $f:L' \longrightarrow K_{sep}$ a morphism of $K$ algebras from $L'$ into the separable closure of of $K$. Is it possible to extend $f$ into a morphism from $L$ to $K_{sep}$?
 A: I think I found an answer.
Consider $L=L'[\theta ]$. If we manage to prove this case, we can then proceed by induction.
First of all an etale algebra is by definition a product of separable extensions of $K$: say $L=F_1 \times \cdots \times F_m$. Now consider $\pi_i$ to be the natural projection of $L$ onto $F_i$.
$ker(f)$ is a maximal ideal, since $\frac{L'}{ker(f)}$ is isomorphic to $im(f)$, which has to be a field.
For all $i$ $ker(\pi_i) \cap L'$ is a maximal ideal of $L'$ for the same reason. Now let us show that there exists $i_0$, such that $ker(\pi_{i_0})\cap L' \subseteq ker(f)$. If it wasn't the case, we could for every $i$ find an $a_i \in ker(\pi_i)$,such that $f(a_i) \neq 0$. Then $f(\prod_i a_i )$ is non zero, since $f$ is a morphism. However $\prod_i a_i$ has to be zero, since all of its components are null, contradicting that $f$ is a morphism.
Now consider the ideal of polynomials $I=\{P\in L'[X]| P(\theta )=0 \}$. Then write $f(I)=\{f(P) \in f(L')[X]|P \in I\}$. Given that $f$ is surjective onto the field $f(L')$, $f(I)$ is an ideal in the polynomial ring $f(L')[X]$. Now let us prove, that $f(I)$ is not the whole $f(L')[X]$.
By contradiction assume it is. Then there exist $a_0, \cdots , a_n \in L'$, such that $f(a_0)+f(a_1)X+\cdots +f(a_n)X^n=1$ and $a_0+a_1 \theta+ \cdots a_n \theta^n=0$. From the first equation we deduce that $f(a_0)=1$ and $f(a_i)=0$ for $i>1$. Since $ker(f)=L'\cap ker(\pi_{i_0})$, we deduce that $pi_{i_0}(a_i)=0$ for $i>1$. Applying that to the second equation, we get that $\pi_{i_0}(a_0)=0$. However since $f$ and $\pi_{i_0}$ have the same kernel, we get that $f(a_0)=0$, which is a contradiction.
Now that we know that the ideal $f(I)$ is not the whole $f(L')$, we know that it is generated by a non constant polynomial $f(Q)$ with $Q\in I$. Since it is non constant, we can take $u$ a root of $Q$ in $K_{sep}$. Then finally we extend $f$ by the formula $f(P(\theta ))=f(P)(u )$.
