Is this inheritance valid for algebraic extensions?

Question

Let $E/F$ be an algebraic extension.

Let $\sigma:F\to E$ be a homomorphism in the category of fields.

Can it be shown that the extension $E/\sigma(F)$ is also algebraic?

Answer 1

Another example, based somehow on Qiaochu’s: Take any field $k$, and let $E=F=k(t_1,t_2,t_3,\cdots)$, a purely transcendental extension of $k$ of countable transcendence degree. Then we can set $\sigma:F\to E$ by $\sigma(t_i)=t_{i+1}$, but identity on $k$. Then $E$ is a transcendental extension of $\sigma(F)$, with transcendence basis $\{t_1\}$.

I suppose that the question to ask is what the situation may be if $\sigma$ is a $k$-morphism when $E$ and $F$ are finitely generated fields over $k$.

Answer 2

No. For example, take $E = F = \mathbb{C}$. I claim there are maps $\sigma : \mathbb{C} \to \mathbb{C}$ making $\mathbb{C}$ a non-algebraic extension of itself. This is because every algebraically closed field of characteristic $0$ and the same cardinality as $\mathbb{C}$ is isomorphic to $\mathbb{C}$, and in particular the algebraic closure of $\mathbb{C}(t)$ has these properties.