For a field extension $L\mid K$, I want to show that the following are equivalent:
- $L|K$ is algebraic.
- For every $E\in \mathfrak{F}(L|K)$, every $K$-algebra homomorphism $\sigma:E\to E$ is an automorphism.
Where $\mathfrak{F}(L|K)$ is the all intermediary subfields of $L|K$.
I can show $1\implies 2$ as below:
Let $\sigma : E \to E$ be a $K$-algebra homomorphism. Since $\sigma$ is a non-zero homomorphism, Ker $\sigma\neq E$. Since $\{0\}$ and $E$ are the only ideals of $E$, it follows that Ker $\sigma = \{0\}$. This implies that $\sigma$ is one-one.\
To show $\sigma$ is an automorphism, it only remains to be shown that $\sigma$ is onto. Let $a \in E$ be any element. Let $$p(X) = a_0+a_1X+\ldots+a_kx^k \in K[X]$$ be the minimal polynomial of $a$ over $K$. Let $b$ be any root of $p(X)$ in $E$. Then $$p(\sigma(b)) = a_0 +a_1\sigma(b)+\ldots+a_k\sigma(b)k = \sigma(a_0 +a_1b+\ldots+a_k b^k) = \sigma(0)=0.$$ Hence, $\sigma(b)$ is also a root of $p(X)$. Let $E_0$ be the subfield of $E$ generated by all roots of $p(X)$ over $K$ that lie in $E$. Thus, $E_0$ is generated over $K$ by a finite set of elements in $E$ that are algebraic over $K$. Hence $E_0|K$ is a finite algebraic extension. Since $\sigma$ maps a root of $p(X)$ to a root of $p(X), \sigma$ maps $E_0$ into $E_0$. Since $$[E_0 : K] = [\sigma(E_0) : \sigma(K)]=[\sigma(E_0):K]$$ and $$[E_0:K]=[E_0:\sigma(E_0)][\sigma(E_0):K]$$ it now follows that $[E_0 : \sigma(E_0)] = 1$ and so $\sigma(E_0) = E_0$. Hence, $a\in E_0 =\sigma(E_0) \subseteq \sigma(E)$. Thus, $\sigma$ is sujective and hence an automorphism.
But how to show the converse?
Suppose $L/K$ is not algebraic. Then $L$ is algebraic over $K(t : t \in S )$ for some set $S$. Take $t \in S$ and let $\sigma : K(t) \rightarrow K(t)$ be a $K$-algebra map. Then $K(t) \stackrel{\sigma}{\rightarrow} K(t) \rightarrow K(t : t \in S)$ extends to a $K$-algebra map $K(s : s \in S) \rightarrow K(s: s \in S)$ sending $s$ to $s$ for $s \in S - \{ t \}$. Then $K(s : s \in S) \rightarrow K(s : s \in S) \rightarrow L$ extends to a map $L \rightarrow L$ in the usual way.
The effect of all this that we have reduced to the case of the extension $K(t) / K$. There are many choices here, but I'll take red_trumpet's approach: Define $K[t] \rightarrow K(t)$ sending $t$ to $t^2$. Then nonzero elements map to nonzero elements, or in other words, nonzero elements map to units since the target is a field. By the universal property of localization, we get a homomorphism $K(t) \rightarrow K(t)$. But $t$ does not occur in the image of this map, so it is not an automorphism.