Let $F$ be an algebraically closed field and $E$ a field. Assume that there are homomorphisms(so monomorphisms)
$i : E \to F$ and $j:F \to E$.
Then $j$ is an isomorphism?
My question origianes from next statement (Gortz, Algebraic Geometry, Corollary 3.36)
Here, the Proposition 3.8. is, :
I'm trying to understand the underlined equality.
By the Proposition 3.8., it suffices to show that there exists a bijection
$$ \{x\in X ; i_x : k \cong \kappa(x) \} \cong \{(x,\iota) ; x\in X, \iota : \kappa(x) \to k \}$$
(where $i_x : k \hookrightarrow \kappa(x)$ is the natural homomorphism (Gortz's book prop.3.33 and p.60 (3.4.2) ) ; i.e, applying the global section functor $\Gamma$ to $\operatorname{Spec}(\kappa(x)) \to X \to \operatorname{Spec}(k)$)
Let's define $\eta : \{x\in X ; i_x : k \cong \kappa(x) \} \to \{(x,\iota) ; x\in X, \iota : \kappa(x) \to k \}$ by $ x \mapsto (x, i_x^{-1})$. I'm trying to show that this map is a bijection.
For showing the surjectivity, I'm trying to show that for $(x, \iota)$, $x$ satisfies that the $i_x : k \hookrightarrow \kappa(x)$ is an isomorphism. If our above question is true, then we may show this. And our question is true? If not, how can we show the underlined statement?
Can anyone help ?
Modified Question : In the above question, if furthermore $i$, $j$ satisfy $i∘j=id_F$, then $j$ is an isomorphism?
If this question is true, then by reffering closed points of a scheme and k-points, we can deduce the underlined equality.
Best Answer
This is false. Consider $\mathbb{C} \rightarrow \mathbb{C}(t) \rightarrow \overline{\mathbb{C}(t)}$.
Since $\overline{\mathbb{C}(t)}$ and $\mathbb{C} $ are both algebraically closed, and have the same characteristic and the same transcendence degree over the prime field, they are isomorphic. However, $\mathbb{C}(t)$ is not isomorphic to $\mathbb{C}$