Understanding Kenji Ueno - Algebraic Geometry 1, lemma 1.23
For a k-Homomorphism $\psi : S\to R$ between k-Algebras, the inverse image $\psi^{-1}(m)$ of a maximal ideal m of $R$ is a maximal ideal of $S$
In the proof, there are two things I don't understand.
1.it states that $\psi $ induces an isomorphism $\widetilde{\psi }:S/\psi^{-1}(m) \to R/m$
Why $\widetilde{\psi }$ is an isomorphism ?
- How we get $R/m \subset S/\psi^{-1}(m) \Rightarrow S/\psi^{-1}(m) $ is a field ?
I understand that $R/m$ is a field sience $m$ is maximal, so if $\widetilde{\psi }$ is isomorphism we automated get that $S/\psi^{-1}(m) $ is a field (and $\psi^{-1}(m)$ is maximal) so I don't really understand the purpose of 2.
Here's what the author is saying in more detail:
Consider the composition $S\overset{\psi}\to R\to R/m$, let's call this $\overline\psi$. The kernel of $\overline\psi$ (you can check) is exactly $\psi^{-1}(m)$, so by the First Isomorphism Theorem you have an induced isomorphism of $k$-algebras
$$\widetilde\psi:S/\psi^{-1}(m)\to\operatorname{Im}(\overline\psi),\quad\quad \widetilde\psi(s+\psi^{-1}(m))=\psi(s)+m.$$
However, I claim that $\overline\psi$ is surjective, meaning that $\operatorname{Im}(\overline\psi)=R/m$ which gives the isomorphism you are asking about.
To see this, you recall that previously given hypotheses (as correctly predicted by KReiser in the comments) are that $(1)$ $k$ is algebraically closed and $(2)$ $R$ and $S$ are finitely generated $k$-algebras.
$(2)$ implies that $R/m$ is a finite extension of $k$, and then by $(1)$ you must have $R/m=k$. [Side note: what this really means is that the natural composition $k\hookrightarrow R\to R/m$ is an isomorphism.]
But since $\overline\psi(\alpha)=\alpha$ for any $\alpha\in k$, and $R/m=k$, you see that every element of $R/m$ is inside $\operatorname{Im}(\overline\psi)$, and so $\overline\psi$ is surjective as claimed.