Inverse image of maximal ideal under $k$-algebra homomorphism is a maximal ideal. 
Let $k$ be an algebraically closed field, let $R,S$
  be finitely generated $k$-algebras without nonzero nilpotent elements.
  Let $f\in \text{Hom}_k(R,S)$, and $M$ a maximal ideal of $S$. Then $f^{-1}(M)$ is a maximal ideal of $R$.

I've thought of this proof:
We know that $R/f^{-1}(M)$ is an integral domain which "contains" $k$.
Furthermore the homomorphism $f$ with the canonic epimorphism $S\to S/M$ induces an homomorphism
$f'=\sigma\circ f:R\to S/M$. By fundamental homomorphism theorem we have a monomorphism $f'':R/f^{-1}(M)\to S/M$.
Abusing notation we can say then:
$$k\subseteq R/f^{-1}(M)\subseteq S/M=k$$
since $S/M$ is a finite extension of $k$. 
Then $R/f^{-1}(M)=k$ and thus $f^{-1}(M)$ is a maximal ideal of $R$.
I have seen other proofs, which extend a bit more, to prove that $R/f^{-1}(M)$ is a field, not necessarily $k$ (but with the hypothesis for $k$ to be algebraically closed). So I have some questions.


*

*Is my attempt of proof correct? Is there any problem in the abuse of the notation?

*I haven't used the fact that the algebras are reduced, but when I found the proposition the first time here, the hypothesis was there. (The proof attempt and the answer seemed too crazy for me so I thought in trying to prove it myself.) In other places it does not appear, like here. Does this hypothesis contribute in any way?

*Does $k$ have to be algebraically closed?
 A: Your proof is basically correct.  To make your abuse of notation rigorous, you should talk in terms of maps instead of inclusions (though your abuse of notation is not uncommon).  The key point is that $S/M\cong k$ as a $k$-algebra, meaning that the ring-homomorphism $k\to S/M$ that makes $S/M$ a $k$-algebra is an isomorphism.  Since the homomorphism $R/f^{-1}(M)\to S/M$ is a homomorphism of $k$-algebras, the composition of the $k$-algebra structure homomorphism $k\to R/f^{-1}(M)$ with $R/f^{-1}(M)\to S/M$ is equal to the $k$-algebra structure homomorphism $k\to S/M$, which we know is an isomorphism.  Since $R/f^{-1}(M)\to S/M$ is injective, this implies that $k\to R/f^{-1}(M)$ must be a bijection, so $R/f^{-1}(M)$ is isomorphic to $k$ as a $k$-algebra as well.
The hypothesis that the algebras are reduced is not necessary.  Indeed, even if you proved it only for reduced algebras, the general case follows immediately, since you could just apply the statement to the induced homomorphism $R/\sqrt{0}\to S/\sqrt{0}$ and use the fact that $R\to R/\sqrt{0}$ induces a bijection on the set of maximal ideals for any ring $R$.
The hypothesis that $k$ is algebraically closed is also unnecessary, but is a bit more difficult to eliminate.  If $k$ is not algebraically closed, you need to use a version of the Nullstellensatz for non-algebraically closed fields, which says that if $K$ is a field extension of $k$ that is finitely generated as a $k$-algebra, then it is finitely generated as a $k$-module.  Applying this to $S/M$, you get that $K=S/M$ is a finite (in particular, algebraic) field extension of $k$, and you have $k\subseteq R/f^{-1}(M)\subseteq K$ (continuing your abuse of notation).  But for any $a\in K$, $k[a]=k(a)$ (since $a$ is algebraic over $k$), which means that $R/f^{-1}(M)$ must be a field.
