$f : A \to B$ be a ring homomorphism, projection map $ p: B \to B/M$, Is $p(f(A))/M$ isomorphic to $A/M$? I am reading a book and I think it uses the following proposition  
Let $f:A \to B$ be a ring homomorphism where $A$ and $B$ are finitely generated $\mathbb{C}$ algebras and let $M$ be a maximal ideal of $B$  and let $p$ be the projection map of $B \to B/M$. Then $p(f(A))/M$ is isomorphic to $A/M.$
The reason I belive this proposition is used is the following. There is an exersice saying if $A$ and $B$ and $p$ and $f$ are as above then prove that $f$ induces a well defined map from $Spec(B)$ to $Spec(A)$ and the solution to this exersice says the following(I am just copying what the solution says right now).We only need to prove that $f^{-1}(M)$ is a maximal ideal whenever $M$ is a maximal ideal.Now note that the composition $g=pf:A \to B/M$ is a homomorphism of $\mathbb{C}$-algebras and its kernel is $f^{-1}(M)$. As a consequence we have a well defined injective map or rings $q:A/f^{-1}(M) \to B/M$. But $A/M$ is again a finitely generated $\mathbb{C}$ algebra and it lies as a $\mathbb{C}$ sub algebra of $B/M= \mathbb{C}$ (Zarisky lemma), since the only non trivial $\mathbb{C}$-subspace of $\mathbb{C}$ is $\mathbb{C}$ we deduce that $A/M= \mathbb{C}$ and that $f^{-1}(M)$ is a maximal ideal.This solution makes sense to me only if we use the proposition I stated so that from the first isomorphism theorem $ A/f^{-1}(M)$ is isomorphic to $g(A)$ which is isomorphic to $A/M= \mathbb{C}$ which is a field and therefore $f^{-1}(M)$ is maximal
Is this true? How can we prove it?
Thanks in advance
 A: First of all I don't understand why you need all these. If we have a ring homomorphism $f: A \rightarrow B,$ then for each prime ideal $P$ of $B,$ $f^{-1}(P)$ is a prime ideal of $A.$ So it induces a well defined map $Spec(B) \to Spec(A).$ But it looks to me (from the proof given) that it was actually map $\text{m-Spec(B)} \to \text{m-Spec(A)}$ where $\text{m-Spec(A)}$ is the set of all maximal ideals of $A.$ In other words, given any maximal ideal $M$ of $B, f^{-1}M$ is a maximal ideal of $A.$
Let $M$ be a maximal ideal of $B.$ Then $f^{-1}M$ is a prime ideal of $A.$ We want to show that it is actually a maximal ideal. Let $p : B \to B/M$ be the canonical map and consider the composite map $g := pf: A \to B/M.$ Then $f^{-1}M = \text{ker}g.$ So we have an induced injective map $A/f^{-1}M \rightarrow B/M.$ Now $B/M = \mathbb{C}$ and the induced map was a $\mathbb{C}$-algebra homomorphism. The only $\mathbb C$-algebra contained in $\mathbb C$ is $\mathbb C$ itself. Thus $A/f^{-1}M = \mathbb C.$ Hence $f^{-1}M$ is a maximal ideal of $A.$
