Why is $k \rightarrow A \rightarrow A / I$ and isomorphism of rings if $I \subset A$ is maximal?

Question

Let $k$ be a algebraically closed field, $A$ a finitely generated $k$-Algebra and $I \subset A$ a maximal ideal. Let $\varphi: k \rightarrow A$ be a ring homomorphism. Why is this combination $$k \rightarrow A \rightarrow A / I$$ an isomorphism?

Answer 1

The general case is not going to be true and here is why: Take $k = \Bbb{C}$, $A = \big(\Bbb{C}(x)\big)[y]$ and $I = (y)$ that is maximal since $\big(\Bbb{C}(x)\big)[y]/(y) \cong \Bbb{C}(x)$ is a field. But now $ \Bbb{C}(x)$ is much bigger than $\Bbb{C}$ (in fact infinite dimensional over $\Bbb{C}$) and so your compositum cannot possibly be an isomorphism.

Your result though is true in the case when $A$ is a finitely generated $k$ - algebra. If $I$ is a maximal ideal then $A/I$ is a finitely generated $k$ - algebra that is a field, and hence by Zariski's lemma is a finite algebraic extension of $k$. Since $k$ is algebraically closed, $A/I = k$ and so your compositum can be viewed as an injective linear map between two $1$ - dimensional vector spaces and thus is an isomorphism.