Integral Ring Extension induces Surjection

Question

Let $\phi:R \subset A$ an integral ring extension.

I want to know why and how to see that in this case the induced morphism $f_{\phi}: \operatorname{Spec}(A) \to \operatorname{Spec}(R)$ is surjective.

Answer 1

The classical proof goes along these lines:

We need a lemma:

Lemma: Let $A$ be an integral domain, $R$ be a subring such that $A$ is integral over $R$. The $A$ is a field if and only if $R$ is.

Suppose first $R$ is local ring, with maximal ideal $\mathfrak m$ , and consider any maximal ideal $\mathfrak n$ of $A$. As $R/\mathfrak n\cap R \hookrightarrow A/\mathfrak n\;$ is injective, the above lemma shows $\mathfrak n\cap R\;$ is a maximal ideal of $R$, hence $$\mathfrak n\cap R=\mathfrak m,$$ i.e. the maximal ideal ideal of $R$ lifts to $\operatorname{Max} A \subset\operatorname{Spec} A$.

Now for the general case: let $\mathfrak p\in\operatorname{Spec}R$, localise at $\mathfrak p$ and consider the commutative diagram: \begin{alignat}{2} R_{\phantom{\mathfrak p}}& \hookrightarrow& A_{\phantom{\mathfrak p}}\\ \downarrow_{\phantom{\mathfrak p}}&&\downarrow_{\phantom{\mathfrak p}}\\ R_{\mathfrak p}& \hookrightarrow&\; A_{\mathfrak p} \end{alignat} By the local case, $\mathfrak p R_{\mathfrak p}$ lifts to a maximal ideal $\mathfrak n $ in $A_{\mathfrak p}$. This ideal is the localisation $\mathfrak q_{\mathfrak p}$ of a prime ideal $\mathfrak q\in\operatorname{Spec} A$, and, since the inverse image of $\mathfrak n =\mathfrak q_{\mathfrak p}$ in $R_{\mathfrak p}$ is $\mathfrak p R_{\mathfrak p}$, hence its inverse image in $R$ is $\mathfrak p$, the inverse image of $\mathfrak q$ in $R$ by the upper map is also $\mathfrak p$.