Atiyah MacDonald Exercise 1.21 iv) Proof of continuity Let $\Phi: A \rightarrow B$ be a surjective ring homomorphism. Let $X = \text{Spec}(A)$ and $ Y = \text{Spec}(B)$, then $\Phi^*=\Phi^{-1}: Y \rightarrow V(Ker(\Phi))$.
I already showed that $\Phi^*$ is bijective and continuous.
Now I want to show that $\Phi^{*-1}=\Phi: V(Ker(\Phi)) \rightarrow Y$ is continuous. For that I wanted to go through the same route as in i) where we showed that $\Phi^*$ is continuous by showing that the inverse image of basic open sets is open.
$P \in \Phi^{-1}(Y_g) \iff \Phi(P) \in Y_g \iff g \not \in \Phi(P) \iff (\forall h \in  A : h \in P \implies \Phi(h)\not=g) \iff (\forall h \in A: \Phi(h) = g \implies h \not \in P) \iff \forall h \in \Phi^{-1}(g): P \in X_h \iff P \in \bigcap_{h\in \Phi^{-1}(g)} X_h$
So I get that that $\Phi^{-1}(Y_g) = \bigcap_{h\in \Phi^{-1}(g)} X_h$ which is an intersection of basic open sets. But couldn't it be potentially an infinite intersection? And an infinite intersection of open sets doesn't need to be open. Can this proof be saved?
 A: Let $\phi \colon A \to B$ be a ring homomorphism, $X = \text{Spec}(A)$, $Y = \text{Spec}(B)$, and $\phi^* \colon Y \to X$ the induced mapping of $\phi$.
The exercise says that if $\phi$ is surjective, then $\phi^*$ is a homeomorphism of $Y$ onto the closed subset $V(\ker \phi)$ of $X$, or in more formal terms, the map
\begin{align*} \newcommand{\q}{\mathfrak{q}}
Y & \stackrel \Phi \longrightarrow V(\ker \phi) \\ \q & \longmapsto \phi^*(\q)
\end{align*}
is a homeomorphism.
First of all, why is $\Phi$ well-defined? That is, given $\q \in Y$, why is $\phi^*(\q)$ an element of $V(\ker \phi)$? Well, that’s easy: applying the inverse image to the inclusion $0 \subseteq \q$ we get that $\ker \phi = \phi^{-1}[0] \subseteq \phi^{-1}[\q] = \phi^*(\q)$.
Second, $\Phi$ is continuous because $\phi^*$ is (due to the universal property of the subspace topology, basically).

Next, you state that the map
\begin{align*} \newcommand{\p}{\mathfrak{p}} V(\ker \phi) & \stackrel \Psi \longrightarrow Y \\ \p & \longmapsto \phi[\p]
\end{align*}
is the inverse of $\Phi$. This is true, but it requires a little more work than you did.
Again, we need to prove that $\Psi$ is well-defined: given $\p \in V(\ker \phi)$, we need to prove that $\phi[\p]$ is an element of $Y$, that is, a prime ideal of $B$.

*

*Since $\phi$ is surjective, the image $\phi[\p]$ is an ideal of $B$.

*Take $x,y \in B$ with $xy \in \phi[\p]$. Then $xy = \phi(p)$ for some $p \in \p$.
Moreover, there are $x’,y’ \in A$ with $\phi(x’) = x$ and $\phi(y’) = y$.
Hence $\phi(x’y’) = \phi(p)$, and that means $x’y’-p \in \ker \phi$.
Finally, as $\ker \phi \subseteq \p$, it follows that $x’y’-p \in \p$, and consecuently $x’y’ \in \p$, from which we can conclude that $x’$ or $y’$ is in $\p$, so that $x$ or $y$ is in $\phi[\p]$.


Now, why is $\Psi$ continuous? If $g \in B$, I will leave to the reader to prove that
$$
\Psi^{-1}[Y_g] = X_f \cap V(\ker \phi),
$$
where $f$ is any element of $A$ with $\phi(f)=g$ ($\phi$ is surjective!).
Hence, the inverse image of any open basic subset of $Y$ is an open subset of $V(\ker \phi)$; so, $\Psi$ is continuous.
Finally, let see that $\Phi$ and $\Psi$ are, indeed, inverses to each other:

*

*As $\phi$ is surjective, $\phi[\phi^{-1}[\q]] = \q$ for any $\q \in Y$; so $\Psi \circ \Phi = \text{id}$.

*Using only the additivity of $\phi$, it can be shown that
$$
\phi^{-1}[\phi[S]] = S + \ker \phi
$$
for any subgroup $S$ of $A$.
Therefore, $\phi^{-1}[\phi[\p]] = \p$ for every $\p \in V(\ker \phi)$; i.e. $\Phi \circ \Psi = \text{id}$.

