A homomorphism induces a continuous map from ${\rm Spec}(A') \to {\rm Spec}(A)$.

Question

Let $A, A'$ be commutative rings with $1 \neq 0$. Let $h : A \to A'$ be such that $h(1) = 1$. Then $f: {\rm Spec}(A') \to {\rm Spec}(A)$ defined by $f(\mathfrak{p}') = h^{-1}(\mathfrak{p}')$ is continous with respect to the spectral topologies.

Answer 1

Hint: The ideal you're looking for is $\mathfrak b' = \mathfrak a^e$ where $\mathfrak a^e$ (the extension of $\mathfrak a$) is the ideal in $A'$ generated by $h(\mathfrak a)$.

Now that you know what the other side should be, showing that $f^{-1}V(\mathfrak a) = V(\mathfrak a^e)$ isn't too bad. Just do both inclusions separately.

Answer 2

Thanks @Jim. I've got it now. $f^{-1} V(\mathfrak{a}) = \{ \mathfrak{p}' \in {\rm Spec(A')} : f(\mathfrak{p}') \in V(\mathfrak{a}) \equiv h^{-1}\mathfrak{p}' \in V(\mathfrak{a}) \equiv \exists \mathfrak{p} \in V(\mathfrak{a}) : h(\mathfrak{p}) = \mathfrak{p}'\} \subset V((h(\mathfrak{a}))$

since $\mathfrak{p} \supset \mathfrak{a} \implies h(\mathfrak{p}) = \mathfrak{p}' \supset (h(\mathfrak{a}))$. The opposite inclusion can be shown more easily.

Answer 3

Your question follows from the following:

• If $I$ is an ideal in $A$, then $I \subset I^{ec}$.
• If $J$ is an ideal in $A'$, then $J^{ce} \subset J$.

These facts can be proved easily from the definition of contracted and extended ideals.

Now, let $I$ be an ideal in $A$. Let $\mathfrak p \in f^{-1}(V(I))$. We have $I \subset \mathfrak p^c$; hence $I^e \subset \mathfrak p^{ce} \subset \mathfrak p$. Thus, $\mathfrak p \in V(I^e)$.

Conversely, if $\mathfrak p \in V(I^e)$, then $I^e \subset \mathfrak p$; hence $I^{ec} \subset \mathfrak p^c$. Since $I \subset I^{ec}$, it follows that $\mathfrak p \in f^{-1}(V(I))$.

Therefore $f^{-1}(V(I)) = V(I^e)$ and $f$ is continuous.