Let $f:A→B$ be a ring morphism and $φ : \operatorname{Spec}B→\operatorname{Spec}A$ its induced continuous map, then prove that $\rm im\ φ ⊆ V(\ker f)$

Question

Today I was asked this question by my teacher:

Let $A$,$\ B$ be two conmutative rings with unity lef $f :A \rightarrow B$ be a ring morphism and $φ : \operatorname{Spec}(B) \rightarrow \operatorname{Spec}(A)$ the continuos function defined by $f$ where if $P \in \operatorname{Spec}(B) \:$ then $φ(P)=f^{-1}(P)$. Knowing this prove: $$\operatorname{im}(φ)\subset V(\ker f)$$ where $\ V(\ker f) = {\{P \in \operatorname{Spec}(A)\text{ and } \ker f \subset P \}}$

I already know that $V(\ker f)$ is homeomorphic to $\operatorname{Spec}(A/\ker f)$.

Answer 1

Strong hint:

This is essentially the correspondence theorem but for rings rather than groups. Namely, ideals in $A$ that contain $\ker f$ are in correspondence with ideals in $B$, under the same action that $\phi$ has. This theorem is introduced earlier in algebra so you’ve probably already seen it.

N.B. It’s homeomorphic and morphism.

Answer 2

I came up with this solution:

Let $P\in \operatorname{Spec}(B)$ lets see that $\phi(P)\in V(\ker (f)) \:$, as $P$ is an ideal we have that $0 \in P\:$ so $f^{-1}(0)\subset f^{-1}(P) \:$ that means that $\ker (f)\subset f^{-1}(P) \:$ so $f^{-1}(P)=\phi(P) \:$, we conclude that $\phi(P)\in V(\ker (f)) \:$

It's pretty simple but I think that is ok.