# ${(F^*)}^{-1}(m)$ is maximal ideal where $m$ is maximal

Our main objective is to interpret $$F:V \to W$$ a morphism as a map $$F:maxSpec \Bbb C[V] \to maxSpec \Bbb C[W]$$, $$V,W$$ are algebraic varieties.

Now from $$F:V \to W$$ using Hilbert Nullstelensatz we have the pullback $$F^*: \Bbb C[W] \to \Bbb C[V]$$. Now indeed given a point $$p \in V$$ we think it as a maximal ideal $$m_v \in maxSpec \Bbb C[V]$$. Then the image of $$p$$ under $$F$$ corresponds to the maximal ideal $$(F^*)^{-1}(m_v) \in maxSpec \Bbb C[W]$$ This is where I have the problem in showing that $$(F^*)^{-1}(m_v)$$ is a maximal ideal in $$\Bbb C[W]$$ which is not true in general.

My attempt: Suppose $$(F^*)^{-1}(m_v)$$ is not maximal then $$\exists m_w \subset \Bbb C[W]$$ maximal s.t $$(F^*)^{-1}(m_v) \subsetneq m_w$$ then $$m_w$$ corresponds to a point $$q \in W$$.

Let us comeback in the pullback part as we should have all the ingredients ready in our hand: for any $$g \in \Bbb C[W]$$ $$(F^*)(g)=g \circ F$$ so for any $$g \in (F^*)^{-1}(m_v)$$ we have $$(F^*)(g)=g \circ F \in m_v$$.

Now can we proceed from here, I think if we can prove $$F^*$$ is surjective then also we are done.

• You should say explicitly what $V, W$ are. Since you mention the Nullstellensatz and $\operatorname{Maxspec}$, my guess is that you mean them to be algebraic varieties (i.e., spaces whose coordinate rings are finitely generated nilpotent-free $\mathbb C$-algebras), but we shouldn't have to guess. – pyon Feb 16 at 8:40
• Okay, let me say it clearly. Thanks for reading the question and the correction. If possible please help. – Gimgim Feb 16 at 9:12

Let $$f\colon A\to B$$ be a $$k$$-algebra homomorphism. If $$J$$ is a maximal ideal of $$B$$ and $$I=f^{-1}(J)$$, then there is an injective $$k$$-algebra homomorphism $$A/I\to B/J$$ where the codomain is a field. If $$B$$ is finite-dimensional over $$k$$, then so are $$B/J$$ and $$A/I$$. Hence $$A/I$$ is an artinian ring which is a domain, hence a field. So $$I$$ is maximal.