0

So this is a very basic question and I am sure that I am just missing something obvious. Given a commutative ring homomorphism $f: A \rightarrow B$, we have an induced map on spectra $f^*: Spec(B) \rightarrow Spec(A)$ via the formula $f^*(\mathfrak{p}):= f^{-1}(\mathfrak{p})$. I just can't see how $f^*$ may not be injective (and hence the fibres nontrivial). It seems to me that the the map definition implies $f^*(\mathfrak{p}) = f^*(\mathfrak{q}) \Rightarrow f^{-1}(\mathfrak{p}) = f^{-1}(\mathfrak{q}) \Rightarrow \mathfrak{p} = \mathfrak{q}$.

szantag
  • 71
  • Why does $f^{-1}(\mathfrak{p}) = f^{-1}(\mathfrak{q})$ imply that $\mathfrak{p} = \mathfrak{q}$? Do you have examples in mind? – Randall Feb 04 '22 at 13:32
  • $f^{-1}(\mathfrak{p})$ is the set of elements that map to $\mathfrak{p}$ under $f$. If it coinsides with the set of elements mapping to $\mathfrak{q}$, how can $\mathfrak{q} \neq \mathfrak{p}$? Aren't they both the image of the same set under the same map? – szantag Feb 04 '22 at 14:16
  • 1
    But the ideals are treated as subsets here. Certainly in a function you can have $f^{-1}(A) = f^{-1}(B)$ yet $A \neq B$ when $A$ and $B$ are subsets of the codomain (not points). – Randall Feb 04 '22 at 14:43
  • I am really sorry, I seem to have some mental block. Wouldn't it mean that there exists $x \in f^{-1}(A)=f^{-1}(B)$ such that $f(x) \in A$ and $f(x) \notin B$ (or vice versa)? But then $x$ wouldn't be in $f^{-1}(B)$ – szantag Feb 04 '22 at 16:21
  • 2
    Dear szantag, consider the following simple (set theoretical) example: $f:{1} \rightarrow {1,2 }$, $f(1)=1$. Then $f^{-1}({1 }) = f^{-1}({1,2 })$. – cos_dm_math21 Feb 04 '22 at 16:25
  • 2
    Here is a counterexample for your statement: Take the inclusion morphism $i:\mathbb{Z} \rightarrow \mathbb{Z}[X]$. Note that the preimages of the zero ideal $(0)$ and ideal $(X)$ are $(0)$. – cos_dm_math21 Feb 04 '22 at 16:37

0 Answers0