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Suppose $f\colon B\to A$ is a ring homomorphism, and $I\subseteq B$ is an ideal. What's the geometric interpretation for the extension $f(I)A$ of the ideal $I$? Especially, I'm interested in the case that $I$ is prime or maximal.

Let's consider a special case. If $k$ is an algebraically closed field, and $X,Y$ are isomorphic to affine lines $\mathbb A_k^1$. Let's consider the morphism $\varphi\colon X\to Y,x\mapsto y=x^2$, which induces the pullback $k$-algebra homomorphism $f\colon B=k[Y]\to A=k[X]$. If $I=(Y-a^2)$ is a maximal ideal in $B$, then $f(I)A=(X^2-a^2)$. If $a\neq0$, $f(I)A$ naturally corresponds to the fiber $\varphi^{-1}(a)$. However, if $a=0$, the case is complicated. It's related to $\varphi^{-1}(0)$, but contains an infinitesimal neighborhood.

I need some general precise statement for this. Any idea? Thanks!

Yai0Phah
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1 Answers1

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The ideal $f(I)A$ is the ideal of the scheme-theoretic fiber $\text{Spec}(A) \times_f \text{Spec}(B/I)$, that is, the restriction of the morphism $\text{Spec}(A) \to \text{Spec}(B)$ to the subscheme $\text{Spec}(B/I)$.

So algebraically, $f(I)A$ is the kernel of the map of $B$-algebras

$$A \to A \otimes_B B/I \cong A/f(I)A.$$

So it is the (ideal of the) preimage, as you said, but it has the correct scheme structure.

Yai0Phah
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Jake Levinson
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    +1: To add to what Jake is saying, take the simplest example, as in the one you wrote down, of a map of smooth affine curves $f:\text{Spec}(B)\to\text{Spec}(A)$. Then, for a prime $\mathfrak{p}$ from $A$, you see that $\mathfrak{p}B$ factors as $\displaystyle \prod_{\mathfrak{P}\in f^{-1}(\mathfrak{P})}\mathfrak{P}^{e(\mathfrak{P}\mid\mathfrak{p})}$. So, this ideal contains the points in the fiber $f^{-1}(\mathfrak{p})$, with the ramification data. This is precisely the geometric information one would expect out of the scheme theoretic fiber, but its origins are also obvious algebraically. – Alex Youcis Jul 14 '14 at 05:27
  • But the fiber of ${}^af$ over a prime ideal $P$ is $\operatorname{Spec}(k(P)\otimes_BA)$, not $\operatorname{Spec}((B/P)\otimes_BA)$, where $k(P)=\operatorname{Frac}(B/P)$? – Yai0Phah Jul 14 '14 at 11:13
  • @FrankScience: Oops, there's some ambiguity in the terminology. What you describe is the "fiber over the generic point of $\text{Spec}(B/P)$". I guess the word "fiber" is most often used to refer to preimages of single closed points (where $P$ is maximal and $k(P) = B/P$, so there's no ambiguity), but in this case I meant it simply as "scheme-theoretic preimage", or, to use possibly better terminology, the fiber product. In other words, it is the preimage of the whole subvariety, not just of the generic point of the subvariety. – Jake Levinson Jul 14 '14 at 14:00
  • In fact I'm not sure about these. I didn't go into these deep theories, but I looked up Atiyah & Macdonald pp 47, ex 21 iv) to find the definition of the fiber. I haven't gone into sheaves, schemes, etc, but only trying to obtain some rough idea on this. Sorry for my ignorance. Thanks! – Yai0Phah Jul 14 '14 at 15:10
  • No worries. AG takes a long time to learn (but is worth it), and it's definitely a good idea to ask a lot of questions. – Jake Levinson Jul 14 '14 at 23:20
  • Further remarks to complete this answer: first, instead of fiber, I prefer to consider the whole morphism: $\operatorname{Spec}(A/f(I)A)\to\operatorname{Spec}(B/I)$, just like the restriction of fiber bundle into a subset. Second, I want to figure out that $A\otimes_BB/I\cong A/f(I)A$ is a corollary of the right exactness of $A\otimes_B-$, though could be checked directly. – Yai0Phah Mar 13 '15 at 11:17