# Hartshorne II.3.10

I'm working on Exercise II.3.10 from Hartshorne and I'm baffled by what should be a relatively simple exercise on schemes. The exercise states

If $$f:X\to Y$$ is a morphism (of schemes), $$y\in Y$$ a point, show that $$\mathrm{sp}(Y)$$ is a homeomorphic to $$f^{-1}(y)$$ with the induced topology.

I've looked up several proofs of this fact and all of the proofs being with the same step: Reducing to case where $$Y$$ is affine. I have a hard time seeing just how one can do this. Many proofs claim that if $$V$$ is an open affine of $$Y$$ containing $$y$$, then $$X_y = (X\times_Y V) \times_V \mathrm{Spec}k(y) = f^{-1}(V)_y$$ The only proofs I've seen, just claim that this is the case from the universal property of fiber products without further elaboration. I can't seem to show that $$f^{-1}(V)$$ satisfies the universal property.

I'd really appreciate if someone could help me fill in the details. A full proof would be the most desirable.

• Your big equation line says $X_y=f^{-1}(V)$ which isn't really what you want (if $Y=V$ is affine, then you're saying $X_y=X$, but $Y$ need not be a single point). Is this a typo, or is that part of your misunderstanding? Mar 31 at 18:51
• Sorry, it's a typo. I'm referring to Proposition 1.16 in Liu's book
– Rdrr
Mar 31 at 18:53

Let $$f : X \rightarrow Y$$ be continuous and injective. Let $$(U_i)_{i \in I}$$ be an open cover of $$Y$$. Suppose that $$f|_{f^{-1}(U_i)} : f^{-1}(U_i) \rightarrow U_i$$ are all homeomorphisms onto their range. Then, $$f$$ is a homeomorphism onto its range.

Proof: To check this, we need to show that the inverse is continuous (as a function $$range(f) \rightarrow X$$). Let $$V \subseteq X$$ be open, and we want to show $$f^{-1}(V)$$ is open in $$range(f)$$. It suffices to show that $$V \cap U_i$$ is open in $$range(f)$$ for each $$i \in I$$. This is true since $$f|_{f^{-1}(U_i)}$$ is a homeomorphism onto its range, and $$U_i$$ being open means that $$V \cap U_i$$ is open in $$range(f)$$ iff it's open in $$range(f) \cap U_i$$. $$\blacksquare$$

Now to schemes : let $$f : X \rightarrow Y$$ be a morphism between schemes, and let $$p \in Y$$ be a point. Let $$V \subseteq Y$$ be an affine open set containing $$p$$ and $$U \subseteq X$$ be an affine open set inside $$f^{-1}(V)$$. Then, make this diagram:

The two squares are fiber products, so the outside rectangle is also a fiber product. Since $$f(U) \subseteq V$$, $$U \times_Y Spec(k(p)) = U \times_V Spec(k(p))$$. This is a fiber product of affine schemes over an affine base, so (assume we already know the result for affine schemes) we know that $$\pi_1^{-1}(U) \rightarrow U$$ is a homeomorphism onto its range. Therefore, it's a local homeomorphism.

The last thing to check is to show that it's injective. I'll use theorem 26.17.5 in the Stacks Project for this. Let $$x \in X$$ with $$f(x) = p$$. The points in $$X \times_Y Spec(k(p))$$ whose image in $$X$$ is $$x$$ are in bijection with prime ideals in $$k(x) \otimes_{k(p)} k(p) = k(x)$$, but $$k(x)$$ is a field so the only prime ideal is $$(0)$$. Therefore, $$\pi_1$$ is injective.

Here is how to show the theorem for affine schemes:

First, note that if $$\phi : A \rightarrow B$$ is a ring homomorphism, and $$I$$ is an ideal of $$A$$, and $$S$$ is a multiplicatively closed subset of $$A$$ , then $$A/I \otimes_A B = B / \phi(I)B$$ and $$S^{-1}A \otimes_A B = \phi(S)^{-1}B$$. (I'll leave the proofs to you, the idea is to find a bilinear map by multiplication, apply the universal property of tensor products, and then find an inverse) .

Now, let $$p \subseteq A$$ be a prime ideal, and let $$\phi' : A/p \rightarrow B / \phi(p) B$$ be the induced ring homomorphism, and construct this diagram (all squares are tensor products):

Both localization and quotients induce morphisms on Spec that are homeomorphisms onto their range. $$R \rightarrow R/I$$ is a homeomorphism onto $$V(I)$$, and $$R \rightarrow S^{-1}R$$ is a homeomorphism onto the prime ideals disjoint from $$S$$.

In our case, $$B \rightarrow \phi'(A/p - \{0\})^{-1} (B / \phi(p) B)$$ induces a homeomorphism onto its range, and its range are the prime ideals $$q$$ of $$B$$ that contain $$\phi(p)$$ and the image of $$q$$ in $$B / \phi(p)$$ doesn't contain anything in $$\phi'(A/p - \{0\})$$. In other words, it doesn't contain anything in $$range(\phi')$$ other than 0.

Now, note that $$q$$ satisfies that iff $$\phi^{-1}(q) = p$$. I'll leave this verification to you as well.

Therefore, $$B \rightarrow \phi'(A/p - \{0\})^{-1} (B / \phi(p) B)$$ is a homeomorphism onto its range, and its range is precisely the prime ideals $$q$$ in $$B$$ such that $$\phi^{-1}(q) = p$$

• You say, assume we already know the result for affine schemes, but that is a large part of my question.
– Rdrr
Apr 1 at 15:43
• I added that part Apr 2 at 2:56