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.

  • $\begingroup$ 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? $\endgroup$
    – KReiser
    Mar 31 at 18:51
  • $\begingroup$ Sorry, it's a typo. I'm referring to Proposition 1.16 in Liu's book $\endgroup$
    – Rdrr
    Mar 31 at 18:53

1 Answer 1


Start with this theorem:

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:

enter image description here

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):

enter image description here

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 $\phi^{-1}(q) = p$

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

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