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The following shows up in the proof of theorem I 4.4 in Hartshorne.

Let $X,Y$ be varieties and $\theta: K(Y) \rightarrow K(X)$ a homomorphism of $k$-algebras. We want to construct a dominant rational map $X \rightarrow Y$. We can assume that $Y$ is affine with coordinate ring $A(Y)$. Let $y_1,\cdots,y_n$ be $k$-algebra generators of $A(Y)$. Then $\theta(y_1),\cdots,\theta(y_n) \in K(X)$ and we can find an open set $U$ of $X$ such that $\theta(y_i)$ is regular on $U$. Then by restricting $\theta$ on $A(Y)$ we obtain a $k$-algebra homomorphism $A(Y) \rightarrow \mathcal{O}(U)$.

So far, i have clear understanding.

But then Hartshorne says that the homomorphism $A(Y) \rightarrow \mathcal{O}(U)$ is injective, and this is where i am stuck. Here is my effort: Let $p(y)$ be inside the kernel. Then $\theta|_{A(Y)}(p(y))=0 \Rightarrow p(\theta(y))=0 \Rightarrow p(\theta(y_1)(P),\cdots,\theta(y_n)(P))=0, \forall P \in X \Rightarrow (\theta(y_1)(P),\cdots,\theta(y_n)(P)) \in \mathcal{Z}(p)$.

The goal should be to show that $p$ actually vanishes on $Y$, which would be true if every point of $Y$ can be written in the form $(\theta(y_1)(P),\cdots,\theta(y_n)(P))$. But why would this be true?

Edit: Additionally, how do we see that the induced morphism of varieties $U \rightarrow Y$ gives a dominant rational map $X \rightarrow Y$?

Manos
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2 Answers2

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While it is technically correct that once we have a morphism $K(Y) \to K(X)$, it is necessarily injective, and hence the map on subrings $A(Y) \to \mathcal O(U)$ is also injective, phrasing things this way is to some extent getting the underlying logic backwards.

The more basic question is: when we have a dominant rational map $X \to Y$, why does it induce a map $K(Y) \to K(X)$? Again, let's assume $Y$ is affine, and replace $X$ by a sufficiently small affine open subset so that the rational map is actually regular on $U$. Then, we are asking: given a dominant morphism $U \to Y$, why does it induce a morphism $K(Y) \to K(U)$ ($= K(X)$)?

The reason is because the assumption of dominance implies that $A(Y) \to A(U)$ ($= \mathcal O(U)$) is injective!

(And an injection of integral domains always induces a map, necessarily injective, of the associated fraction fields.)

Why is this map injective? Because any element in the kernel would vanish identically on the image of $U$ (since the map if just pull-back of functions), and hence on the closure of the image of $U$ (by def'n of the Zariksi topology). But since our morphism is dominant, the image of $U$ is dense in $Y$. Thus any element in the kernel vanishes on all of $Y$, i.e. vanishes identically, i.e. the kernel is trivial, i.e. the map $A(Y) \to A(U)$ is injective.

[Indeed, this reasoning shows that $A(Y) \to A(U)$ being injective is equivalent to $U \to Y$ having dense image, i.e. being dominant, because if the image is not dense, then there will be some function on $Y$ that vanishes on the image but not on all of $Y$, again by def'n of the Zariksi topology. See my answer here for an elaboration on this point.]

Community
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Matt E
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  • Very interesting elaboration. You mention in your other answer that "...restricting to a dense subset is injective for continuous functions." Do you mean that literally? i.e. if $f: X \rightarrow Y$ is a continuous map of topological spaces, and $U$ is dense in $X$, then $f|_U$ is a $1-1$ mapping? (i do understand though, that as you explain, $A(Y) \rightarrow A(U)$ will be injective.) – Manos Aug 09 '13 at 18:24
  • @Manos: Dear Manos, I think you misunderstood; what I mean is that the map $f \mapsto f_{|U}$ is injective on the space of functions (of which $f$ is an element), at least if the target is Hausdorff (or in other related contexts; note that in algebraic geometry the spaces are seldom Hausdorff, but the diagonal copy of $X$ in $X \times X$, for any variety $X$, is closed, so the same argumetn applies). I am not saying anything about whether the function $f$ is injective when restricted to $U$, which obviously depends a lot on the nature of $f$! Regards, – Matt E Aug 09 '13 at 20:28
  • P.S. Just to be precise, I mean that if $f|U = g|U$, then $f = g$ (when $U$ is dense in $X$). – Matt E Aug 09 '13 at 20:29
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I believe he is using the fact that the homomorphism $A(Y)\to K(X)$ factors through $K(Y)$, which by assumption is the fraction field of the domain $A(Y)$, as $Y$ is taken to be affine. Since $K(Y)\to K(X)$ is necessarily injective, the composition is injective. Then, he claims there exists $\mathcal O(U)$, a subring of $K(X)$, such that the map $A(Y)\to K(X)$ factors through $\mathcal O(U)$.

Andrew
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    Actually $A(Y)$ can be viewed as a subring of $K(Y)$, so i don't have a problem there. I think that what i was missing is that $K(Y) \rightarrow K(X)$ is injective because $K(Y)$ is a $k$-algebra that is also a field. Is that right? – Manos Aug 09 '13 at 16:57
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    Dear @Manos, yes, we use the basic fact that a ring homomorphism whose domain is a field is either trivial or injective. – Andrew Aug 09 '13 at 17:02
  • Yes, i was aware of this fact, i just did not make the connection, since $K(Y)$ is a field. Something else: how do we see that the induced morphism of varieties $U \rightarrow Y$ is dominant? Indeed we have that $A(Y) \rightarrow \mathcal{O}(U)$ is injective but i am not sure about what that implies. – Manos Aug 09 '13 at 17:06
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    Dear @Manos, well this is another question really, but I see it's not explained in Hartshorne. The idea is $f:X\to Y$ induces $f^:A(Y)\to A(X)$, and $f^(u)=0$ iff $u(f(x))=0$ for all $x\in X$, i.e., $u$ vanishes on (the closure of) $f(X)$. Assuming $f^*$ is injctive, we get the result (this is explained more clearly in Shafarevich. p.31). – Andrew Aug 09 '13 at 17:18