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On The Arithmetic of Elliptic Curves, Joseph H. Silverman, the definition of rational map is given by:

Let $V_1$ and $V_2 \subseteq \mathbb{P}^n$ be projective varieties. A rational map from $V_1$ to $V_2$ is a map of the form $$\varphi : V_1 \rightarrow V_2, \qquad \varphi = [f_0,\ldots,f_n]$$

where the functions $f_0,...,f_n ∈ K(V_1)$ have the property that for every point $P ∈ V_1$ at which $f_0,...,f_n$ are all defined, $$\varphi (P) = f_0(P),...,f_n(P).$$

And Hartshorne, the definition is:

Let $X$ and $Y$ be varieties. A rational map $\phi: X \to Y$ is an equivalence of pairs $(U, \phi_U)$ where $U$ is a nonempty open subset of $X$, and $\phi_U$ is a morphism of $U$ to $Y$, and where $(U, \phi_U)$ and $(V, \phi_V)$ are considered equivalent if $\phi_U$ and $\phi_V$ agree on $U \cap V$.

I wonder if these two definitions are equivalent? So far I can see the first definition satisfies Hartshorne's definition, but how to see if Hartshorne's also agrees with the first definition?

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6666
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1 Answers1

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Let's get a handle on the assumptions baked in to each text's use of the word "varieties". Silverman assumes projective and integral, while Hartshorne assumes just integral. So if you fail to enforce "projective" on Hartshorne's definition, then the definitions are not equivalent.

If one assumes that Hartshorne's varieties are in fact projective, then the two definitions coincide. Starting from Hartshorne's definition, we may find a maximal domain of definition $U$ for our morphism: take the union of the open sets in all the pairs. Now I claim that there is a unique (honest) map $f:U\to V_2$ which represents our rational map $\phi:X\to Y$. Consider two pairs $(V,g),(V',g')$ which are representatives of $\phi$. Then $g,g'$ agree on a dense open subset $W\subset V\cap V'$, and as $V_1$ is reduced and $V_2$ separated, we get that they actually agree on all of $V\cap V'$. This means we can glue together $g,g'$ along $W$ to get an honest morphism $f:U\to V_2$. Now we can pick an affine open subsets $W_2\subset V_2$ and $W_1\subset f^{-1}(W_2) \subset V_1$ which gives us a map of coordinate algebras $k[W_2]\to k[W_1]$ which we can then extend to a map on fraction fields $k(W_2)\to k(W_1)$ and recover the $f_i$ from Silverman's definition.

Warning: in general, you can't pick a single formula with which to write down the $f_i$ on the whole domain of definition. Consider the example from Ted's answer here: on the variety $V(xz-yw)\subset\Bbb P^3$, there's a rational map given by $\frac xy$ when $y\neq 0$ and $\frac wz$ when $z\neq 0$. So this rational map is defined away from $V(y,z)$, and it's represented by $f_i=\frac xy = \frac wz$ because these terms are equal in $k(X)$, but they're not the same formula.

KReiser
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  • Also if $Y$ is covered by finitely many affines I think there is a way to check if the rational map has a representative which is regular at $p$: assuming it is, once we know the affine charts around $p,\phi(p)$ we have a rational map of affine varieties ie. finitely many rational functions and it reduces to search for representatives that are regular at $p$, doing so repeatedly may give a maximal open where the rational map is a morphism – reuns Dec 28 '19 at 02:35
  • If we consider the affine case, then can we always pick a single formula? – 6666 Dec 28 '19 at 05:21
  • No, take the same example from the post but in affine space. – KReiser Dec 28 '19 at 05:51
  • If we consider the affine case, then you don't need to extend from the affine to the projective case in your proof, does that imply the maps $f_i\in K[V_1]$ instead of $k(V_1)$? – 6666 Dec 29 '19 at 16:02
  • I don't understand your comment. I'm saying consider the variety $V(xz-yw)\subset \Bbb A^4_k$ and the rational function represented by $x/y$ when $y\neq 0$ and $w/z$ when $z\neq 0$. Could you clarify your last comment? – KReiser Dec 29 '19 at 19:56
  • I am sorry about the confusion, I was referring "Now we can pick an affine open subsets $W_2\subset V_2$ and $W_1\subset f^{-1}(W_2) \subset V_1$ which gives us a map of coordinate algebras $k[W_2]\to k[W_1]$ which we can then extend to a map" I was thinking, if we switch to affine instead of projective varieties for $V_i$, it seems we don't need to extend, then $f_i\in K[V_1]$ and we don't need to extend to $K(V_i)$? – 6666 Dec 30 '19 at 01:28
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    No, you still need to do this - there's no guarantee that $f^{-1}(W_2)$ is affine, nor does it have to be all of $V_1$. In particular, $W_1$ is chosen to lie inside $U$ (maybe this choices was obscured a little by the notation, but it's in there: $W_1$ is an affine open subset of $f^{-1}(W_2)\subset U$), which is the domain of definition of our rational map, which is in general not equal to $V_1$. – KReiser Dec 30 '19 at 01:55