1
$\begingroup$

Hartshorne defines a rational map $\phi: X \rightarrow Y$ between two varieties to be an equivalence class of pairs $\langle U, \phi_{U} \rangle$ of open subsets and morphisms $\phi_{U}: U \rightarrow Y$ on them, with equivalence relation given by $$\langle U, \phi_{U} \rangle \sim \langle V, \phi_{V} \rangle \iff \phi_{U} \mid_{U \cap V} = \phi_{V} \mid_{U \cap V}.$$

Lemma I.4.1 right before this definition gives us that two morphisms $\phi, \psi: X \rightarrow Y$ which agree on open $U \subset X$ will actually be equal.

So, this makes me think that the equivalence relation in the definition of rational map above is really just equality, and so a rational map would be exactly one map on an open subset of $X$? But I realize this could be wrong since the lemma describes its two morphisms to have domain all of $X$ instead of just quasi-(affine, projective) varieties.

Also for regular functions: if two regular functions $f: U \rightarrow k$, $g: V \rightarrow k$ agree on the overlap $U \cap V$, then by denseness of $U \cap V$ (since it is open in an irreducible space $X$) I would assume $f = g$ on all of $X$. Is this correct thinking? There are many exercises that deal with these types of situations, and given my possible misconception above they all seem very trivial, so I must be missing something here.

I would like some help to clear up my confusions.

$\endgroup$
  • 1
    $\begingroup$ You cannot add two functions defined in different open sets. What you do is restrict both to the intersection of their domains and sum that. For this to make any sense, you have to at least identify every function defined on an open set to anybody it's restrictions to open subsets contained there. That is what the definition does. $\endgroup$ – Mariano Suárez-Álvarez Jul 8 '17 at 20:55
  • 1
    $\begingroup$ About your last paragraph: if f and g are defined on U and V, repsectively, it makes no sense to say that f and g agree on the whole of X, as they are simply not defined there. At most, what can happen is that they agree on the intersection of their domains. $\endgroup$ – Mariano Suárez-Álvarez Jul 8 '17 at 20:56
  • $\begingroup$ I think I see now. So because $f$ and $g$ do not have to agree on the overlap, we need the equivalence relation above to identify pairs that do agree ? But I still don't see why we can just say "a rational map is a morphism from an open subset of $X$ to $Y$". $\endgroup$ – Freddie Jul 8 '17 at 20:57
  • $\begingroup$ @MarianoSuárez-Álvarez Second Comment: Okay I see. But in the affine case, can I say "...all of $X$", since the regular functions are really the polynomials in the coordinate ring? In this respect, it seems that if $f = g$ on $U \cap V$, then $f - g$ vanishes on $U \cap V$, and so $f - g$ vanishes on the closure of $U\cap V$, which would be $X$. $\endgroup$ – Freddie Jul 8 '17 at 21:01
  • 1
    $\begingroup$ A function which isn't defined at a point cannot vanish there: it simply does not make sense, as I said. $\endgroup$ – Mariano Suárez-Álvarez Jul 8 '17 at 21:05
3
$\begingroup$

I think your confusion is that when we write "a rational map $\phi: X \to Y$", then $\phi$ need not be defined on all of $X$, but only on an open subset $U \subset X$.

For example, on the variety $xz-yw = 0$, the formula $x/y$ defines a rational function at the points where $y \ne 0$, and also the formula $w/z$ defines a rational function at the points where $z \ne 0$. But when both $y\ne 0$ and $z \ne 0$, we have $x/y = w/z$ on the variety. So all in all, we get a rational function which is defined at any point where either $y\ne 0$ or $z \ne 0$, but there is no single formula that defines it at all such points.

$\endgroup$
  • $\begingroup$ Alright. So the definition of rational map above allows us to sort of piece together these functions on open sets that agree on the overlaps. And a rational function need not be defined on all of $X$, and so Lemma 4.1 doesn't necessarily apply. As I said in the comments above, I will have to review some definitions. Thank you for your help. $\endgroup$ – Freddie Jul 8 '17 at 21:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.