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A rational map is a class of equivalence on $$F=\{f:U \rightarrow Y \mid \mbox{morphisms with } U\subset X \mbox{ open subvariety}\}$$ given by $$f\sim g \iff f_{| (\operatorname{Dom}f \cap \operatorname{Dom}g)}= g_{|(\operatorname{Dom}f \cap \operatorname{Dom}g)}$$ In this way, William Fulton says in his book Algebraic Curves that "a rational map can be defined as a function $f$ from an open subvariety $U$ of $X$ to $Y$ such that $f$ cannot be extended to a morphism from any larger open subset of $X$ to $Y$." I really think that a I get this, but I can't find an example of a rational map $f$ defined on a proper subvariety $U\subsetneq X$.

Esteban G.
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2 Answers2

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The example of Eric Wofsey is kind of "artificial" because one could extend this map using the projective space as target variety. More generally, if $Y$ is projective and $X$ is a smooth curve, then any rational map $f : X \to Y$ will extends.

On the other hand, if $P = [0:0:1]$ and $X = \Bbb P^2 \backslash \{P\}$ the map $$ \pi : X \to \Bbb P^4, [x:y:z] \mapsto [x^2:y^2:xy:xz:yz] $$ can't be extended. So this is a "true" rational map.

Edit : let met add a counter-example of a morphism $X \to Y$ which can't be extended where $C$ is projective but not smooth. We take $X = \{(x,y,z) \in \Bbb P^2 : zx^2 = y^3\}$ and $Y = \Bbb P^1$. We have a bijective map $f : \Bbb P^1 \to C, (t,s) \mapsto (t^3, t^2s, s^3)$. On the other hand, we have a well-defined inverse defined on the smooth locus of $C$ (i.e $C \backslash \{(0:0:1)\}$) defined as $g(x,y,z) = (x,y)$. If one could extend $g$ into a regular map, then $f,g$ would be inverse and we would have an isomorphism $C \cong \Bbb P^1$ but $C$ is not smooth.

  • Hi! I got some questions with your answer but maybe, instead of give me the answer you could show me a book to read it or something like that. 1) Why is every rational map from a curve into a proyective space extendable? And 2) Why is your map $\pi$ not extendable into P? Those are by sure noob questions, so thank you in advance! – Esteban G. Jun 07 '17 at 15:46
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    Not noob at all.

    1. This is because in local coordinates, your map will be locally something like $f/g$. But writing $f = a_nz^n + ...$ and $g = b_m z^m + \dots $ you see that the limit is $a_n/b_m z^{n-m}$ which exists all the time : it is $[1:0] = "\infty"$ if $n < m$, $a_n/b_m$ if $n=m$ and $0$ else.

    2. for any $[x:y] \in \Bbb P^1$, we have that $\lim_{t \to 0} \pi(tx,ty, 1) = [0 : 0 : 0 : x : y ]$. This shows that this map can't be extended.

    3. Reference : Rick Miranda, Algebraic curves and Riemann surfaces is a good place. But the book by Fulton is also excellent !

     –  Jun 07 '17 at 16:00
  • @EstebanGutiérrez : I was a bit sloppy for 1) as the target space might be $Y \subset \Bbb P^n$ instead of $\Bbb P^1$, but the idea is the same : write your projective map $f : X \to \Bbb P^n$ as $[f_0 : f_1 : \dots : f_n]$, and do Taylor around the point, and take the lowest degree term, this will give you a well defined value in $\Bbb P^n$. –  Jun 07 '17 at 16:06
  • Latest question! Let $f$ as defined for Eric Wofsey. Using what you said I could think on $Y$ as a subset of $\mathbb{P}^1$ so now what is not really clear is, what exactly $f_0$ and $f_1$ should I choose? My attempt was using $f_0=f$ and $f_1=0$ so I have the new function $f:\mathbb{A}^1\rightarrow\mathbb{P}^1$ defined as $f(x)=[1/x:0]$, and the thing now is: Should this function actually be $f(x)=[1:0]$ identically? – Esteban G. Jun 07 '17 at 16:39
  • Here one should take $f_0 = 1$ and $f_1 = x$. The map you obtain is therefore $f(x) = [1:x]$. Now, let $f_{new}(x) = [1:x]$ and $f_{old}(x) = 1/x$. We have a "natural" inclusion $i : \Bbb A^1 \to \Bbb P^1, x \mapsto [x:1]$. And moreover we can see that $i \circ f_{old} = f_{new}$ (so really, $f_{new}$ extends $f_{old}$ as desired), as $i \circ f_{old}(x) = i(1/x) = [1/x:1] = [1:x] = f_{new}(x)$. Don't hesitate to ask question if this is not clear. –  Jun 07 '17 at 16:49
  • It is now SO CLEAR hahaha. Thank you so much! I have to dissert in a couple of days some examples of rational and birrational maps in my class, and now that I understand how this works it will be a lot easier. Answer Accepted! – Esteban G. Jun 07 '17 at 17:05
  • Thanks ! Glad I helped :) –  Jun 07 '17 at 17:06
  • "... if $Y$ is projective and $X$ is a curve, then any rational map $f : X \to Y$ will extend" Only true if $X$ is smooth. – Georges Elencwajg Jun 08 '17 at 07:14
  • Dear Georges, you are right, I will edit accordingly my answer. Thanks you ! –  Jun 08 '17 at 08:55
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Take $X=Y=\mathbb{A}^1$ and define $f(x)=1/x$. This is defined on $U=\mathbb{A}^1\setminus\{0\}$ but cannot be extended to $0$.

Eric Wofsey
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  • I thought this kind of maps could work but I couldn't show that this is in fact a rational map by just using the definition. Could you help with that? (At least I can see the continuity) – Esteban G. Jun 06 '17 at 23:40
  • What is your definition of a morphism? – Eric Wofsey Jun 06 '17 at 23:41
  • $f$ is a morphisms if $f$ is continuous and for all open set $U$ de $Y$, if $f \in \Gamma(U,O_Y)$ then $f(g) \in \Gamma(f^{-1}(U),O_X)$. – Esteban G. Jun 07 '17 at 00:02
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    That definition takes a lot of work to verify directly, but there are some general theorems you can use to make things easier. For instance, do you know that for any variety $V$, a function $V\to \mathbb{A}^1$ is a morphism iff it is a regular funciton on $V$? In this case, you then just have to verify that $1/x$ is a regular function on $\mathbb{A}^1\setminus{0}$. – Eric Wofsey Jun 07 '17 at 00:27
  • Are not regular maps just polynomials restricted in this case? – Esteban G. Jun 07 '17 at 00:46
  • No. I don't know what exact definitions you're working with, but a regular function on an open subset of an affine variety is given by a quotient of polynomials (where the denominator does not vanish on the domain), not necessarily a polynomial. So in this case, $1/x$ is a regular function on any open subset of $\mathbb{A}^1$ where $x$ does not vanish. – Eric Wofsey Jun 07 '17 at 01:02
  • Thank you so much for your time, I really apreciate your answer. I definitively have to keep studying. – Esteban G. Jun 07 '17 at 01:08