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Let $f$ be a rational function on a variety $V$. Let $U = \{P\in V; f \textrm{ is defined at }P\}$. Then $f$ defines a function from $U$ to $k$. Show that this function determines $f$ uniquely. So a rational function may be considered as a type of function, but only on the complement of an algebraic subset of V, not on V itself.

I thought as follows: if there is an other $g\in k(V)$ such that $g(P)=f(P)$ for all $P\in U\cap \widehat{U}$ where $\widehat{U}=\{Q\in V;g \textrm{ is defined at }P\}$ then need conclude that $f=g$ in $k(V)$. I am sure?

Bruno Rodrigues dos Santos
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    Please write down the problem in question to make the question as self-contained as possible. – Nils Matthes Feb 10 '15 at 14:31
  • It would also be great if you added what your thoughts are on th eproblem, what you tried and so on. Copying problems out of books is not generally well-received here! – Mariano Suárez-Álvarez Feb 10 '15 at 22:21
  • Sim. Estou certo. E, de fato, isso é verdade. Com efeito, $U\cap\widehat{U}$ é denso em $V$ e o complementar de $V(ad-bc)$ em $V$ é aberto em $V$, onde $g=\frac{c}{d}$. Portando $f=g$, ou seja, a função determina $f$ unicamente. – Bruno Rodrigues dos Santos Feb 11 '15 at 17:33

1 Answers1

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Maybe this will do,

Let $f=\overline{a}/\overline{b}, a, b\in k[X_1, \ldots, X_n]$ (by $\overline{a}$ we denote the residue of a polynomial in $k[X_1, \ldots, X_n]$ in $\Gamma(V)$) and $g=\overline{c}/\overline{d}\in k(V), c, d\in k[X_1, \ldots, X_n]$ such that $g(P)= f(P)$ for all $P\in U$. We assume that $U= \{P\in V\mid g \text{ is defined at } P\}$ (I do not know if that is necessary to assume). In order to prove that $f\equiv g$ it suffices to show that $V(ad-bc)\supseteq V$.

We have that $a=bf, c=gd$ as rational functions and hence $a(P)=f(P)b(P)$ and $P\in U$ and $c(P)= g(P)d(P)$ for all $P\in U$. So, if $P\in U, a(P)d(P)- c(P)b(P)= f(P)b(P)d(P)- g(P)d(P)b(P)= b(P)d(P)(f(P)-g(P))= 0$, thus $V(ad-bc)\supseteq U$. Now if $P\in V\setminus U$, $f$ and $g$ are not defined on $P$ so $b(P)= d(P)= 0$ and thus, $a(P)d(P)- b(P)c(P)=0$, i.e. $V(ad-bc)\supseteq V$.

Jim Rohn
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