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In The book Basic Algebraic Geometry 1 by Shaferevich page 37 (1st paragraph) it says that if $\phi$ is a rational function (elements of quotient field of the ring of regular functions on $X$, $k[X]$) on an affine variety $X$ (irreducible closed subset of some $\mathbb{A}^n$) and $\phi =0$ on some non-empty open subset $U$ of $X$ the $\phi=0$ on $X$. My questions are:

(i) When it says that $\phi =0$ on $X$ does it mean that $\phi$ is zero whenever it is well-defined (because a rational function may not be well-defined on each point of X) or $\phi$ will be defined at every point of $X$ and $\phi=0$ on $X$.

(ii) To prove the statement it says that if $\phi\neq 0$ on $X$ any expressiom $\phi=\frac{f}{g}$ with $f,g\in k[X]$ gives a representation of $X$ as a union $X=X_1\cup X_2$ where $X_1=X-U$ and $X_2=Z(f)$ which will be a contardiction to the irreducibility of $X$. I could not understand the decomposition. The decomposition basically says that $Z(f)\supset U$ but I don't see how. Because it may happent that at some point of $U$, $g$ is zero.

Here $Z(f)=\{x\in X: f(x)=0\}$

Thank you very much.

1 Answers 1

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Well, the statement is that $\phi\equiv 0$ on $U$, not "on that part of $U$ where it is well-defined". As a result, $g$ cannot be zero at any point of $U$, and thus $\phi\equiv 0$ on $U$ implies that $f\equiv 0$ on $U$. It thus follows that $U\subseteq Z(f)$. This, in turn, implies that $\phi \equiv 0$ on the entirety of an irreducible $X$.

I'm not sure about this, but wouldn't be surprised if not assuming that $\phi$ is well-defined on the entire $U$ could work as well by removing the singularities (which form a subvariety of one less dimension). The conclusion would then still be that $\phi\equiv 0$ on $X$ and hence on $U$. This is just a gut-feeling without proper proof, though.

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    My concern is that when we have one particular form $\frac{f}{g}$ of $\phi$ such that $g=0$ at some point of $U$ but $\phi$ has another representation $\frac{f_1}{g_1}$ such that at that point $g_1\neq 0$ and $f_1=0$ so that $\phi=0$ at that point. – 2017-01-31
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    @Shafer: check your cross multiplication- if $g=f_1=0, g_1\neq 0$ at a point then $f$ is necessarily also $0$ at that point. In fact, the order of vanishing of $f$ at that point is greater than the order of vanishing of $g$, as $f=\frac{f_1g}{g_1}$ and $g_1\neq 0$. So nothing bad is happening. – 2017-02-01
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    @KReiser Thank you for your answer. So this means for any irreducible closed $X$ of affine space there there is no rational function which is $0$ on some nonempty open subset of $X$ but is not defined on some point of $X$. – 2017-02-08