# The regular functions on an open subset $U$ of an affine variety $X$ are the polynomials on $A(X)$.

This is an exercise of Gathmann's Algebraic Geometry.

Show that the regular functions on an open subset $U$ of an affine variety $X$ are the polynomials on $A(X)$, if $A(X)$ is a UFD and $U$ is the complement of an irreducible subvariety of codimension at least 2 in $X$.

There was an example showing that regular functions on $U=A^2\backslash\{0\}$ are polynomials. I am trying to use the same idea in that example.

Let $V\subsetneq V_0 \subsetneq V_1 \subset X$ be a chain of irreducible varieties.

Let $\phi$ be a regular function on $U$.

My idea is: if we can find two irreducible functions $f_1, f_2$ such that $V\subsetneq V(f_1), V\subsetneq V(f_2)$, then

$\phi=\frac{g_1}{f_1^n}$ on $D(f_1)$ and $\phi=\frac{g_2}{f_2^m}$ for some $g_1,g_2\in A(X)$ and $m,n\in N$.

($D(f)$ here means the distinguished open subset of $f$ in $X$, i.e., the set where $f$ is not equal to $0$ in $X$.)

Suppose $f_1$ does not divide $g_1$, $f_2$ does not divide $g_2$.

Then on $D(f_1)\cap D(f_2)$, we have $g_1f_2^m=g_2f_1^n$. Since $V(g_1f_2^m-g_2f_1^n)$ is closed, we also have $g_1f_2^m=g_2f_1^n$ on the closure of ${D(f_1)\cap D(f_2)}$ which is $X$. Then because of the irreducibility of $f_1,f_2$ and $A(X)$ is a UFD, we can prove that $m=n=0$. That proves that $\phi$ is a polynomial.

Sorry for the long post.

I'm not sure whether this is on the right track. Also I don't know how to show that I can find the irreducible $f_1, f_2$. Or they are probably not really irreducible, but product of linear factors. But how to state that in the proof? Any help would be appreciated!

• Where is codimension 2 argument being applied here? Commented Nov 27, 2016 at 1:38
• @user45765: It has codimension at least $2$ so the chain $V\subsetneq V_0 \subsetneq V_1 \subset X$ exists. Commented Nov 27, 2016 at 10:32
• I think there's a small issue here. Why is the closure of $D(f_1) \cap D(f_2)$ equal to $X$. This is true if $X$ is irreducible, because that set is open. But it is not the case here Commented Nov 15, 2020 at 9:12
• @leducquang $A(X)$ is assumed to be a UFD, in particular it is an ID hence $\mathcal{I}(X)$ is a prime ideal and hence $X$ is irreducible by asusmption. Commented May 16, 2022 at 22:05

$$(0)\subset I_1 \subsetneq I_0\subsetneq I$$
Then there exists non-zero irreducible functions $f_1, f_2 \in I$.
Let $V_1=V(f_1)$ and $V_2=V(f_2)$. Then $V\subsetneq V_1$ and $V\subsetneq V_2$. The rest of my above argument then follows.