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This is Exercise 3.12(b) of Gathmann's 2021 notes of Algebraic Geometry.

Let $Y$ be a non-empty irreducible subvariety of an affine variety $X$ and set $U=X\setminus Y$, Exercise 3.12(b) asks for an example that if $A(X)$ is not assumed to be a UFD, then there exists $Y$ such that $\text{codim}_XY\geq 2$ and $\mathscr{O}_X(U)\neq A(X)$, where $\mathscr{O}_X(U)$ is the ring of regular functions on $U$ and $A(X)$ is the coordinate ring of $X$.

I tried $X=V(x_1x_4-x_2x_3)\subset \mathbb{A}^4$, for which $A(X)$ is known as a non-UFD in a preceding exercise. The only (up to a permutation) irreducible subvariety of codimension $2$ that I find is $Y=V_X(x_1,x_2,x_3)$, since $V(x_1,x_2)$ is irreducible of codimension $1$. Now that

$$U = D(x_1)\cup D(x_2)\cup D(x_3),$$

hence elements in $\mathscr{O}_X(U)$ are the gluing of regular functions on $D(x_i)$'s, $i=1,2,3$. To find an element in $\mathscr{O}_X(U)$ that is not in $A(X)$, I considered $\frac{x_3}{x_1}\in \mathscr{O}_X(D(x_1))$ and $\frac{x_4}{x_2} \in \mathscr{O}_X(D(x_2))$. They do agree on $D(x_1)\cap D(x_2)$, however, I failed finding an extension of them to $D(x_3)$.

Similar difficulty appears when I tried considering $X=V(x_1x_5- x_2x_4, x_2x_6 -x_3x_5, x_1x_6-x_3x_4) $ and $X= V(x_1x_2x_3-x_4x_5x_6)$.

Is there any simple example for this question? Thanks in advance for any help.

EDIT: Also, the base field $K$ is always assumed to be algebraically closed, so the example better satisfies this condition...

Shana
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  • Does https://math.stackexchange.com/a/144054 help? – Ben Oct 05 '22 at 19:00
  • @Ben Yes! It seems that $X=V(x_1,x_2)\cup V(x_3,x_4)\subset \mathbb{A}^4$ and $Y= V(x_1,x_2,x_3,x_4)$ would work. Thanks! – Shana Oct 06 '22 at 08:37

1 Answers1

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Seemingly the simplest example: Let $X = V(x_1,x_2)\cup V(x_3)\subset \mathbb{A}^3$ and $Y =V(x_1,x_2)\cap V(x_3) =V(x_1,x_2,x_3)$, then $Y$ is a point, hence have codimension $2$ in the plane $V(x_3)$. $U=X\setminus Y$ is the disjoint union of open sets $X\setminus V(x_1,x_2)$ and $X\setminus V(x_3)$, hence the regular functions on $U$ are pairs of regular functions on $ X\setminus V(x_1,x_2)$ and $X\setminus V(x_3)$. Since $\mathscr{O}_X(X\setminus V(x_3))=A(X)_{x_3}$, it follows that $\mathscr{O}_X(U)\neq A(X)$.

Shana
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