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In particular, I am interested in seeing an example of a rational function defined on a integral variety $X$ that is regular on the complement of a codimension $2$ closed set, but so that there is no extension to a regular function defined on all of $X$.

Elle Najt
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    Consider $X=\mathrm{Spec}\mathbb{C}[x^4,x^3y, xy^3,y^4]$ and consider the rational function $x^2y^2$. – Mohan Nov 28 '15 at 00:06
  • @Mohan Wow, what a great example! Let me make sure I understand: obviously (because of degree) $x^y^2$ is not regular, but it is rational ($(x^3 y)^2 / x^4 = (x y^3)^2 / y^4 = x^2 y^2$). Moreover, it is regular in $D(x^4)$ and $D(y^4)$, which means that the locus of non-regularity is contained in $V(x^4, y^4)$. But \mathbb{C} [ x^4, x^3 y, x y^3, y^4] / (x^4, y^4) \cong \mathbbC [s,t] / (st)$, hence is codimension 3.

    But something that bothers me about this example is that it is not "geometrical" - not yet to my eyes anyway.

     – Elle Najt Nov 28 '15 at 01:44
  • Recall that a ring is normal if and only if it satisfies Serre conditions $R_1,S_2$. If it is $S_2$, then in general, the rational function would be not regular on codimension one subset. So, it is is better to have something which is $R_1$ but not $S_2$ and the above example is a standard one for that. So, it is as geometrical as it can get. – Mohan Nov 28 '15 at 05:05

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