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I am reading Gathmann's Notes on algebraic geometry and am confused with what I see as a contradiction. For an affine variety $X\subset\mathbb{A}^n$ and $P\in X$ a point, he defines the local ring of $X$ at $P$ to be $\mathcal{O}_{X,P}:=\{f/g\mid f,g\in A(X), g(P)\neq 0\}$, where $A(X)=k[x_1,...,x_n]/I(X)$ is the coordinate ring. Then he defines the ring of regular functions on an open set $U$ to be $\mathcal{O}(U)=\bigcap_{P\in U}\mathcal{O}_{X,P}$. He then makes the point that this does not mean that we can always write a regular function as the quotient of two polynomials in $k[x_1,...,x_n]$. For example, if we let $X\subseteq\mathbb{A}^4$ be the variety defined by the equation $xw=yz$, and let $U\subset X$ be the open subset of all points in $X$ where $y\neq 0$ or $w\neq 0$, then the function $x/y$ is defined for all points where $y\neq 0$ and $z/w$ isdefined for all points where $w\neq 0$. By the equation of $X$, these functions coincide where they are both defined. So we have $x/y=z/w$, giving rise to a regular function on $U$. Yet there is no representation of this function on all of $U$.

Now, the above makes sense to me. But I am confused by the next example given on page 21 of the notes. The example is to compute $\mathcal{O}_{k^2}(U)$ where $k=\mathbb{C}$ and $U=k^2-\{0\}$. He says by the definition given above that any element $\varphi\in\mathcal{O}_{k^2}(U)$ is globally the quotient $\varphi=f/g$ of two polynomials $f,g\in k[x,y]$. But why is this the case? Why doesn't the predicament above apply here? I thought by definition that a regular function is only given locally by the quotient of polynomials. I must be missing something.

ponchan
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    Are you familiar with Hartog's theorem from complex analysis in several variables? Understanding this extension property that's unique to higher dimensional holomorphic functions may answer your question. – Sidhanth Raman Jan 18 '21 at 18:51
  • @SidhanthRaman No I'm not. The author makes it seem that this should follow from the definition, but I don't see how. – ponchan Jan 18 '21 at 18:55
  • Your definition of $O(U)$ is the intersection of locally regular functions which by definition are of the form of the quotient of polynomials. – Youngsu Jan 18 '21 at 20:33
  • Proposition 2.1.10, lying between the two statements you mention, which explains that $\mathcal{O}_X(X_f)=(\mathcal{O}_X(X))_f$ turns this in to the classical problem about checking the global sections of $U$, the affine plane without it's origin, via the cover by $D(x)$ and $D(y)$, which has been well-covered here. I think this is pretty closed to being a duplicate of that question - what do you think? – KReiser Jan 18 '21 at 20:59

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