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On page 158 of Andreas Gathmann's notes on Algebraic Geometry, http://www.mathematik.uni-kl.de/~gathmann/class/alggeom-2002/main.pdf, it states that given some projective plane curve $X$ of degree $d$ with coordinates $x_0,x_1,x_2$ not containing the point $(0:0:1)$, it states that $$\mathcal{O}_X(U_0 \cap U_1)=\left\{\dfrac{x_2^i}{x_0^jx_1^k}:0\le i \le d-1 \textrm{ and }i=j+k\right\},$$ $$\mathcal{O}_X(U_0)=\left\{\dfrac{x_2^i}{x_0^jx_1^k}:0\le i \le d-1,k\le0,\textrm{ and }i=j+k\right\},$$ and $$\mathcal{O}_X(U_1)=\left\{\dfrac{x_2^i}{x_0^jx_1^k}:0\le i \le d-1 , j\le 0,\textrm{ and }i=j+k\right\},$$ where $U_0=\{x_0\ne0\}$ and $U_1=\{x_1\ne0\}$.

How do we know that the regular rings can be written in this way? In other words, how do we know that any regular function in the above regular rings can be written as a sum of monomials with $x_2$ in the numerator? For instance, if our curve was $x_1^3x_0=x_2^4-x_2^2x_0^2$, I think the rational function $r=\dfrac{x_1}{x_2-x_0}$ lies in $\mathcal{O}_X(U_0 \cap U_1)$, and although it isn't too hard to see that we can simplify $r$ using the equation of the curve $\left(r=\dfrac{x_2^3}{x_1^2x_0}+\dfrac{x_2^2}{x_1^2}\right)$, it's not obvious that we can always do this (what if the denominator of $r$ did not share a common factor with the curve?).

On a separate note, is it true that $\mathcal{O}_X(U_0)+\mathcal{O}_X(U_1)$ is generated by monomials?

Edit: Perhaps a better example would be the curve $x_1^3x_0=x_2^2(x_2+x_0)^2$. We know that the rational function $\dfrac{x_2(x_2+x_0)}{x_1x_0}$ lies in $\mathcal{O}(U_0)$, but I can't think of a way to write it as a sum of monomials in $\mathcal{O}(U_0)$ (unless I'm mistaken, I believe neither $\dfrac{x_2^2}{x_1}$ nor $\dfrac{x_2}{x_1x_0}$ are in $\mathcal{O}(U_0)$, since they are not regular at the point $(1:0:-1)$).

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This is essentially projective Nullstellensatz.

We start by calculating regular functions on $U_0 \cap U_1$. Recall regular functions on $U_0 \cap U_1$ are quotients of homogeneous polynomials $f/g$ of the same degree with $g$ not vanishing on $U_0 \cap U_1$. That means the vanishing of $g$ is contained in $X - (U_0 \cap U_1) \subset Z(x_0x_1)$. So the irreducible factors of $g$ must be (associate to) $x_0$ or $x_1$ by projective Nullstellensatz. So a typical regular function on $U_0 \cap U_1$ looks like $\frac{f}{x_0^jx_1^k}$. After breaking up $f$ into monomials, swapping out terms higher than $x_2^d$ for lower powers of $x_2$ (since $f$ has a $x_2^d$ term) and absorbing any $x_0$ and $x_1$ terms into the bottom, we see that $\mathcal O_X(U_0 \cap U_1)$ is generated over $k$ by the terms Gathmann claims.

The same ideas will work for calculating regular functions on $U_0$. This time the vanishing of $g$ lies in $X - U_0 \subset Z(x_0)$. So a typical regular function on $U_0$ looks like $\frac{f}{x_0^j}$. Again break $f$ into monomials and swap out terms higher than $x_2^d$ for lower terms in $x_2$. Again we can absorb any $x_0$ terms into the bottom. Then the terms Gathmann gives generate $\mathcal O_X(U_0)$. The reason he puts $x_1^k$ on the bottom is to make it easy to calculate $H_1$.

I am fibbing a little bit. Really we should be working with a representative of $g$ in $S(X)$ and that representative will lie in $Z(x_0x_1) \cap Z(f)$ (in the first case) so the representative of $g$ will look like $x_0$ or $x_1$ plus some extra polynomial in $I(X) = (f)$. But $I(X)$ is already modded out so this isn't a problem.

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  • $\begingroup$ This might be a dumb question, but why is the vanishing of $g$ contained in $\{(0:1:0),(1:0:0)\}$? For example, if $(1:1:0)$ is on the curve, couldn't $g$ also contain $x_0-x_1$? $\endgroup$
    – whetham
    Commented Aug 1, 2016 at 18:03
  • $\begingroup$ @whetham The bottom cannot vanish on $U_{01}$ by definition of regular function on $U_{01}$. So the vanishing is contained in the complement. $\endgroup$
    – hwong557
    Commented Aug 1, 2016 at 18:10
  • $\begingroup$ See the definition on page 41 of Gathmann's notes. $\endgroup$
    – hwong557
    Commented Aug 1, 2016 at 18:13
  • $\begingroup$ Thank you! However, how would this apply to $\mathcal{O}_X(U_0)$ and $\mathcal{O}_X(U_1)$? $\endgroup$
    – whetham
    Commented Aug 1, 2016 at 18:47
  • $\begingroup$ @whetham Sorry I realized I've made a big mistake! My very first claim is flat out wrong. I'll make some edits when I go home tonight (if you haven't figured it all out already) but it should still come down to the Nullstellensatz. Sorry! $\endgroup$
    – hwong557
    Commented Aug 1, 2016 at 19:12

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