Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

consider the plane cones $s=\langle(1,0),(1,n)\rangle$ and $t=\langle(1,0),(1,-n)\rangle$. This produce a toric variety obtained glueing $k[x,xy^n]$ and $k[x,xy^{-n}]$ along $k[x]$. There is a more "compact" way to see this variety as $k[u_1,\dots,u_n]/I$ for some ideal $I$.

Thanks

share|improve this question
    
I'm not sure what the question is here. Could you clarify? –  Zev Chonoles Oct 13 '11 at 20:06

1 Answer 1

According to the conventions I know of, you are doing this wrong.

You have two cones, $s$ and $t$. You have to compute their dual cones $s^\vee$ and $t^\vee$, and $(s\cap t)^\vee$ and glue the spectra of $\mathbb C[s^\vee\cap\mathbb Z^2]$ and $\mathbb C[s^\vee\cap\mathbb Z^2]$ along their open set $\mathbb C[(s\cap t)^\vee\cap\mathbb Z^2]$.

Let me take $n=3$.

  • Then $s=\langle(1,3),(1,0)\rangle$ so $s^\vee=\langle(3,-1),(0,1\rangle$ and you can easily check that $s^\vee\cap Z^2$ is generated as a semigroup by $a=(0,1)$, $b=(1,0)$ and $c=(3,-1)$ subject to the relation $b^2=ac$. It follows that $A=\mathbb C[s^\vee\cap\mathbb Z^2]$ is the algebra generated by $a$, $b$ and $c$ subject to the relation $b^2-ac=0$.

  • Likewise, $B=\mathbb C[t^\vee\cap\mathbb Z^2]$ is the algebra generated by $e=(0,-1)$, $f=(1,0)$ and $g=(3,1)$ subject to the relation $f^2-egc=0$.

  • Finally, $(s\cap t)^\vee$ is the cone $\langle(0,1), (1,0), (0,-1)\rangle$ so that $(s\cap t)^\vee\cap\mathbb Z^2$ is the semigroup generated by $x=(1,0)$, $y=(1,0)$ and $y^{-1}=(0,-1)$, and $C=\mathbb C[(s\cap t)^\vee\cap\mathbb Z^2]$ is $\mathbb C[x, y, y^{-1}]$.

  • The inclusion of $s^\vee$ into $(s\cap t)^\vee$ induces the map $i:A\to C$ such that $a\mapsto y$, $b\mapsto x$ and $c\mapsto x^3y^{-1}$, and we can check that we can identify $C$ with the localization of $A$ at $a$. Similarly, the inclusion of $t^\vee$ into $(s\cap t)^\vee$ induces the map $j:B\to C$ with $e\mapsto y^{-1}$, $f\mapsto x$ and $g\mapsto x^3y$, and $C$ can be identified with the localization of $B$ at $e$.

But your toric variety is not affine —its fan is not a cone and its faces— so you will not be able to write is as the spectrum of an algebra of the form $k[u_1,\dots,u_n]/I$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.