# figure out a simple toric variety

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

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

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 $k[u_1,\dots,u_n]/I$.

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