I'm trying to establish an explicit description of the open subsets of affine toric varieties given by faces of the underlying cone.


For a rational convex polyhedral cone $\sigma\subseteq N_\mathbb R$ over a lattice $N$, we know that the semigroup algebra $S_\sigma=\sigma^\vee\cap M$, where $M$ is the dual lattice of $N$, is finitely generated by some $m_1,\dots,m_k\in M$, thus $$ U_\sigma =\operatorname{Spec}(\mathbb C[S_\sigma]) = \operatorname{Spec}(\mathbb C[\chi^{m_1},\dots,\chi^{m_k}]) $$ and we find an explicit (coordinate) description of this affine variety by looking at the $\mathbb C$-algebra homomorphism given by \begin{align} \varphi : \mathbb C[x_1,\dots,x_k] &\longrightarrow \mathbb C[\chi^{m_1},\dots,\chi^{m_k}], \\ x_i &\longmapsto \chi^{m_i} \end{align} And let $U_\sigma = \mathbf V(\ker \varphi) \subseteq \mathbb C^k$.

Now let $\tau = \sigma \cap H_m$ be the face given by some supporting hyperplane $H_m$ with $m\in S_\sigma$. From this description we have the dual $\tau^\vee = \operatorname{Cone}(\sigma^\vee \cup \{-m\})$ containing $\sigma^\vee$, then for the semigroups we have $S_\tau = S_\sigma + \mathbb Z(-m)$, so all that happened is we made one of the elements of $S_\sigma$ invertible in $S_\tau$. Thus on the level of $\mathbb C$-algebras we have the localization $$ \mathbb C[S_\tau] = \mathbb C[S_\sigma]_{\chi^m} = \mathbb C[\chi^{m_1},\dots,\chi^{m_k}, \chi^{-m}], $$ where $m\in S_\sigma = \mathbb N\{m_1,\dots,m_k\}$.


How can we relate $U_\tau$ in coordinates to $U_\sigma=\mathbf V(\ker \varphi)\subseteq \mathbb C^k$ as given above?

I guess we should look at the situation in $\mathbb C^{k+1}$ first, since we have $k+1$ generators now and project down to $\mathbb C^k$ afterwards, but I'm not sure how to do this.

My thoughts

Consider \begin{align} \psi : \mathbb C[x_1,\dots,x_k,y] &\longrightarrow \mathbb C[\chi^{m_1},\dots,\chi^{m_k},\chi^{-m}], \\ x_i &\longmapsto \chi^{m_i},\\ y &\longmapsto \chi^{-m}. \end{align} By treating $\mathbb C[x_1,\dots,x_k]$ as a subalgebra of $\mathbb C[x_1,\dots,x_k,y]$ we have $\ker \varphi\subseteq \ker\psi$. Since $m\in S_\sigma$, we have $m=\sum_{i=1}^k c_i m_i$ for some $c_i\in\mathbb N$. This relation yields $$ \chi^m = (\chi^{m_1})^{c_1} \cdots (\chi^{m_k})^{c_k}, $$ so $x_1^{c_1}\cdots x_k^{c_k} y-1\in\ker\psi$ and thus $$ \ker \psi \supseteq \ker\varphi + \langle x_1^{c_1}\cdots x_k^{c_k} y-1\rangle. \tag{1}\label{1} $$ Do we have equality here?

Continuing this line of thought, we have $$ U_\tau = \mathbf V(\ker \psi)\subseteq \mathbb C^{k+1}. $$

The projection $\mathbb C^{k+1}\to\mathbb C^k$ that sets $y=0$ is injective on $U_\tau$: Every point $(x_1,\dots,x_k,y)\in U_\tau$ is a solution of $x_1^{c_1}\cdots x_k^{c_k} y=1$, thus $y=x_1^{-c_1}\cdots x_k^{-c_k}$ is uniquely determined. So we can identify $U_\tau$ with it's image in $\mathbb C^k$.

If we had equality in \eqref{1}, we could conclude $$ U_\tau = \left\{\, (x_1,\dots,x_k)\in U_\sigma\,\big|\, x_1^{c_1}\cdots x_k^{c_k}\neq 0 \right\}, $$ which is an open subset of $U_\sigma$.

Is this reasoning correct and do we infact have equality in \eqref{1}?


1 Answer 1


Let $R$ be a commutative ring, and let $f$ be a non-zero element. Write $R_f$ for the localization of $R$ at $f$. Then Spec($R_f$) is naturally contained in Spec($R$): it is the locus where $f$ is non-zero. This is just the geometrical version of the commutative ring theory statement that the prime ideals of $R_f$ are the prime ideals of $R$ that don't contain $f$.

Edited to add: in response to the OP's first comment, let me try to say a bit more.

We do have equality in (1), because $R_f=R[y]/(fy-1)$.

Projection from Spec($R$) x $\mathbb C$ to Spec($R$) corresponds to the inclusion of rings of $R$ into $R[y]$. To get the geometric map, we intersect prime ideals of $R[y]$ with $R$. A prime ideal in $R[y]$ lying on the hypersurface defined by $fy-1=0$ restricts to a prime ideal in $R$ such that $f$ is non-zero. In particular, this holds for maximal ideals, i.e., points.

  • $\begingroup$ I knew about that, but I'm trying to give the geometrical statement a precise coordinate-wise description. $\endgroup$
    – Christoph
    Mar 25, 2014 at 8:19

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