In the book "Toric Varieties" by Cox-Little-Schenck, Proposition 1.3.8 is left as an exercise. It essentially characterizes what the normalization of an affine toric variety is. The construction is as follows:

Pick $V=\operatorname{Spec}(\mathbb{C}[\mathsf{S}])$ where $\mathsf{S}$ is an affine semigroup. Let $\mathscr{A} = \{m_1,\ldots,m_s\}\subseteq \mathsf{S}$ be such that $\mathbb{N}\mathscr{A} = \mathsf{S}$, and let $M = \mathbb{Z}\mathscr{A}$ be the lattice of characters of the torus of $V$. They pick the rational cone given by $\sigma := \operatorname{Cone}(\mathscr{A})\subseteq N_{\mathbb{R}}$, and the affine semigroup given by $\mathsf{S}_\sigma := \sigma^\vee \cap M$. They claim that the obvious inclusion $\mathbb{C}[\mathsf{S}]\hookrightarrow \mathbb{C}[\mathsf{S}_\sigma]$ induces a morphism at the level of varieties, $\operatorname{Spec}(\mathbb{C}[\mathsf{S}_\sigma]) \to \operatorname{Spec}(\mathbb{C}[\mathsf{S}])$ which is a normalization map.

What I tried to do so far is to prove using the plain definition of normalization map, that if $\alpha$ is in the fraction field $\operatorname{Frac}(\mathbb{C}[\mathsf{S}])$ and satisfies a monic polynomial equation with coefficients in $\mathbb{C}[\mathsf{S}]$ then it has to be an element of $\mathbb{C}[\mathsf{S}_{\sigma}]$.

I suspect, however, that if one could prove first that the normalization of an affine toric variety is again an affine toric variety (without necessarily being precise of who the lattice, torus and action are), then the characterization of normal toric varieties given by Theorem 1.3.5 could be used somehow. Any help would be highly appreciated.


2 Answers 2


We define the saturation of $S$ to be $$\overline{S} = \left \{p \in M \mid kp \in S \ \text{for some} \ k \in \mathbb{N}^* \right \}.$$ This is another description of $\sigma^{\vee} \cap M$ (and highly nontrivial to prove). At first, I am going to show that $\mathbb{C}[S] \hookrightarrow \mathbb{C}[\overline{S}]$ is indeed a normalization.

  • $\mathbb{C}[\overline{S}]/\mathbb{C}[S]$ is an integral extension: since sum and product of integral elements are again integral, we just need to prove that each "elementary" character $\chi^p \in \mathbb{C}[\overline{S}]$ is integral over $\mathbb{C}[S]$, but this is obvious since if $kp \in S$ then $\chi^p$ is a root of the polynomial $X^k - \chi^{kp} \in \mathbb{C}[S][X]$.
  • $\mathrm{Frac}(\mathbb{C}[\overline{S}])$ is integrally closed because $\overline{S}$ is saturated.
  • $\mathrm{Frac}(\mathbb{C}[S]) = \mathrm{Frac}(\mathbb{C}[\overline{S}])$ (an important remark here is $M =\mathbb{Z}S$; if $M \neq \mathbb{Z}S$ we may consider a counterexample: $M = \mathbb{Z}, S = 2\mathbb{N}, \overline{S} = \mathbb{N}$, hence $\mathbb{C}[S] = \mathbb{C}[x^2]$ and $\mathbb{C}[\overline{S}] = \mathbb{C}[x]$ are both integrally closed but have different fields of fractions). In fact, we are going to prove a stronger result, that is, $$\mathrm{Frac}(\mathbb{C}[S]) = \mathrm{Frac}(\mathbb{C}[M]).$$ Let $x = (\sum_{p \in I}a_p \chi^p)/(\sum_{p \in I}b_p \chi^p) \in \mathrm{Frac}(\mathbb{C}[M])$ for some finite index set $I \subset M$. For each $p \in I$, we express $$ p = \sum_{i=1}^s n_i^p m_i,$$ where $n_i^p \in \mathbb{Z}$ and $\left \{m_1,...,m_s \right \}$ is a generating set of $S$ (as a semi-group). Choose any element $q = \sum_{i=1}^s n_i m_i$ with $n_i \geq \left |n_i^p \right| \ \forall p \in I$ and multiply both numerator and denominator by $\chi^q$ to see that $$x = \frac{\sum_{p \in I} a_p \chi^{p+q}}{\sum_{p \in I} b_p \chi^{p+q}} \in \mathrm{Frac}(\mathbb{C}[S]).$$

From the two results above, we conclude that $\mathbb{C}[S] \hookrightarrow \mathbb{C}[\overline{S}]$ is a normalization. We are left to prove that $\overline{S} = \sigma^{\vee} \cap M = \mathrm{Cone}(S) \cap M$. The inclusion $\overline{S} \subset \sigma^{\vee} \cap M$ is obvious so we will be done once we prove $\mathrm{Cone}(S) \cap M \subset \overline{S}$. This is much more difficult! Let $(e_1,...,e_n) \in \mathbb{M}^n$ to be a $\mathbb{Z}$-basis of $M$. Pick any $x \in \mathrm{Cone}(S) \cap M$, then there exists $a_1,...,a_s \in \mathbb{R}_{\geq 0}$ (not $\mathbb{Z}_{\geq 0}$!) such that $$x = a_1m_1 + \cdots + a_s m_s \in M.$$ We express $m_i = b_{i1}e_1 + \cdots + b_{in}e_n$ for some $b_{ij} \in \mathbb{Z}$. The above expression of $x$ shows that we have a system of equations $$\sum_{i=1}^s b_{ij}a_i = z_j \in \mathbb{Z} \ \forall \ j = \overline{1,n}.$$ This system of equation has at least one real solution, say, $(a_1,...,a_s)$. By this post, it has a rational solution; that is, $x$ can be expressed in terms of $m_i$ with rational coefficients. Finally, we multiply $x$ with some natural number to kill the denominator and hence $x \in \overline{S}$, finish our proof.


Hint: use the fact that for a variety over a field, the normalization is the maximal finite birational morphism $W \to V$; it seems you have found a normal variety with a finite birational morphism $ V' \to V$, so...

  • $\begingroup$ Thank you very much for the answer! Yes, but that is where I get stuck, I don't know how to conclude the maximality. $\endgroup$ Mar 17, 2021 at 20:07
  • $\begingroup$ Maybe I am missing some point, but it seems to me that a normal map "bounds from above" how big should be the normalization. In other words, you have the universal property that for (well behaved) maps from a normal variety to $V$, they factor through the normalization. This tells you that a (well behaved) map from a normal variety to $V$ is "bigger" than the normalization, which is maximal, so it must be maximal itself. See this question for this point of view: mathoverflow.net/questions/46/… $\endgroup$ Mar 18, 2021 at 10:48
  • $\begingroup$ Another point of view, which is maybe more elementary, is the following:math.stackexchange.com/questions/2553201/… $\endgroup$ Mar 18, 2021 at 10:51

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