I'm reading the book Toric Varieties by Cox, Little and Schenk, and have a small question about the proof of their Proposition 1.1.11 (on page 16). The key part of the proposition is this:

Claim: Let $I\subseteq\mathbb{C}[x_1,\ldots,x_s]$ be a prime ideal generated by (pure) binomials. Then $V(I)\subseteq\mathbb{C}^s$ is an affine toric variety.

The idea of the proof (slightly reformulated to simplify the notation) is this:

  • Use the fact that $I$ is both prime and binomial, to conclude that $V(I)\cap(\mathbb{C}^*)^s$ is a torus (since it's both an irreducible subvariety and a subgroup of $(\mathbb{C}^*)^s$).

  • Suppose that the dimension of $V(I)\cap(\mathbb{C}^*)^s$ is $n$, and pick an isomorphism $(\mathbb{C}^*)^n\to V(I)\cap(\mathbb{C}^*)^s$.

  • Form the composition $\Phi\colon (\mathbb{C}^*)^n\to V(I)\cap(\mathbb{C}^*)^s\hookrightarrow (\mathbb{C}^*)^s$, and note that it's an algebraic group homomorphism (concretely, this means that every component is given by a Laurent polynomial). Let $Y$ be the Zariski closure of $\mathrm{Im}(\Phi)$ in $\mathbb{C}^s$.

  • Observe that $V(I)=Y$.

  • Since we have already shown (see Proposition 1.1.8 on page 13) that every affine variety that arises as the closure of the image of an algebraic group homomorphism $(\mathbb{C}^*)^n\to(\mathbb{C^*})^s$ is an affine toric variety, the desired conclusion follows.

It's the second to last step that confuses me.

Since $\mathrm{Im}(\Phi)=V(I)\cap(\mathbb{C}^*)^s$, this essentially boils down to showing that $\overline{V(I)\cap(\mathbb{C}^*)^s}=V(I)$. The inclusion "$\subseteq$" is easy, because $V(I)$ certainly is a Zariski closed subset that contains $V(I)\cap(\mathbb{C}^*)^s$. But do we know that $V(I)$ is the smallest such subset?

Edit: I think I figured it out! See below.

  • $\begingroup$ What happens when $s=2$ and $I=(x_2)$? Then $V(x_2) \cap (\mathbb C^*)^2 = \varnothing$? $\endgroup$
    – user5826
    Commented Jul 6, 2022 at 21:33

1 Answer 1


If I'm not mistaken, this was easier than I thought. Simply note that $V(I)\cap(\mathbb{C}^*)^s$ is Zariski open in $V(I)$. Since $V(I)$ is irreducible, and every Zariski open subset of an irreducible affine variety is dense, it follows that the Zariski closure of $V(I)\cap(\mathbb{C}^*)$ in $V(I)$ is $V(I)$. And this, in turn, implies that the Zariski closure in $\mathbb{C}^s$ is $V(I)$, which is what we wanted to prorve.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .