Is $V(X+Y-Z)$ a toric variety?
问题内容
I'm reading CLS's Toric Varieties right now, and something is confusing me.
By Theorem 1.1.17, an affine variety $V$ is toric iff $I(V)$ is toric, i.e. prime and generated by binomials.
Now the variety $V = V(X+Y-Z) \subset \mathbb{C}^3$ seems to me to be toric simply because it's isomorphic to $\mathbb{C}^2$. Via the isomorphism $\phi: \mathbb{C}^2 \rightarrow V, (x,y) \mapsto (x,y,x+y)$, I think one should get the open torus (?) $T_V = \phi((\mathbb{C}^*)^2) = V\setminus V(XY) = \{(x,y,x+y)\ |\ x,y\in\mathbb{C}^*\} \subset V$ with the inherited action $(\mathbb{C}^*)^2 \curvearrowright T_V$ and everything working fine.
On the other hand, the ideal $I(V) = (X+Y-Z) \subset \mathbb{C}[X,Y,Z]$ of $V$ is clearly prime, but I just cannot see how one would be able to generate it using only binomials.
Where is my mistake here?
回答 (1)
Theorem 1.1.17 asserts the existence of an embedding for which the defining ideal is toric, not that the ideal of every embedding of a toric variety must be toric. Consider the map $\phi:\mathbb C^2\longrightarrow V$ where $(x,y)\longmapsto (x,y,x+y)$, which is an isomorphism. Therefore, the open subset $T=\phi\left( (\mathbb C^*)^2 \right)=\left\{ (x,y,x+y)\mid x,y\neq 0 \right\}$ is isomorphic to the torus $(\mathbb C^*)^2$. Its action extends to all of $V$ by the formula $(\lambda,\mu)\cdot(x,y,x+y)=(\lambda x,\mu y,\lambda x+\mu y)$, and hence $V$ is indeed an affine toric variety
However, our embedding $V\subset\mathbb C^3$ is not toric with respect to the coordinates $X,Y,Z$. On the open torus, the restrictions of the coordinate functions are $X|_T=x$, $Y|_T=y$, $Z|_T=x+y$. The functions $x$ and $y$ are characters of the torus, whereas $x+y$ is a sum of two characters rather than a character. In a toric embedding, the coordinate functions must be given by characters, that is, by monomials, so in these coordinates one obtains the trinomial $X+Y-Z$ rather than a binomial relation. It is worth noting that the ideal $I(V)=(X+Y-Z)$ in these coordinates indeed contains no nonzero binomials. If $X^aY^bZ^c-\lambda X^{a'}Y^{b'}Z^{c'}\in I(V)$, then after substituting $Z=X+Y$ one would obtain $\lambda X^{a'}Y^{b'}(X+Y)^{c'}$. Since $X$, $Y$, and $X+Y$ are pairwise nonassociate irreducible elements of the ring $\mathbb C[X,Y]$, uniqueness of factorization implies that all the exponents coincide and $\lambda=1$. Hence such a binomial is zero
To obtain a toric ideal, it suffices to choose different coordinates $u=X$, $v=Y$, $w=Z-X$. Then the equation $X+Y-Z=0$ becomes $v-w=0$. Hence $V\simeq V(v-w)\subset\mathbb C^3$, where $(v-w)$ is a prime binomial toric ideal. The corresponding monomial parametrization is $(s,t)\longmapsto(s,t,t)$