I hope my question is not too trivial, but unfortunately i'm just starting to study toric varieties.


Let's take $X$ a lattice and $\sigma\subset X^\*$ a strongly convex rational polyhedral cone, so that, given $k$ an integrally closed field of char $0$, $P=Spec(k[\sigma^\*\cap X])$ is a toric variety on which acts the torus $T=Hom(X,k^\*)$.


Now i know that $T$ is normal and hence reduced, but i've read that it is true even more: for each point $p$ of $P$ its stabilizer is connected and reduced. Can you explain me why or give me a reference?


Thank you very much



Reduced stabilizers of torus action on toric variety
I can at least explain how this works if $k = \mathbb{C}$. As explained in Fulton's intro book, closed points in the affine toric variety $P$ are in one-to-one correspondence with semigroup homomorphisms $u: \sigma^\* \cap X \rightarrow \mathbb{C}$, where $\mathbb{C}$ is regarded as a semigroup under multiplication, and $u(0) = 1$. Closed points of the torus are just group homomorphisms $t : X \rightarrow \mathbb{C}^\*$. Given $u \in P$ and $t \in T$ the torus action is defined by sending $u$ to the new semigroup homomorphism $u'$ defined by $u'(x) = u(x)t(x)$. From this formula we can see that $t$ stabilizes $u$ if and only if $t(x) =1$ for all $x$ such that $u(x) \neq 0$. 


Now, you can easily show that the set of all $x \in \sigma^\* \cap X$ such that $u(x) \neq 0$ has to be a saturated sub-semigroup $S \subseteq \sigma^\* \cap X$, where saturated means that $nx \in S$ for $n$ a natural number implies $x \in S$. $t(s) = 1$ for all $s \in S$ if and only if $t(x) =1$ for all $x$ in the sublattice $Y$ spanned by $S$, which will also have to be saturated. So we can identify the stabilizer of $u$ with the subgroup of $T$ defined by $t(y) =1$ for all $y \in Y$. Since $Y$ is saturated, we can pick a basis $y\_1, \dots, y\_k$ of $Y$ and extend it to a basis of $X$ by adding $x\_1, \dots, x\_j$. Then the functions $z^{y\_1}, \dots, z^{y\_k}, z^{x\_1}, \dots, z^{x\_j}$ can be taken as coordinates on $T$, and the stabilizer of $u$ is just the subgroup defined by $z^{y\_1}=\cdots= z^{y\_k} =1,$ which is clearly connected. 


As far as being reduced, the multiplication map $T \rightarrow P$ defined by $t \rightarrow t \cdot u$ corresponds to the ring morphism $r: \mathbb{C}[\sigma^\* \cap X] \rightarrow \mathbb{C}[X]$ defined by $z^x \rightarrow u(x)z^x$. The ideal defining $u \in P$ is the ideal generated by the binomials $z^x-u(x)$ as $x$ ranges over all elements of $\sigma^\* \cap X$. Under $r$ these get mapped to $u(x)z^x-u(x)$, so we just get $z^x-1$ if $u(x) \neq 0$ and zero otherwise. This is clearly the radical ideal that defines the stabilizer subgroup. 


