An exercise from Fulton's 'Introduction Toric Varieties' asks the reader to prove that if $\tau \subset \sigma$ is a rational polyhedral subcone of a rational polyhedral cone $\sigma$, so that the induced morphism of affine toric varieties $U_{\tau} \to U_{\sigma}$ is an open embedding, then $\tau$ is a face of $\sigma$. By a face, I mean a polyhedral subcone entirely contained in the topological boundary of $\sigma$. There are other equivalent definitions (like $x + y \in \tau$ implies $x, y \in \tau$), which are fair game.
I was having significant trouble with this problem. Most of my attempts have been trying to show that if $\tau$ is not a face (ie. it contains an element of the interior of $\sigma$), then the induced map $U_{\tau} \to U_{\sigma}$ is never an open embedding.
I am pretty stuck here though. I tried computing some examples, and noticed that a lot of the time, if $U_{\sigma} = \operatorname{Spec} A$, then $U_{\tau} = \operatorname{Spec} A[u, u^{-1}]$ for some $u$, and tried to argue that the map $A \hookrightarrow A[u, u^{-1}]$ never induces an open embedding under some assumptions on $u$. I can't seem to find the right $u$, and I doubt that this is the right method generally, so I was hoping that someone could offer any help or hints.
(Without conditions on $u$, this is definitely not right. For example, if $u \in k[x]$ is $x$, then this is just the inclusion $\mathbb{G}_m \hookrightarrow \mathbb{A}^1$).
Fulton mentions that this can be done by analyzing the torus action on the associated toric varieties, but the part of the book I am on has not developed that yet, so I am hoping to solve this another way. Again, any hints or ideas are greatly appreciated!
Thanks!
The following works if $\sigma$ and $\tau$ have the same dimension.
The cone $\tau$ is defined by intersecting $\sigma$ with half spaces. I will show that if $\tau = \sigma \cap H$, where $H \subset \mathbb R^n$ is a single half space, then $U_{\tau} \to U_\sigma$ is not injective. The more general case follows from that by considering the sequence $\sigma = \tau_0 \supset \tau_1 \supset \dotsc \supset \tau_k = \tau$, where in each step $\tau_{i+1} = \tau_i \cap H_i$ for some half space $H_i$. Then $U_\tau \to U_{\tau_{k-1}}$ is not injective, so $U_\tau \to U_\sigma$ is not injective either. This reduction step goes wrong if $\tau$ has a smaller dimension than $\sigma$, because then $\tau$ could be a face of $\tau_{k-1}$.
By assumption, $\tau \subset \sigma$ is not a face, so on the dual cones we get, $$\sigma^\vee \subset \tau^\vee = \sigma^\vee + \alpha \mathbb R_{\geq 0},$$ where $\alpha \in \mathbb Z^n$ is non-negative on $H$. We may also assume that $\lambda \alpha \notin \mathbb Z^n$ for $0 < \lambda < 1$, i.e. $\alpha$ is minimal with that property. So if we set $A = \mathbb C[\sigma^\vee \cap \mathbb Z^n]$, which is the coordinate ring of $U_\sigma$, then the coordinate ring of $U_\tau$ is $$\mathbb C[\tau^\vee \cap \mathbb Z^n] = A[x^\alpha],$$ where I use multi index notation $x^\alpha = x_1^{\alpha_1} \dotsm x_n^{\alpha_n}$. Since we are dealing with monomials, $x^\alpha$ satisfies minimal a relation $$x^\alpha \cdot x^\beta = x^\gamma\tag{1}$$ for $\beta, \gamma \in \sigma^\vee$. I claim that $\beta$ and $\gamma$ cannot be zero. If $\beta = 0$, then $\alpha = \gamma \in \sigma^\vee$. If $\gamma = 0$, then $\alpha = -\beta$. Since $\sigma$ and $\tau$ have the same dimension, the maximal linear subspaces contained in $\sigma^\vee$ and $\tau^\vee$ have the same dimension and are contained in each other. Hence they are the same. As $\alpha, -\alpha \in \tau^\vee$ also $\alpha \in \sigma^\vee$. Contradiction.
From (1) we see that $A[x^\alpha]$ is isomorphic to the quotient ring $$A[\xi] /(\xi x^\beta - x^\gamma).$$ So we may think of $U_\tau$ as a closed subset of $U_\sigma \times \mathbb A^1$, and the map $U_\tau \to U_\sigma$ is the projection. Now as $\beta, \gamma \neq 0$, we see that $\{0\} \times \mathbb A^1$ is contained in $U_\tau$, and is contracted to $0 \in U_\sigma$. Hence $U_\tau \to U_\sigma$ is not injective.