Faces of cones give localizations of affine toric varieties

I'm trying to establish an explicit description of the open subsets of affine toric varieties given by faces of the underlying cone.

Background

For a rational convex polyhedral cone $\sigma\subseteq N_\mathbb R$ over a lattice $N$, we know that the semigroup algebra $S_\sigma=\sigma^\vee\cap M$, where $M$ is the dual lattice of $N$, is finitely generated by some $m_1,\dots,m_k\in M$, thus $$U_\sigma =\operatorname{Spec}(\mathbb C[S_\sigma]) = \operatorname{Spec}(\mathbb C[\chi^{m_1},\dots,\chi^{m_k}])$$ and we find an explicit (coordinate) description of this affine variety by looking at the $\mathbb C$-algebra homomorphism given by \begin{align} \varphi : \mathbb C[x_1,\dots,x_k] &\longrightarrow \mathbb C[\chi^{m_1},\dots,\chi^{m_k}], \\ x_i &\longmapsto \chi^{m_i} \end{align} And let $U_\sigma = \mathbf V(\ker \varphi) \subseteq \mathbb C^k$.

Now let $\tau = \sigma \cap H_m$ be the face given by some supporting hyperplane $H_m$ with $m\in S_\sigma$. From this description we have the dual $\tau^\vee = \operatorname{Cone}(\sigma^\vee \cup \{-m\})$ containing $\sigma^\vee$, then for the semigroups we have $S_\tau = S_\sigma + \mathbb Z(-m)$, so all that happened is we made one of the elements of $S_\sigma$ invertible in $S_\tau$. Thus on the level of $\mathbb C$-algebras we have the localization $$\mathbb C[S_\tau] = \mathbb C[S_\sigma]_{\chi^m} = \mathbb C[\chi^{m_1},\dots,\chi^{m_k}, \chi^{-m}],$$ where $m\in S_\sigma = \mathbb N\{m_1,\dots,m_k\}$.

Question

How can we relate $U_\tau$ in coordinates to $U_\sigma=\mathbf V(\ker \varphi)\subseteq \mathbb C^k$ as given above?

I guess we should look at the situation in $\mathbb C^{k+1}$ first, since we have $k+1$ generators now and project down to $\mathbb C^k$ afterwards, but I'm not sure how to do this.

My thoughts

Consider \begin{align} \psi : \mathbb C[x_1,\dots,x_k,y] &\longrightarrow \mathbb C[\chi^{m_1},\dots,\chi^{m_k},\chi^{-m}], \\ x_i &\longmapsto \chi^{m_i},\\ y &\longmapsto \chi^{-m}. \end{align} By treating $\mathbb C[x_1,\dots,x_k]$ as a subalgebra of $\mathbb C[x_1,\dots,x_k,y]$ we have $\ker \varphi\subseteq \ker\psi$. Since $m\in S_\sigma$, we have $m=\sum_{i=1}^k c_i m_i$ for some $c_i\in\mathbb N$. This relation yields $$\chi^m = (\chi^{m_1})^{c_1} \cdots (\chi^{m_k})^{c_k},$$ so $x_1^{c_1}\cdots x_k^{c_k} y-1\in\ker\psi$ and thus $$\ker \psi \supseteq \ker\varphi + \langle x_1^{c_1}\cdots x_k^{c_k} y-1\rangle. \tag{1}\label{1}$$ Do we have equality here?

Continuing this line of thought, we have $$U_\tau = \mathbf V(\ker \psi)\subseteq \mathbb C^{k+1}.$$

The projection $\mathbb C^{k+1}\to\mathbb C^k$ that sets $y=0$ is injective on $U_\tau$: Every point $(x_1,\dots,x_k,y)\in U_\tau$ is a solution of $x_1^{c_1}\cdots x_k^{c_k} y=1$, thus $y=x_1^{-c_1}\cdots x_k^{-c_k}$ is uniquely determined. So we can identify $U_\tau$ with it's image in $\mathbb C^k$.

If we had equality in \eqref{1}, we could conclude $$U_\tau = \left\{\, (x_1,\dots,x_k)\in U_\sigma\,\big|\, x_1^{c_1}\cdots x_k^{c_k}\neq 0 \right\},$$ which is an open subset of $U_\sigma$.

Is this reasoning correct and do we infact have equality in \eqref{1}?

Let $R$ be a commutative ring, and let $f$ be a non-zero element. Write $R_f$ for the localization of $R$ at $f$. Then Spec($R_f$) is naturally contained in Spec($R$): it is the locus where $f$ is non-zero. This is just the geometrical version of the commutative ring theory statement that the prime ideals of $R_f$ are the prime ideals of $R$ that don't contain $f$.
We do have equality in (1), because $R_f=R[y]/(fy-1)$.
Projection from Spec($R$) x $\mathbb C$ to Spec($R$) corresponds to the inclusion of rings of $R$ into $R[y]$. To get the geometric map, we intersect prime ideals of $R[y]$ with $R$. A prime ideal in $R[y]$ lying on the hypersurface defined by $fy-1=0$ restricts to a prime ideal in $R$ such that $f$ is non-zero. In particular, this holds for maximal ideals, i.e., points.