Let $A$ be an abelian group and $a_1,\ldots ,a_n$ be some elements of $A$ and $Q$ be the finitely generated abelian monoid generated by $a_1,\ldots ,a_n$. We define a map from $Z^n$ to $A$ by sending the basis vectors $e_i$ to $a_i$. Let $L$ be the kernel of this map. Let $k[Q]$ be the $k$-algbera with $k$-basis $t^a$, where $a\in Q$ with multiplication defined as $t^a\cdot t^b=t^{a+b}$. Let $S=k[x_1,\ldots ,x_n]$ and $I_L$ is the ideal generated by $\{\underline{x}^u-\underline{x}^v|u,v\in\mathbb N^n,u-v\in L\}$. I need to show that $S/I_L$ is isomorphic to $k[Q]$ as a $k$-algebra.
First we define a map $\phi$ from $S$ to $k[Q]$ sending $x_i$ to $t^{a_i}$. Clearly this is a surjective $k$-algebra homomorphism and $I_L$ is contained in $\ker\phi$. Now we want to show this map is injective that is $\ker\phi=I_L$. As a hint the following is given but I could not figure it out:
For each $a\in Q$ consider the vector space $S_a$ whose basis consists of the monomials $\underline{x}^u$ which maps to $t^a$ under $\phi$. Image of $S_a$ in the quotient $S/I_L$ has dimension $1$ over $k$ as all the basis vector have same image. The canonical map $S/I_L\rightarrow k[Q]$ is therefore an isomorphism of vector spaces and hence $k$-algebra isomorphism.
I can see that each $S_a$ has dimension $1$ over $k$ but from this how can I conclude that $\ker\phi\subset I_L$.
Thank you in advance.