Maximality of an ideal for showing that an algebra is in fact a field

Question

I have an algebra $A$ over the field $F$, with the finite dimensionality $n$ as a vector space over $F$. I can also assume that $A$ is an integral domain. Assuming that $v_1,...,v_n$ is a spanning list of vectors and that $v_1=1$, I believe I can represent $A$ as $F[v_1,v_2,...,v_n]/I$ where $$I=\left(v_1-1,\ v_iv_j\ \forall\ 2\leq i,j\leq n\right).$$ Now, I'd hope to show that $A$ is a field by proving the maximality of $I$, but I can't figure out how to show that.

Answer 1

Here's a different way to show $A$ is a field. We will simply show that if $a\in A$ is nonzero, then $a$ has a multiplicative inverse in $A$. Consider the $F$-linear map $M_a:A\to A$ given by $M_a(b) = a\cdot b$. It is an injective $F$-linear map between $F$-vector spaces of the same (finite) dimension, and so it is also surjective. Therefore, $M_a(b) = 1$ for some $b\in A$, i.e., $a\cdot b = 1$.