Let $R$ be a commutative ring containing a field $K$ such that $R$ viewed as a $K$-Vector space has finite dimension. We then get a pretty neat thing:
Observation Let $r\in R$. The left multiplication map $$ \lambda_r: R \to R,\quad x\mapsto r\cdot x $$ is $K$-linear.
This enables us to observe some properties of ring elements by the behavior of their according left multiplication map. Note that if $r\in K$, the map is diagonal in every $K$-linear Basis of $R$.
Lemma $r$ is invertible if and only if it is not a zero divisor.
Proof If $r$ is not a zero divisor, $\lambda_r$ must have trivial kernel, i.e. it is injective. By finite dimensionality, it is also surjective, hence $1\in \mathrm{Im} \lambda_r$. Therefore, we can find an $s\in R$ such that $1=\lambda_r (s) = rs$. On the other hand, if $r$ is invertible, it cannot be a zero divisor, since for every $s\in r$, $0=rs$ implies $0 = r^{-1}rs = s$.