The following problem appeared on a problem set I'm working on: "Let $A$ be a finite-dimensional associative algebra with $1$ over a field $k$. Show that an element $a$ in $A$ has a multiplicative inverse if $a$ is not a zero divisor" (emphasis mine). I'd appreciate if y'all could check my solution to see that it's correct.
Let $a \in A$ not be a zero divisor. Then the map $T : A \to A$ given by $T : x \mapsto ax$ is an injective linear transformation from a finite-dimensional vector space $A$ to itself, as $\ker(T) = \{ x \in A : ax = 0 \} = \{ 0 \}$ by the assumption that $a$ isn't a zero divisor. Thus $T$ must also be onto, so in particular there exists a unique $b \in A$ such that $ab = 1$. Let us fix this $b$.
We know that $b$ is a right-inverse to $A$, so must show that it's also a left-inverse, i.e. that $ba = 1$. First, we wanna show that a right-inverse to $b$ exists, for which it'll suffice to show that $b$ is not a zero-divisor. If it were the case that $bx = 0$, then we'd have $a (bx) = (ab) x = 0$. But $(ab) x = 1x = x$, so $x = 0$. Thus we know that $b$ is not a zero-divisor, so there exists a unique $c \in A$ such that $bc = 1$.
It now remains to show that $a = c$. This follows as \begin{align*} a(bc) & = (ab) c \\ = a 1 & = 1 c \\ = a & = c . \end{align*} Thus $ab = ba = 1$, so $b$ can rightly be called a multiplicative inverse to $a$.
Is my solution correct? Thanks!
