Left divisors are right divisors in an algebra where every element is algebraic

Question

Let $A$ be an algebra with $1$ over a field $K$ which is algebraic over $K$. Show that if $a$ is a left divisor, then $a$ is a right divisor.

I tried to use a similar approach as in Product of algebraic numbers So let $K[a,b]$ be the subalgebra generated by $a$ and $b$ where $ab=0$, $b\neq 0$. Am I right that this is again spanned by $\{b^ia^j∣0≤i<m,0≤j<n\}$, where $a$ is the root of an $n$-degree polynomial over $K$ and $b$ is the root of an $m$-degree polynomial over $K$? If we have $ab$ appearing the whole term is getting zero, so we can just have $b^ka^l$'s and again higher powers can be expressed by using the polynomial equations.

Now I considered the map from $K[a,b]$ to itself, sending $x$ to $ax$ which is not injective, so not surjective because of the finite dimensionality. But how can I proceed from here? (I would like to avoid Artinian rings)

Answer 1

Yes, you can apply the same logic to conclude that $K[a,b]$ is Artinian, and Artinian rings also have the property that every element is either a unit or a (two-sided) zero divisor. See this for example.

To try to short-circuit with a more specific argument like the one in your previous post, if $a$ as a left homomorphism $x\mapsto xa$ is one-to-one (it would have to be if it is not a right zero divisor) then by the same logic as last time it is also onto, and then there necessarily exists a $c$ such that $1=ca$.

Then $ab=0$ turns into $b=0$ after left multiplying with $c$. So this would contradict the any assumption that $a$ is a left zero divisor.