if A is a finite dimensional algebra over a field K,then every left zero divisors in A are right zero divisors?

Question

I have thought about some examples and it seems to be right.I,think it is an interesting question,but I aren't able to prove it.....

Answer 1

I'll assume that $A$ is unital.

Let's look at the contrapositive: every element which is not a right zero-divisor is not a left zero-divisor. I claim that if $a$ is not a right zero-divisor, then $a$ is a unit (so certainly not a left zero-divisor). Consider the map $\mu:x\mapsto xa$. Regarding $A$ as a vector space over $K$, it is a linear map. It is injective from $A$ to $A$, since its kernel consists of $x$ with $xa=0$, and there aren't any non-zero solutions for this. As $A$ is finite-dimensional over $K$, then $\mu$ must be surjective: there is $b$ with $\mu(b)=1$, that is $ba=1$.

We can repeat this argument with $b$ replacing $a$, since $xb=0$ implies $xba=0$ implies $x=0$. There is $c$ with $cb=1$. Then $c=cba=a$ and $ab=1$; $a$ is a unit with inverse $b$.