Product of algebraic numbers Let $A$ be an algebra with $1$ over a field $K$ which is algebraic over $K$. Show that if $ab=1$, then also $ba=1$.  
Any hints on that?
 A: Suppose $a$ is the root of an $n$-degree polynomial over $K$ and $b$ is the root of an $m$ degree polynomial over $K$.
Then every element of $K[a,b]$ (meaning the subring of $A$ generated by $K$, $a$ and $b$) can be reduced to the form $\sum \alpha_{ij}b^ia^j$ where $i< m$ and $j< n$ and $\alpha_{ij}\in K$.  Of course, all the $ab$'s you might find vanish right away, and any higher powers of $b$ or $a$ will reduce to polynomials of lower degree via the relations introduced by their polynomials.
Then apparently $\{b^ia^j\mid 0\leq i< m, 0\leq j< n\}$ is a $K$ generating set for $K[a,b]$, so $K[a,b]$ is a finite dimensional $K$ algebra.
By finite dimensionality, it is a left and right Artinian ring. In such rings, $xy=1$ implies $yx=1$, as discussed, for example here. 
There are many arguments floating around the site for you to choose from. In this case you might observe that the condition that $ab=1$ implies $b$ is an injective right $K$-linear map from $K[a,b]\to K[a,b]$, and use finite dimensionality to conclude it is surjective, and hence has to have a right inverse.
Update

Sorry for being picky: I would like to avoid Artinian rings and use the last argument

In that case, I would go this route. If $ab=1$, then as homomorphisms (acting by multiplication on the left) $b$ is one-to-one and $a$ is onto. Since $K[a,b]$ is finite dimensional, $b$ is necessarily also onto. That means $ba$ a composite of onto maps, and hence is onto as well.
Therefore there must exist $x\in K[a,b]$ such that $bax=1-ba$.  But multipying this on the left with $a$ you get $ax=a-a=0$, so that $bax=1-ba$ turns into $0=1-ba$. Of course then $1=ba$.
wait but how did you know to...
The last paragraph seems a little mystical: how did I know to look at $1-ba$?  My thinking was this: "$ba$ is an idempotent, meaning that $(ba)^2=ba$. An idempotent homomorphism is like a projection and a projection that's onto has to be $1$, right?"  I was pretty sure that a nontrivial idempotent homomorphism could not map onto nonzero elements of its own kernel, and so I immediately looked at $1-ba$ since I know it's in the kernel: $ba(1-ba)=ba-(ba)^2=0$.
