# 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?

• Surely $a$ and $b$ lie in some finite-dimensional subalgebra of $K$. In an Artinian ring, $ab=1$ implies $ba=1$. Jun 25, 2018 at 14:02
• @LordSharktheUnknown Good hint... could you please consider posting it as a hint-answer? Nothing is served by leaving the question without an answer, and it is a little awkward when a third person parrots someone else's hint-comment as an answer. Jun 25, 2018 at 14:23
• I'm no longer sure this is a valid approach. If someone else can salvage it, good luck to them! @rschwieb Jun 25, 2018 at 17:39
• @LordSharktheUnknown So be it... I'm not sure what the concern was, but I hope you will tell me if I blundered into it in my solution. Jun 25, 2018 at 20:43

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: $ba(1-ba)=ba-(ba)^2=0$.

• Good job. I'm not at my sharperst today, so it took me a while to realize that your description of $K[a,b]$ works even without commutativity because the relation $ab=1$ allows you to do simplify products like $b^{i_1}a^{j_1}b^{i_2}a^{j_2}\cdots$. Jun 25, 2018 at 20:51
• I guess that the subring of $A$ generated by $K$,$a$ and $b$ means the same as the subalgebra generated by a und b? Jun 26, 2018 at 9:57
• @mathstackuser Here they amount to the same thing, yes. But you only need the ring structure to draw the conclusion. Jun 26, 2018 at 10:33
• Thanks! Sorry for being picky: I would like to avoid Artinian rings and use the last argument, which I think goes the other way round. I think you can not show injectivity first, because by multiplying with $b$ from the right $1$ and $ba$ both are mapped to $b$ (because $bab=b$). But we can show surjectivity first ($b^{i}a^{j+1}$ is mapped to $b^{i}a^{j}$) and then injecticity follows by finite dimensionality, thus $1=ba$ because they are both mapped to $b$? Jun 26, 2018 at 13:13
• @mathstackuser I don't think it has to be that hairy. I updated my answer with the argument I was referring to. Jun 26, 2018 at 13:50