4
$\begingroup$

Let $A$ be an (not necessarily unital) algebra over $\mathbb{R}$ or $\mathbb{C}$. If $a\in A$ is an element such that $ab=0$ for all $b\in A$ (or equivalently, $aA=\{0\}$), can we then conclude that $a=0$?

At first sight, it looks like a trivial statement. However, I am not able to prove or counterprove it.

It is clearly true for unital algebras (take $b=1$). I was also able to prove that this statement is true for (complex) C*-algebras (a certain class of algebras):

If $A$ is a C*-algebra, then $A$ admits at least one approximate unit $(u_{\lambda})_{\lambda\in\Lambda}$. By assumption we have $au_{\lambda}=0$ for all $\lambda\in\Lambda$. Taking the limit on both sides yields $a=0$.

Any suggestions would be greatly appreciated. It feels like I'm missing something trivial...

$\endgroup$
1
  • $\begingroup$ It is not true for Banach Algebras, I do not have a counter-example in mind, but here is a fact: If it was true, Murphy's book wouldn't state exercise 2.1 as it does $\endgroup$ May 22, 2020 at 21:32

2 Answers 2

8
$\begingroup$

We cannot conclude this. Indeed, on any vector space $V$ over any field, one can define multiplication by assigning $vw=0$ for all $v,w\in V$.

Of course, if your algebra $A$ is unital, with unit $e$, then the conclusion does hold. Indeed, $a=ae=0$.

Moreover, there is a more elementary proof that this is true for $C^*$-algebras

Let $A$ be a $C^*$-algebra, and suppose $a\in A$ satisfies $ab=0$ for all $b\in A$. Then $$0=\|aa^*\|=\|a\|^2.$$ Thus $\|a\|=0$, and therefore $a=0$.

$\endgroup$
3
$\begingroup$

Hopefully, I am not saying something stupid.

Pick your favourite algebra $A$.

Let $a \notin A$ be any element, and define $$B= \mathbb{F} a \oplus A$$ where $\mathbb{F}$ is your field.

Now, $B$ becomes an algebra, under the obvious addition and multiplication defined as $$(\alpha a+b)(\beta a +c)=bc$$

And clearly $aB=0$.

$\endgroup$
1
  • $\begingroup$ @DavidC.Ullrich Ty, fixed. That was a typo, by construction $aa=0$ too. $\endgroup$
    – N. S.
    May 23, 2020 at 5:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.