0
$\begingroup$

Let $R$ be a non commutative ring. We will say that an element of $R$ is isolated if it is zero divisor and nothing nonzero annihilates it at the same time on both sides.

Note that there are many classes of rings that do not contain isolated elements.

$\bullet$ If $R$ is an integral ring, it does not contain isolated elements.

$\bullet$ If $R$ is a nilpotent ring, it does not contain isolated elements.

$\bullet$ If $R$ is a reversible ring, it does not contain isolated elements.

$\bullet$ If $R$ is a finite unital ring, it does not contain isolated elements.

Do you know of any indecomposable ring that has no isolated elements and is neither reversible, nor integral, nor nilpotent, nor finite unital?

Improve this question
$\endgroup$
1
  • $\begingroup$ Hi: would you mind clarifying your title if you meant "not finite unital" as you said in the body? I can't be sure what you meant. $\endgroup$
    – rschwieb
    Commented Sep 22, 2022 at 17:34

1 Answer 1

Reset to default
4
$\begingroup$

Let $F$ be a field, and let $$ R:=F\langle x,y\, :\, x^2=xy=y^2=0\rangle. $$ Notice that $\{1,x,y,yx\}$ is an $F$-basis for $R$ as an $F$-vector space.

This ring is not reversible since $xy=0$ but $yx\neq 0$.

If by "integral" you mean a domain, then this ring is clearly not a domain (since it isn't reversible).

It is not nilpotent, since it contains $1\neq 0$.

It is not finite if $F$ is an infinite field.

This ring is indecomposable, as the only idempotents are the trivial ones. (To see this, first note that the constant term of an idempotent $e$ must be idempotent, as this is a graded ring, graded by degree. After replacing $e$ by $1-e$ if necessary, we may assume the constant term is $0$. But then $e$ is nilpotent, and the only nilpotent idempotent is $0$.)

Every element in this ring without $1$ in its support is annihilated, from both the left and the right, by $yx$. As no element with $1$ in its support is a zero-divisor, there are no isolated elements.

Improve this answer
$\endgroup$
3
  • 1
    $\begingroup$ There's some ambiguity: the title asks for "not unitary" whereas the body says "not finite unital." This is a great answer for the body question, at least (which I guess is probably the intended one.) $\endgroup$
    – rschwieb
    Commented Sep 22, 2022 at 17:33
  • 2
    $\begingroup$ @rschwieb Good point. If they don't want it unital, they can add a new variable $z$, commuting with $x$ and $y$, and take the nonunital subring generated by $x,y,z$. This ring won't be unital, since it is graded by degree. It won't be nilpotent since $z$ isn't nilpotent. We didn't add any new idempotents, so it is still indecomposable. $\endgroup$
    – Pace Nielsen
    Commented Sep 22, 2022 at 17:50
  • 2
    $\begingroup$ Alternatively, just replace $F$ with $2\mathbb{Z}$, and adjust things accordingly. $\endgroup$
    – Pace Nielsen
    Commented Sep 22, 2022 at 17:51

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .