4
$\begingroup$

Question. Is there an example of a ring $R$ (commutative or not) without unity and an element $x \in R$ such that for every $y \in R$ there exists a $z \in R$ such that $y = x z$?

In other words, is there an example of a ring without unity which has an element which divides every other element of the ring on the left? ("On the left" is of course arbitrary.) In a ring with unity, such an element would have a right inverse, and in a commutative ring with unity it would be a unit. What about the non-unital case?

If $R$ and $x \in R$ are an example, then in particular there exists a $z \in R$ such that $x = x z$. I am very unfamiliar with non-unital rings, so I don't know what this implies about $x$.

$\endgroup$
  • $\begingroup$ If it helps, every rng $I$ is an ideal of a ring (with unity); e.g. the ring $\mathbb{Z}\oplus I$ whose elements are $\mathbb{Z} \times I$, where addition is componentwise, and multiplication is $(m+i) (n+j) = mn + (mj + ni + ij)$ $\endgroup$ – Hurkyl Jun 7 '16 at 8:17
  • $\begingroup$ In what context did you find this question? $\endgroup$ – Hurkyl Jun 7 '16 at 8:37
  • $\begingroup$ @Hurkyl I thought of this question while grading algebra homework. Thank you for your remark about embedding rngs as ideals in rings, by the way! $\endgroup$ – Artem Mavrin Jun 7 '16 at 8:41
  • $\begingroup$ Note that if we weaken the requirement to "$a$ divides some power of $b$, for every $b$" then there are examples - e.g. $\mathbb{Q}/\mathbb{Z}$. $\endgroup$ – Noah Schweber Jun 8 '16 at 2:16
2
$\begingroup$

Let $a\in R$ be such an element. Then there exists in particular an element $e\in R$ with $ae=a$, and an element $\bar a$ with $a\bar a=e$.

If the multiplication is commutative, this implies that $e$ is unity: For $b\in R$ with $b=ac=ca$, say, we find $$be=cae=ca=b.$$

$\endgroup$
0
$\begingroup$

Let $S$ be any semigroup with the property that $xS = S$. e.g.the one with two elements $x$ and $y$ satisfying $x^2 = yx = x$ and $y^2 = xy = y$.

Then, the semigroup rng $R = \mathbb{Z}[S]$ will satisfy $xR = R$; e.g. if

$$ \sum_i m_i s_i $$

is an element of $R$, then we can select $t_i$ so that $x t_i = s_i$ and get

$$ x \sum_i m_i t_i = \sum_i m_i (x t_i) = \sum_i m_i s_i $$

So the only remaining question is if we can arrange so that $\mathbb{Z}[S]$ does not have a unit.

I speculate that $\mathbb{Z}[S]$ is a (unital) ring if and only if $S$ is a monoid, but just to get an example, it's easy to check that if $x$ is a left identity but not a two-sided identity of $S$ then $x$ is a left unit but not a two-sided unit of $\mathbb{Z}[S]$.

Furthermore, any quotient algebra $Q$ of $\mathbb{Z}[S]$ also satisfies $xQ = Q$ (although the quotient could potentially contain a unit).

In fact, every example arises (up to isomorphism) this way — if $R$ satisfies the condition $xR = R$, then $R$ is itself a semigroup (satisfying $xR = R$ as a semigroup!) and we can express it as a surjection $\mathbb{Z}[R] \to R$.


For reference, $\mathbb{Z}[S]$ is the free module with basis $S$ equipped with the algebra structure whose multiplication operation given by applying the semigroup operation to basis elements. (and extended to all elements by distributivity)

$\endgroup$

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.