# Element of a ring without unity which divides every other element

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 $x$.

• 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 $(m+i) (n+j) = mn + (mj + ni + ij)$ – user14972 Jun 7 '16 at 8:17
• In what context did you find this question? – user14972 Jun 7 '16 at 8:37
• @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! – Artem Mavrin Jun 7 '16 at 8:41
• Note that if we weaken the requirement to "$a$ divides some power of $b$, for every $b$" then there are examples - e.g. $\mathbb{Q}/\mathbb{Z}$. – Noah Schweber Jun 8 '16 at 2:16

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 $$be=cae=ca=b.$$

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)