What is the following type of ring? Does it have a name?

Suppose $(R,\cdot,+,0,1)$ is a ring with another binary operation, $\circ$ with the properties:

$$(a\circ b)\circ c=a\circ(b\circ c)\\ (a+b)\circ c=(a\circ c)+(b\circ c)\\ (a\cdot b)\circ c = (a\circ c)\cdot (b\circ c)\\ 0\circ c = 0; 1\circ c=1$$

Many of these rings have a left-right $\circ$-identity.

Foundational Example 1: Given a ring $S$, the set $S^S$ of all functions $S\to S$ with point-wise addition and multiplication and composition for $\circ$:

$$(f+g)(s)=f(s)+g(s)\\(fg)(s)=f(s)g(s)(f\circ g)(s)=f(g(s))$$

This has a left-right $\circ$ identity, $I(s)=s$.

Every algebra $R$ of this sort with a right $\circ$ identity in this class is isomorphic to a sub-algebra of some $S^S$, specifically, with $S=R$. This is because if $f_a=c\mapsto a\circ c$ then if $f_a=f_b$ then $a=f_a(X)=f_b(X)=b$ where $X$ is the right identity.

Foundational example 2: Given a commutative ring $S$, $S[x]$ with polynomial composition. This case has a left-right $\circ$ identity, $x$.

Continuous functions: Given a topological ring, the set of continuous functions $S\to S$ satisfies this condition, with the identity function a left-right $\circ$-identity.

Entire functions: The ring of entire functions $\mathbb C\to\mathbb C$ is an integral domain, containing (an isomorphic image of) $\mathbb C[x]$, with $\circ$ being standard function composition.

Less well-known example: The set of all functions $f:\mathbb Z\to\mathbb Z$ with the condition: $$f(m)\equiv f(n)\pmod{m-n}$$ for all $m,n\in\mathbb Z$.

This ring is an odd integral domain that contains an isomorphic image of $\mathbb Z[x]$, since every integer polynomial function is of this sort, but there are more elements that are non-polynomials (and some non-integer polynomial functions, too.) It has $I(n)=n$ as a left-right $\circ$ identity.

The condition can be seen as requiring some "smoothness" of the functions in all the $p$-adic metrics. They can be seen as matching a certain strong uniform continuity/Lipshitz continuity in all the $p$-adics.In particular, for any $p$, we have that $f$ can be extended to the $p$-adic numbers uniquely to make it continuous, and $\|f(a)-f(b)\|_p\leq \|a-b\|_p$ for all $p$-adic integers $a,b$. Indeed, it is essentially a "bounded Lipshitz" criterion - this doesn't work in general, but does work in non-Archimedian valuation rings.

Related: If $(S,\|\|_S)$ is a non-Archimedian valuation ring with bounded valuation $\|s\|_S\leq 1$ (think the $p$-adic integers, or the rational integers under the $p$-adic norm.) Then we can take the ring $R$ of functions $f:S\to S$ with the property:

$$\|f(r)-f(s)\|_S\leq \|r-s\|_S$$

This is a ring of the above sort.

If we take $S=(\mathbb Z,\|\|_p)$, the integers with the $p$-adic norm, this set $R_p$ is just all functions $f:\mathbb Z\to\mathbb Z$ such that $p^k\mid f(n+p^k)-f(n)$ for all $n\in\mathbb Z$ and $k>0$.

Then the original ring at the beginning of this section can be seen as $\bigcap_p R_p$. The intersection of sub-algebras is always a sub-algebra.

Near misses - without multiplicative unit: The ring of meromorphic functions $\mathbb C\to\mathbb C$ is almost of this form, but when $c(x)=C$ is constant and $h(x)$ is not defined at $C$, we don't get a meromorphic function. However, if we relax the condition, and not require $R$ to have a multiplicative unit, we can take the ring of meromorphic functions $f$ with $f(0)=0$. Then the set of such functions is such a ring.

A related near miss: The sub-algebra of $S[[x]]$ of formal power series over a ring $S$, with no constant term. Again, no multiplicative unit, but everything else satisfied.


For any $f\in R$, $R_{f}=\{a\circ f\mid a\in R\}$ is a sub-algebra. The condition that $0\circ f=0,1\circ f=1$ shows that $0,1\in R_f$. Indeed, $R_f$ is a sub-algebra, because $(a\circ f)\circ (b\circ f)=(a\circ f\circ b)\circ f$. More generally, given $g\in R$ and $a\in R_f$ then $g\circ a\in R_f$. So $R_f$ is sort of like a $\circ$-right-ideal.

For example, the sub-algebra of even entire functions is $R_{x^2}$.

The subring $R_0=\{a\circ 0\mid a\in R\}$ has the property that if $r\in R_0$ then $r\circ f = (a\circ 0)\circ f = a\circ(0\circ f)=a\circ 0=r$ for any $f$.

Note that $R_0\subseteq R_f$ for any $f$. Indeed, if $r\in R_0$ then $r=a\circ 0 = a\circ(0\circ f)=(a\circ 0)\circ f$. Now, if $f\in R_0$, we get $R_0=R_f$ because $a=a\circ f =a\circ(f\circ 0) = (a\circ f)\circ 0$.

On the other hand, if $a=a\circ f$ for all $f$, then $a=a\circ 0$ so $a\in R_0$.

Basically, $R_0$ is really "all" of the constant functions. Also, for $f\in R$ and $r\in R_0$, $f\circ r\in R_0$.

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    $\begingroup$ For the first near miss, do you mean meromorphic functions? $\endgroup$ – Matt Samuel Jan 17 '16 at 23:35
  • $\begingroup$ Yep, too long since I took complex analysis. Thanks @MattSamuel $\endgroup$ – Thomas Andrews Jan 17 '16 at 23:37
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    $\begingroup$ The analogous concept for functions on a group would be a near-ring and there's some non-negligible amount of literature on them. I'd imagine that there's some name for what you describe (but I don't know what it is). $\endgroup$ – Milo Brandt Jan 19 '16 at 15:05
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    $\begingroup$ Your two foundational examples appear in this answer on MO to a question vaguely related to yours. No name was given to those rings, though. $\endgroup$ – Pierre-Guy Plamondon Jan 19 '16 at 15:26
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    $\begingroup$ Ive been to a talk where someone defined a name for the field versions (He called them thrields). Considering the speaker probably did some research into this name too, and still had to define a name suggests there isnt one. If you google "WIMP thrields", you should be able to find this. If you do some more searching, you might find more on thrields. $\endgroup$ – mdave16 Jul 26 '17 at 9:20

This appears to be very similar to a composition ring, except that the axioms on Wikipedia don't include the axioms $0\circ c=0$ and $1\circ c=1$ (and indeed, don't require an unital ring, so there may not be an $1$), but require a commutative ring, which you don't require.

However, $0\circ c=0$ follows from $0\circ c = (0+0)\circ c = (0\circ c) + (0\circ c)$, so at least that axiom is superfluous.

For $1\circ c$ this doesn't work, since in a ring you don't need to have multiplicative cancellation. So this is a true extra condition to a composition ring; indeed, the linked Wikipedia üage contains an example that explicitly contradicts this axiom (namely, $f\circ g=0$ for all $f,g\in R$).


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