Why is "the number of a rings of order $n$" a multiplicative function?

Question

The OEIS sequence A037234 (also A027623) counts the number of (not necessarily unital) rings of order $n$. On the OEIS page (as pointed out by @Smiley1000) it says that the sequence is multiplicative. Why is this so? The number of groups of order $n$ does not exhibit this property... is it unique to rings? Does the multiplicativity have any useful implications?

Answer 1

Any torsion abelian group $A$ is a direct sum of its $p$-primary components, $A=\oplus_p A_p$. For a prime $p$, the component $A_p$ is the subgroup of all elements of order a power of $p$. This is a good group theory exercise.

Thus if $R$ is a finite ring of order $n$, we have $R=\oplus R_p$ as a direct sum of the underlying additive groups of the ring $R$ (note: the notation $R_p$ is in conflict with other things in ring theory like localization).

It is a quick argument to show that if $x\in R_p$ and $y\in R_q$ for distinct primes $p,q$ then $xy\in R_p\cap R_q=0$, hence the primary components are ideals, hence the direct sum is also a direct product of (non-unital) rings.

Thus, every ring of order $n=\prod p^v$ is uniquely a direct product of rings of orders $p^v$ (in the non-unital world).

Answer 2

Multiplicativity holds for the number of rngs of order $n$ to which the linked OEIS A037234 sequence refers to. But it also holds for the number of rings of order $n$ (A037291), as well as the number of abelian groups of order $n$ (A000688). The proof for rngs and rings is a rather straight forward adaptation of the proof for finite abelian groups.

Let $R$ be a rng with $nm$ elements, where $n,m$ are coprime. By Lagrange's Theorem for the additive group of $R$ we have $mnx = 0$ every $x \in R$. For the moment, let us assume only this, i.e. $mnx = 0$ for all $x \in R$, which means that the additive group of $R$ is of exponent $mn$. Define $R_m := \{x \in R : mx = 0\}$ and $R_n := \{x \in R : nx = 0\}$. Clearly, these are ideals of $R$. These are in particular subrngs of $R$.

I claim that $R = R_m \times R_n$ as rngs (internal direct product*). Choose integers $u,v$ with $un + vm=1$. If $x \in R_m \cap R_n$, then $mx=0$ and $nx=0$, hence $x = 1x = unx + vmx = 0 + 0 = 0$ (we have not used a unit in $R$, which might not exist, but rather the $\mathbb{Z}$-module structure). This shows $R_m \cap R_n = 0$. If $x \in R$, then $x = unx + vmx$ with $unx \in R_m$ (since $mnx = 0$) and likewise $vmx \in R_n$. This proves the claim $R = R_m \times R_n$.

Two rngs $R,R'$ of exponent $mn$ are isomorphic if and only if $R_m \cong R'_m$ and $R_n \cong R'_n$ as rngs. The direction $\Rightarrow$ is immediate from the definitions of $R_m$ and $R_n$, and the direction $\Leftarrow$ follows from the direct product decomposition.

This shows that the number of finite rngs of a given exponent up to isomorphism is multiplicative; in fact we have just recycled the proof for finite abelian groups of a given exponent. In our case, though, we look at finite rngs of order $n$ up to isomorphism. So we also need to check that

1. If $R_m$ has order $m$ and $R_n$ has order $n$, then $R$ has order $mn$.
2. If $R$ has order $mn$, then $R_m$ has order $m$ and $R_n$ has order $n$.

The claim 1) is an immediate consequence of $R = R_m \times R_n$. The part 2) is more interesting (and is currently also missing in the other answers). But it can again be recycled from the additive groups.

Assume that $R$ has order $mn$. By Lagrange's Theorem, the order of $R_m$ divides $mn$, and since $m,n$ are coprime it is then of the form $kl$ for divisors $k \mid m$, $l \mid n$ which are also coprime. Assume that $l \neq 1$. Then $l$ has a prime divisor $p$. By Cauchy's Theorem, there is an element $x \in R_m$ which has order $p$ in the additive group of $R_m$. We have $mx = 0$ and $px = 0$, and since $m,p$ are coprime it follows $x=0$, a contradiction. This shows $l=1$. Thus, the order of $R_m$ divides $m$. In the same way we see that the order of $R_n$ divides $n$. The product of the orders is the order of $R_m \times R_n = R$, which is $mn$. So the orders must be $m$ resp. $n$.

If $R_m$ and $R_n$ have a unit, then of course also $R = R_m \times R_n$ has a unit. And if $R$ has a unit, then $R_m \times R_n$ has a unit, which easily implies that $R_m$ and $R_n$ have a unit. Two rings are isomorphic if and only if their underlying rngs are isomorphic. So in the unital case, there is even an isomorphism of rings $R \cong R_m \times R_n$. This implies that the statement is also true for rings of a given order.

*Let $R$ be a rng with two subrngs $A,B$. We call $R$ the internal direct product of $A$ and $B$ if $ab=0$ and $ba=0$ for all $a \in A$, $b \in B$ and every element of $R$ can be written as $a+b$ for $a \in A$, $b \in B$. Equivalently, $A$ and $B$ are ideals of $R$ with $A \cap B = 0$ and $R = A + B$. In this case, the multiplication in $R$ is given by $(a+b)(a'+b') = aa' + bb'$, and we see that the map $(a,b) \mapsto a+b$ from the external direct product $A \times B$ to $R$ is an isomorphism of rngs.

Let me mention that there is no internal direct product for rings. If $R,S$ are rings, then $R$ is not a subring of $R \times S$, in the sense that the canonical additive map $R \to R \times S$ is not a ring homomorphism.