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I believe that in the category of $(R,S)$-bimodules where $R$ and $S$ are rings with identity, then ${}_R R \otimes_\mathbb{Z} S_S$ is a free object on one generator in this category.

But, if $R, S$ are rings without identity, does the category of $(R,S)$-bimodules contain a free object on one generator?

user02262019
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    Closely related : https://math.stackexchange.com/questions/3129450/do-categories-of-bimodules-have-enough-projectives – Arnaud D. Apr 16 '19 at 02:13
  • Isn't the category of $(R,S)$-bimodules the same thing as the category of left $R\otimes_\mathbb ZS^{op}$ modues even when they don't have identity? Doesn't that reduce it to the case of ordinary modules over rings (possibly without identity)? – rschwieb Apr 16 '19 at 14:47

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The answer is "obviously" yes, and there are a bunch of ways to see it.

I'll sketch a few of them, getting into more or less detail.

1) Universal algebra : an $(R,S)$-bimodule is just a specific type of algebra (in the sense of universal algebra) defined by equations, therefore the category of $(R,S)$-bimodules is a variety, and as such it has all free objects.

2) Category theory : It's quite easy to see that the forgetful functor from the category of $(R,S)$-bimodules to the category of sets preserves limits (just check products and equalizers), and satisfies the solution-set condition (you can't generate much more than $\aleph_0 |R||S||X|$ elements starting with a set of $X$ elements), therefore it has a left adjoint, therefore the category has all free objects

3) Ring theory : First of all, as pointed out in the comments and in my answer to Do categories of bimodules have enough projectives?, an $(R,S)$-bimodule is the same as a left $R\otimes_\mathbb{Z} S^{op}$-module, thus we can just worry about left $R$-modules for a general nonunital ring $R$. Then a free module on $x$ should have $x$, $x+x =: 2x$, $3x$ etc. and $rx$ for all $r\in R$ with no more relations. So you should get $nx +rx$ for $n\in \mathbb Z, r\in R$.

This can make us think of the general way to embed any ring in a unital one : $\mathbb{Z}\times R$ with usual addition, and as product $(n,r)(m,s) = (nm,ns+mr+rs)$ where $ns$ and $mr$ are not to be understood as products in $R$ but as the obvious action of $\mathbb{Z}$ on $R$.

Then this new ring (call it $\tilde{R}$) is unital (check that $(1,0)$ is a unit) and $R$ embeds in it via $r\mapsto (0,r)$. Thus it is an $R$-module, and I claim it is free on $(1,0)$. Indeed, let $M$ be a left $R$ -module and let $x\in M$. Then let $f:\tilde{R}\to M$ be any $R$-morphism with $(1,0)\mapsto x$. Then $(n,0)=(1,0)+...+(1,0)\mapsto n\cdot x$ and $(0,r) = (0,r)(1,0)= r\cdot (1,0) \mapsto r\cdot x$, therefore $(n,r) \mapsto n\cdot x + r\cdot x$; thus this $R$-morphism, if it exists, is unique.

Now define $f(n,r) = n\cdot x + r\cdot x$. This is a well-defined application, and it satisfies $f(1,0) = x$. Moreover it is very clearly a group morphism. Finally, $r\cdot (m,s) = (0,r)(m,s) = (0,mr+rs)$ and $f(r\cdot (m,s)) = f(0,mr+rs) = (mr+rs)\cdot x= (mr)\cdot x + r\cdot (s\cdot x) = (r+...+r)\cdot x + r\cdot (s\cdot x) = r\cdot x +... + r\cdot x + r\cdot (s\cdot x) = r\cdot (x+...+x) + r\cdot (s\cdot x) = r\cdot (m\cdot x)+ r\cdot (s\cdot x) = r\cdot (m\cdot x + s\cdot x) = r\cdot f(m,s)$, thus $f$ is also an $R$-morphism, which proves the existence; and concludes the proof that $\tilde{R}$ is free on $(1,0)$.

Maxime Ramzi
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  • I was under the impression that only when $R$ and $S$ are unital rings is an $(R,S)$-bimodule the same as a left $R \otimes_\mathbb{Z} S^{op}$-module. Is there a reason this applies also to rings without unit? – user02262019 Apr 16 '19 at 15:17
  • Well does the proof use neutral elements at any point ? An $(R,S)$-bimodule is just a pair of morphisms $R\to End(M), S^{op}\to End(M)$ that are compatible, in that the obvious map $R\times S^{op}\to End(M)$ is bilinear, which means precisely that it factors through $R\otimes S^{op}$. – Maxime Ramzi Apr 16 '19 at 15:41