A definition of "algebra" (that is, an associative algebra, in the sense of ring theory) generally requires a commutative base ring. But there are cases where it's reasonable to consider algebras over general unital rings.

For example, suppose $R$ is a commutative ring and consider $R^2$. If we multiply ordered pairs coordinatewise, $R^2$ has an obvious $R$-algebra structure: it's an $R$-module in which the abelian group structure is extended to a ring whose multiplication is bilinear, or equivalently (*) $r \cdot \alpha \beta = (r \cdot \alpha) \beta = \alpha (r \cdot \beta)$ (writing $\cdot$ for the $R$-module action).

Now let $R$ be a general unital ring. $R^2$ is a left $R$-module, but the ring multiplication is not generally $R$-bilinear because (*) fails: $r \cdot \alpha \beta \neq \alpha (r \cdot \beta)$. So the definition of an algebra suggested above does not work. On the other hand, $R^2$ does have a ring structure and a module structure (on each side), and those structures seem compatible, so there should be some good way to refer to $R^2$ as an algebra over $R$.

So here's a definition: a "two-sided algebra" $M$ over a ring $R$ is a bimodule $_R M _R$ (including the requirement that $(r \cdot \alpha) \cdot s = r \cdot (\alpha \cdot s)$) with a ring structure on the abelian group such that $r \cdot \alpha \beta = (r \cdot \alpha) \beta$ and $\alpha \beta \cdot r = \alpha (\beta \cdot r)$, for all $\alpha, \beta \in M, r \in R$. (The term "bialgebra" would be nicer but seems well established with a different meaning.)

There's an obvious two-sided algebra structure on $R^2$. If the left and right actions in a two-sided algebra $M$ are the same, then $r \cdot \alpha \beta = \alpha \beta \cdot r = \alpha(\beta \cdot r) = \alpha(r \cdot \beta)$, so (*) is satisfied and the ring multiplication on $M$ is bilinear. So in the case of $R^2$, the two-sided algebra structure over an arbitrary ring generalizes the algebra structure over a commutative ring.

Other good examples are general product rings $R^I$, matrix rings $M_n(R)$, and polynomial rings $R[x]$. The two-sided algebras over $R$ (which we could call $_R \text{Alg}_R$) seem to form a category much like $R$-algebras in the commutative case: it has finite products, an initial object $R$, and a terminal object $0$.

Finally, my questions:

  1. Is the above correct?

  2. Are there any other approaches to this problem?

  • $\begingroup$ You forgot to specify a ring structure on the underlying abelian group on $M$. $\endgroup$ Commented Jun 28, 2014 at 16:17

2 Answers 2


Algebra objects or monoid objects can be defined in any monoidal category. When $R$ is a commutative ring, then the category of left $R$-modules has a monoidal structure given by $\otimes_R$, and algebras in that category coincide with $R$-algebras. But the category of left $R$-modules has no "natural" monoidal structure when $R$ is not commutative - this is the conceptional reason why one usually assumes $R$ to be commutative (of course it has lots of monoidal structures, for example the one given by the categorical product, but this has no interesting algebras).

However, if $R$ is any ring, then the category of $R$-bimodules is a monoidal category (although not symmetric monoidal ...). According to the general definition, an algebra in that monoidal category is an $R$-bimodule $_R A _R$ equipped with a homomorphism $\mu : {}_R A \otimes_R A _R \to {}_R A _R$ of $R$-bimodules and a homomorphism $\eta : {}_R R_R \to {}_R A _R$ of $R$-bimodules such that certain diagrams commute. Equivalently, we have a ring structure $(U(A),*,1)$ on the underlying abelian group $U(A)$ of $A$ such that for all $r \in R$ and $a,b \in A$:

  • $r \cdot (a * b) = (r \cdot a) * b$
  • $(a * b) \cdot r = a * (b \cdot r)$
  • $a * (r \cdot b) = (a \cdot r) * b$
  • $r \cdot 1 = 1 \cdot r$

The last two axioms are missing in your definition. We obtain a variety of algebras in the sense of universal algebra and hence limits & colimits exist.

Notice that we really need the axiom $r \cdot 1 = 1 \cdot r$ in order to ensure that $R$ is the initial $R$-algebra.

Caution. If $S$ is a commutative ring with underlying noncommutative ring $R$ (usually, one writes $S=R$, which is confusing here), then any $S$-algebra yields an $R$-algebra, but not every $R$-algebra is of this form! In fact, we only get those $R$-algebras for which $r \cdot a = a \cdot r$ holds.

See MO/21899 for a discussion about definitions of algebras over non-commutative rings.

  • 1
    $\begingroup$ I'm still trying to digest the monoidal category concept, but the list of axioms looks like what I need, thank you. Can we say more about what a "natural" monoidal structure is, and why the bimodules have one but the left modules don't? $\endgroup$
    – Hew Wolff
    Commented Jul 9, 2014 at 2:31
  • $\begingroup$ Doesn't the last axiom follow from the third? $$r \cdot 1 = 1 \ast (r \cdot 1) = (1 \cdot r) \ast 1 = 1 \cdot r$$ $\endgroup$
    – seldon
    Commented Sep 9, 2021 at 9:05
  • 1
    $\begingroup$ It seems that you are right. $\endgroup$ Commented Sep 9, 2021 at 12:13

There is a simpler definition which is equivalent to the category of monoid objects in the category of $R$-bimodules. nLab calls them rings over $R$, and describes them with the "coslice category" or "under category" $R/\operatorname{Ring}$. An object in this category is a ring homomorphism $h: R \to A$ for some $A$, and a morphism from $h$ to $h': R \to A'$ is a ring homomorphism $A \to A'$ making the triangle commute.

Given a monoid object $A$ in the category of $R$-bimodules, define a corresponding object $h \in R/\operatorname{Ring}$ by $h(r) = r \cdot 1_A$. In the other direction, given $h$, define an $R$-bimodule structure on $A$ by $r \cdot a = h(r)a$, $a \cdot r = ah(r)$. The details are tedious, but these mappings extend easily to functors which give an isomorphism of categories.


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