MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Given a noncommutative ring $R$, and two (left) $R$-modules $M$ and $N$, how does one define a left action on the the vector space tensor product $M \otimes N$? Multiplying on just the first factor of the tensor product seems a little unnatrual, but I can't see what else to do.

share|cite|improve this question
You usually do this when $R$ also has a coalgebra structure $R \to R \otimes R$. – Angelo Sep 22 '12 at 14:30
What do you mean by the tensor product of two left R-modules? Normally when R is noncommutative, you need a left R-module and a right R-module, and then the tensor product doesn't have an R-structure. – anon Sep 22 '12 at 14:30
With a definition like $r \triangleright (m \otimes n) = \sum_i r_i m \otimes r'_i n$, where $\Delta(r) = \sum_i r_i \otimes r_i'$? – Ago Szekeres Sep 22 '12 at 14:39
I'd agree with Angelo, you need an algebra map $R \to R\otimes R$. Without more information on $R$ the only two such maps are $r \mapsto r \otimes 1$ and $r \mapsto 1 \otimes r$, which make $M \otimes N$ into a left module by multiplying the $M$ or $N$ factor, respectively. This usually isn't a reasonable thing to do, I think. – Peter Samuelson Sep 22 '12 at 16:08
@Ago: because both constructions ignore the fact that one of $M$ and $N$ is also a module. What do you mean by "vector space tensor product" if $R$ is not an algebra over a field? – Qiaochu Yuan Sep 22 '12 at 18:47
up vote 13 down vote accepted

Let $R, S$ be two (unital and associative to be safe) algebras over a commutative ring $k$ and let $M, N$ be respectively a left $R$-module and a left $S$-module. Then we can define the tensor product $M \otimes_k N$ by the usual universal property, and it is naturally a left $R \otimes_k S$-module by functoriality. If $M, N$ are both left $R$-modules, then $M \otimes_k N$ is a left $R \otimes_k R$-module, so defining a left $R$-module structure on the tensor product which is compatible with its $k$-module structure is tantamount to giving a morphism of $k$-algebras

$$\Delta : R \to R \otimes_k R.$$

Note that any such morphism gives a left $R$-module structure to the tensor product of two left $R$-modules, but it won't behave the way you expect it to with respect to multiple tensor products unless $\Delta$ is coassociative and cocommutative. I will be ignoring this.

Now, the ring $R \otimes_k R$, by its universal property, naturally admits two morphisms $R \to R \otimes_k R$, namely the two inclusions $i_1, i_2 : R \to R \otimes_k R$, which correspond to the two $R$-module structures described by Peter Samuelson in the comments. If you don't want to just provide $\Delta$ as extra data, then your question can be reinterpreted as follows:

What other natural morphisms $R \to R \otimes_k R$ are there?

Suppose that $\Delta_R : R \to R \otimes_k R$ is a family of morphisms which is natural in $R$; that is, it defines a natural transformation. Applying the forgetful functor to $\text{Set}$, we get a natural transformation from the forgetful functor $U : k\text{-Alg} \to \text{Set}$ to the functor

$$U^2 : k\text{-Alg} \ni R \mapsto U(R \otimes_k R) \in \text{Set}.$$

The functor $U$ is representable by the $k$-algebra $k[x]$, so by the Yoneda lemma, natural transformations $U \to U^2$ can be naturally identified with elements of

$$U^2(k[x]) \cong U(k[x, y]).$$

More concretely, any polynomial in two variables $f(x, y) = \sum f_{ij} x^i y^j \in k[x, y]$ over $k$ induces a natural map of sets

$$R \ni r \mapsto \sum f_{ij} r^i \otimes r^j \in R \otimes_k R$$

and these are the only such natural maps. Now the question reduces to the following:

Which of the maps $r \mapsto \sum f_{ij} r^i \otimes r^j$ are always morphisms of $k$-algebras?

The short answer is that it depends a lot on $k$. For starters, letting $R = k[x], r = x, R \otimes_k R = k[x, y]$ and requiring compatibility with scalar multiplication gives

$$cr \mapsto c f(x, y) = f(cx, cy) \in k[x, y]$$

for all $c \in k$. Comparing coefficients of $x^i y^j$ gives

$$(c - c^{i+j}) f_{ij} = 0$$

for all $i, j$ and for all $c$. This suggests the possibility of other natural morphisms $R \to R \otimes_k R$ in characteristic $p$. In particular, if $k = \mathbb{F}_p$ then we have $c^{p^r} = c$ for all $r \ge 0$, hence $f$ can have components of degree $p^r$ and some of these give rise to genuine natural morphisms such as

$$R \ni r \mapsto r^p \otimes 1 \in R \otimes_k R.$$

Assume for simplicity that $k$ is an integral domain of characteristic $0$ (these hypotheses can probably be considerably weakened). Then the above condition implies that $f$ is homogeneous and linear, so our desired morphism can only have the form

$$R \ni r \mapsto f_{10} r \otimes 1 + f_{01} 1 \otimes r \in R \otimes_k R.$$

Compatibility with multiplication now requires (taking $R = k[x, y], R \otimes_k R = k[x, y, z, w]$)

$$xy \mapsto f_{10} xy + f_{01} zw = (f_{10} x + f_{01} z)(f_{10} y + f_{01} w)$$

and comparing coefficients gives

$$f_{10}^2 = f_{10}, f_{01}^2 = f_{01}, f_{10} f_{01} = 0.$$

Our hypotheses then imply that exactly one of $f_{10}$ or $f_{01}$ can be equal to $1$ and the other must be equal to $0$. (If $k$ is not an integral domain then other things can happen; in general a choice of $f_{01}, f_{10}$ above is equivalent to a direct product decomposition $k = k_1 \times k_2$ with respect to which $f_{10} = (1, 0), f_{01} = (0, 1)$.) In other words, when $k$ is an integral domain of characteristic $0$ then the only natural morphisms $R \to R \otimes_k R$ are

$$R \ni r \mapsto r \otimes 1 \in R \otimes_k R$$


$$R \ni r \mapsto 1 \otimes r \in R \otimes_k R$$

and the corresponding natural left $R$-module structures on $M \otimes_k N$ are given by multiplication on the first resp. the second factor. In particular,

If $k$ is a field of characteristic $0$, then the only natural morphisms $R \to R \otimes_k R$ are the two obvious ones.

Group algebras are highly misleading here: any group algebra $k[G]$ (more generally any monoid algebra $k[M]$) is naturally equipped with a comultiplication given by extending

$$\Delta : k[G] \ni g \mapsto g \otimes g \in k[G] \otimes_k k[G]$$

and $\Delta$ induces the usual tensor product of representations of $G$ over $k$. It is worth mentioning that this map is not defined using any of the structure maps of the group $G$; more generally, for any set $S$ there is a canonical diagonal map $\Delta : S \to S \times S$ given by sending $s$ to $(s, s)$, and this induces a canonical coalgebra structure on the free module $k[S]$ given by extending

$$k[S] \ni s \mapsto s \otimes s \in k[S] \otimes_k k[S].$$

But a generic $k$-algebra is not a free vector space over any distinguished set.

share|cite|improve this answer
Well put: explaining, in effect, what "reasonable" should be, which was part of the question's "problem", and then decisively answering. – paul garrett Sep 22 '12 at 21:56
It's nice to see that the intuition that the two obvious morphisms $R \to R\otimes_k R$ are the only "natural" ones (in characteristic 0) can be formalized by removing the quotation marks :) – Peter Samuelson Sep 23 '12 at 4:35

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.