# For $M\otimes_R N$, why is it so imperative that $M$ be a right $R$-module and $N$ a left $R$-module?

I was reading about the construction of a tensor product of any right $R$-module $M$ and left $R$-module $N$ over a ring $R$.

Why is it required that $M$ and $N$ be right and left modules, respectively? How does the construction not work otherwise?

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Well, if they were both left modules, say, you'd get stuff like $r(sx)\otimes y = sx \otimes ry = x \otimes s(ry) = x \otimes (sr)y = (sr)x \otimes y$, which is problematic if $R$ is not commutative. – Omar Antolín-Camarena Mar 8 '12 at 4:19

Try tensoring a left $R$-module $M$ with a left $R$-module $N$. This tensor product should be comprised of "tensors" $m\otimes n$ (sorry for denoting them by the same notation as usual tensors; they don't really deserve this) satisfying $rm\otimes n = m\otimes rn$ for all $r\in R$.

Now, for any $r,s\in R$, we must have $rsm\otimes n=sm\otimes rn=m\otimes srn$, but at the same time $rsm\otimes n=m\otimes rsn$ (because we can move the $rs$ as a whole across the tensor sign). Thus, $m\otimes srn=m\otimes rsn$. In other words, $m\otimes\left(rs-sr\right)n=0$. Similarly, $\left(rs-sr\right)m\otimes n=0$.

But this means that our fake tensor product $M\otimes N$ does not depend on the left $R$-modules $M$ and $N$, but only on the left $R / \left(\mathrm{Ab} R\right)$-modules $M / \left(\left(\mathrm{Ab} R\right) M\right)$ and $N / \left(\left(\mathrm{Ab} R\right) N\right)$, where $\mathrm{Ab} R$ is the commutator ideal of $R$ (that is, the ideal generated by all differences of the form $rs-sr$ with $r,s\in R$). And it is actually exactly the tensor product of these two $R / \left(\mathrm{Ab} R\right)$-modules over the commutative ring $R / \left(\mathrm{Ab} R\right)$. This tensor product, of course, can be reinterpreted as a tensor product of a left module with a right module (since over a commutative ring, modules can be switched from left to right at will).

So the notion of the tensor product of a right module with a left module is more reasonable and completely encompasses the notion of a "tensor product" of two left modules.

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So unless $M$ is an $(R,R)$-bimodule, it only makes sense to have the relation $mr\otimes n=m\otimes rn$ one the construction is finished? – Tents Mar 8 '12 at 4:32
I don't understand this question, sorry. – darij grinberg Mar 8 '12 at 22:35

Abstractly, the tensor product is the construction such that there is a natural isomorphism $\newcommand{\op}{\textrm{op}} \newcommand{\Hom}{\textrm{Hom}}$ $$\Hom_{(S, T^\op)}(A \otimes_R B, C) \cong \Hom_{(S, R^\op)}(A, \Hom_{T^\op}(B, C))$$ where $A$ is a $(S, R^\op)$-bimodule, $B$ is a $(R, T^\op)$-bimodule, and $C$ is a $(S, T^\op)$-bimodule. (You can take $S = T = \mathbb{Z}$ if this helps. I write $(S, R^\op)$-bimodule to mean an object which has a left $S$-action and a right $R$-action simultaneously in a compatible way.)

Now, although I will use language suitable for bimodules here, what I say is also applicable to:

• sets with a monoid or group action,
• bifunctors $\mathcal{A} \times \mathcal{B}^\op \to \textbf{Set}$, where $\mathcal{A}$ and $\mathcal{B}$ are small categories, and more generally,
• bifunctors of categories enriched over a suitable monoidal category.

Firstly, let us examine the hom objects, since tensor products are defined in terms of them. If $A$ and $B$ are left $R$-modules, then they are also abelian groups, and there is a natural monomorphism $$\Hom_R(A, B) \hookrightarrow \Hom(A, B)$$ Indeed, observe that the $R$-actions of $B$ and $A$ (respectively) induce a $(R, R^\op)$-bimodule structure on $\Hom(A, B)$, as follows: if $f : A \to B$ is a homomorphism of abelian groups, we define $r \cdot f \cdot s : A \to B$ by $$(r \cdot f \cdot s)(m) = r f(s m)$$ It is easily checked that this makes $\Hom(A, B)$ into a $(R, R^\textrm{op})$-bimodule. But what is $\Hom_R(A, B)$? It is the subgroup of those homomorphisms $f : A \to B$ such that $$r \cdot f \cdot 1 = 1 \cdot f \cdot r$$ Thus, we get an equaliser diagram of the form $$\Hom_R(A, B) \longrightarrow R \otimes \Hom(A, B) \mathrel{\rlap{\lower{0.5ex}{\longrightarrow}}\raise{0.5ex}{\longrightarrow}} \Hom(A, B)$$ where the upper map is given by the left $R$-action of $A$: $$r \otimes f \mapsto 1 \cdot f \cdot r$$ and the lower map is given by the left $R$-action of $B$: $$r \otimes f \mapsto r \cdot f \cdot 1$$

More generally, we need to form a special kind of limit called an ‘end’: $$\Hom_R(A, B) = \int_R \Hom(A, B)$$ In the case of two functors $A, B : \mathcal{R} \to \textbf{Set}$, this amounts to the well-known formula $$\textrm{Nat}(A, B) \cong \int_{r : \mathcal{R}} \textrm{Hom}(A r, B r)$$

Now, what about tensor products? They are formed in an essentially dual fashion. We take it for granted that there is a tensor product of abelian groups satisfying $$\Hom(A \otimes B, C) \cong \Hom(A, \Hom(B, C))$$ and we want to understand how to get the corresponding constructions over $R$. Let $A, B , C$ be bimodules as in the first paragraph. First, consider the hom-object $\Hom(A, \Hom_{T^\op}(B, C))$. Because $\Hom$ is left exact in the second variable, we can extract the end to obtain $$\int_{T^\op} \Hom(A, \Hom(B, C)) \cong \Hom(A, \Hom_{T^\op}(B, C))$$ Ends commmute with ends, so we can replace the outer $\Hom$ with $\Hom_{(S, R^\op)}$ as follows: $$\int_S \int_{R^\op} \int_{T^\op} \Hom(A, \Hom(B, C)) \cong \Hom_{(S, R^\op)}(A, \Hom_{T^\op}(B, C))$$ We have assumed that there is a tensor-hom adjunction for abelian groups, so we can replace the integrand on the left hand side to obtain $$\int_S \int_{R^\op} \int_{T^\op} \Hom(A \otimes B, C) \cong \Hom_{(S, R^\op)}(A, \Hom_{T^\op}(B, C))$$ and we know that $$\Hom_{(S, T^\op)}(A \otimes B, C) \cong \int_S \int_{T^\op} \Hom(A \otimes B, C)$$ so we obtain $$\int_{R^\op} \Hom_{(S, T^\op)}(A \otimes B, C) \cong \Hom_{(S, R^\op)}(A, \Hom_{T^\op}(B, C))$$ and now, applying the Yoneda lemma, we may push the end into the first argument of the integrand to get the coend formula for the tensor product: $$A \otimes_R B = \int^R A \otimes B$$ Unpacking the definition of coend, this means there is a coequaliser diagram of the form $$A \otimes R \otimes B \mathrel{\rlap{\lower{0.5ex}{\longrightarrow}}\raise{0.5ex}{\longrightarrow}} A \otimes B \longrightarrow A \otimes_R B$$ where the upper arrow is the map given by the right $R$-action of $A$: $$a \otimes r \otimes b \mapsto a r \otimes b$$ and the lower arrow is the map given by the left $R$-action of $B$: $$a \otimes r \otimes b \mapsto a \otimes r b$$ Thus, we see that the requirement that $A$ be a right $R$-module and $B$ be a left $R$-module is forced on us by the fact that $\Hom(B, -)$ has a natural right $R$-module structure when $B$ is a left $R$-module!

Moreover, it is clear from this construction that $\Hom_R$ and $\otimes_R$ are just two dual ways of neutralising two opposite $R$-actions:

• $\Hom_R(B, C)$ extracts the universal $R$-equivariant subobject of $\Hom(B, C)$, while
• $A \otimes_R B$ extracts the universal $R$-equivariant quotient of $A \otimes B$.
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I don't understand the possible confusion. The tensor product of $M$ And $N$ is defined to be the object that takes $R$-biadditive maps $M\times N\to G$ to additive maps $M\otimes_R N\to G$. In other words, the entire function of a tensor product, its entire universal reason for being is to be a machine that turns $R$-biadditive maps into additive ones. With this said it's clear that the notion of $R$-biadditive maps are important, no? What are $R$-biadditive maps? They're maps $f:M\times N\to G$ (where $G$ is some abelian group) wiht the property that they are additive in each slot and are "balanced" in the sense that $f(mr,n)=f(m,rn)$ fora ll $r\in R$ and $(m,n)\in M\times N$. To even say this last sentence, to make sense of $f(mr,n)=f(m,rn)$ we need $M$ to be a right $R$-module and $N$ a left $R$-module.

Now, if $R$ happens to be commutative so that distinguishing between left and right modules is just a formality we can really take the tensor product of any two modules.

This all said, there is no problem in just taking to abelian groups $M$ and $N$ and forming the quotient $\mathbb{Z}[M\times N]/S$ where $S$ is the subgroup generated by making $\otimes:M\times N\to\mathbb{Z}[M\times N]$ bilinear. That said, you are secretly just computing the tensor product of two $\mathbb{Z}$-modules.

The moral of the story is that we need some kind of $R$ structure on $M$ and $N$ since tensor products were born out of a desire to transfer one $R$ property to another ($R$-bilinear to $R$-linear for commutative $R$).

EDIT: This is to address KCd's remarks. He is most likely right about me misinterpreting your question, for that I apologize.

So, why exactly can we not consider the tensor product of two left $R$-modules? Let's first see why the question that tensor products attempt to answer doesn't make sense if we consider two left $R$-modules, say. In this case it's clear that the analogue we are looking for is additive in each variable maps $f:M\times N\to R$ such that $f(rm,n)=f(m,rn)$. The issue with this that if $R$ is not commutative we start getting screwy things--things we probably didn't anticipate. For example,

$$f((rs)m,n)=f(r(sm),n)=f(sm,rn)=f(m,srn)$$

which if this is to be true for all $r,s$ strongly suggests that there is some kind of commutativity thing going on here. Of course, if our rings are commutative then everythings fine and we don't care if $M$ or $N$ are left or right $R$-modules, but if our ring isn't commutative then we seem like we might have issues. For example, a let's assume that our $M,N$ are actually algebras over $R$. A very natural map then is to consider the multiplication map $M\times M\to M$--our hope is that this would fit the kind of map we'd like to consider, but we see that if this map were "$R$-biadditive" in this new sense then we'd have $rs=(rs1)1=1(sr1)=sr$ for all $s,r$ so that $M$ would need to be commutative.

This all said, I guess you could theoretically make the same construction with both left $R$-modules, I guess the problem is what the construction would correspond to. It's easy to think of $R$-biadditive maps with a left and right $R$-modules (for example think about the map $\text{Mat}_{n\times m}(R)\times\text{Mat}_{m\times n}(R)\to \text{Mat}_n(R)$ given by matrix multiplication--this is is a $\text{Mat}_m(R)$-biadditive map) but I can't imagine (off the top of my head) where the kind of maps we'd be discussing when we have two left $R$-modules where $R$ is noncommuative (since any map would have to satisfy that weird "commutative condition")--and of course, if $R$ is commutative then thinking about them as left $R$ modules and doing it the way you suggest is the same as thinking about them as a left and right $R$-modules respectively.

Moral of the story: You could make such a construction, but why it would be meaningful is not at all clear.

About why anyone would care about biadditive maps I don't think I can beat the explanation given by Gowers here.

I hope that I was able to help a little bit.

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Concerning the first paragraph, it is not at all clear why a notion (like R-biadditive maps) is important just because it is related to a universal mapping problem. But the more significant misunderstanding in this answer is that you didn't highlight what goes wrong if $M$ and $N$ are, say, both left $R$-modules. My sense of the intent behind the question is why one should want to take $M$ and $N$ to be $R$-modules on different sides in general. – KCd Mar 8 '12 at 4:21
@KCd Thank you for pointing that out. I'm sorry that I initially misinterpreted things. I'm not as confident about my answer now that I have understood the question, but hopefully I was able to add something. P.S. Thanks for the great notes. – Alex Youcis Mar 8 '12 at 4:48