Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Suppose that $M$ and $M'$ are $\mathbb{K}$-algebras where $\mathbb{K}$ is a field.

Now suppose that $N$ is an $M$-module and $N'$ is an $M'$-module, then I can view both of them as $\mathbb{K}$-modules which gives me the abelian group $N\otimes_\mathbb{K} N'$ .

I want to show that $N\otimes_\mathbb{K} N'$ is an $M \otimes_\mathbb{K} M'$-module.

The operation I considered taking is $n \otimes n' \cdot m \otimes m'= mn \otimes n'm' $, but I have two questions :

1- I can't figure out how to prove that this operation is well-defined .

2- I am having a hard time proving all the axioms; so maybe my operation is wrong to begin with.

Any help would be greatly appreciated !

share|improve this question
    
Two questions: are $M$ and $M'$ commutative? Do you know the universal property of tensor product? –  Kevin Carlson Jan 23 at 13:41
    
@KevinCarlson I believe that we can give $M$ and $M'$ the structures of either right or left $\mathbb{K}$-modules right? –  CeCe Jan 23 at 13:45
    
@KevinCarlson Yeah I do. Employing it on the other hand... I just need a push in the right direction to get me started. –  CeCe Jan 23 at 13:48
1  
Of course, since right and left $\mathbb{K}$-modules are the same thing. I was concerned about right and left in your structure of $N$ as an $M$-module. –  Kevin Carlson Jan 23 at 13:48
1  
Never mind: from your notation I take in that $N$ is a left $M$-module and $N'$ is a right $M'$-module. –  Kevin Carlson Jan 23 at 13:51

3 Answers 3

up vote 3 down vote accepted

Please let us change the notation. $K$ is an arbitrary commutative ring, $A,A'$ are $K$-algebras, $M$ is a left $A$-module and $M'$ is a left $A'$-module. Since the scalar multiplication $A \times M \to M$ is $K$-bilinear, it lifts to a $K$-linear map $\alpha : A \otimes_K M \to M$. Define $\alpha' : A' \otimes_K M' \to M'$ in the same way. Then we can endow $M \otimes_K M'$ with the structure of a left $A \otimes_K A'$-module in a canonical way via $$(A \otimes_K A') \times (M \otimes_K M') \xrightarrow{\otimes} (A \otimes_K A') \otimes_K (M \otimes_K M')$$ $$\cong (A \otimes_K M) \otimes_K (A' \otimes_K M') \xrightarrow{\alpha \otimes \alpha'} M \otimes_K M'$$

In other words, we have $$(a \otimes a') \cdot (m \otimes m') = a \cdot m \otimes a' \cdot m'$$ by definition. Notice that we don't have to prove well-definedness since we have constructed the action from $A \otimes_K A'$ by using the general associativity and symmetry isomorphisms for tensor products above. There is no need to repeat them in this special case, although this is certainly possible, but cumbersome.

Now let us check the module axioms. The action is already $K$-bilinear by construction, so that we only have to prove $1 \cdot x=x$ and $r \cdot (s \cdot x)=(r \cdot s) \cdot x$ for $x \in M \otimes_K M'$ and $r,s \in A \otimes_K A'$. We may assume $x = m \otimes m'$ and $r = a \otimes a'$, $s = b \otimes b'$ are pure tensors (because they generate all tensors), but then the equations follow easily from the corresponding module axioms for $M$ and $M'$: $$1 \cdot x = (1 \otimes 1) \cdot (m \otimes m') = 1 \cdot m \otimes 1 \cdot m' = m \otimes m'= x.$$ $$r \cdot (s \cdot x) = (a \otimes a') \cdot (b \cdot m \otimes b' \cdot m') = a \cdot (b \cdot m) \otimes a'\cdot (b' \cdot m) $$ $$= (a \cdot b) \cdot m \otimes (a'\cdot b') \cdot m = (a\cdot b \otimes a'\cdot b') \cdot (m \otimes m') = (r \cdot s) \cdot x.$$

Everything here works in an arbitrary symmetry monoidal category with two monoid objects $A,A'$ and two left module objects $M,M'$ over them (and this is also the reason why in reality no calculations are necessary at all). For example, if $G,G'$ are two groups and which act on two sets $M,M'$ (from the left), then $G \times G'$ acts on $M \times M'$ via $(g,g') \cdot (m,m') := (g \cdot m,g' \cdot m')$.

share|improve this answer
    
Might need some adjustment to fit the OP's suggestion that one of them is a right module. –  rschwieb Jan 23 at 14:54
    
The construction doesn't work if, say, $M$ is a right $A$-module. Or let's say then $M \otimes M'$ is a left $A^{op} \otimes A'$-module, or equivalently an $(A',A)$-bimodule. –  Martin Brandenburg Jan 23 at 15:01
    
@MartinBrandenburg Thanks a billion. I have a tiny question though, the isomorphism $(A \otimes_K A') \otimes_K (M \otimes_K M') \cong (A \otimes_K M) \otimes_K (A' \otimes_K M')$? is the one that takes $(a\otimes a') \otimes (m \otimes m') \rightarrow (a \otimes m)\otimes (a' \otimes m')$, right? I ask cause I'm not quite familiar with tensor of tensors ! –  CeCe Jan 23 at 17:25
1  
Yes. The isomorphism follows from the general isomorphism 1) $A \otimes (B \otimes C) \cong (A \otimes B) \otimes C$ (associator), 2) $A \otimes B \cong B \otimes A$ (symmetry). Apply this a couple of times. –  Martin Brandenburg Jan 23 at 17:27
    
@MartinBrandenburg I have one more question if that's ok. If i suppose that $M$ is projective; is it true then that $M \otimes_K M'$ is isomorphic to a direct sum of copies of $M'$? –  CeCe Jan 26 at 4:00

I think it' a little strange to use "M" for algebras, so I hope you don't mind if I shift these letters around a bit.

Promoting $M$ and $M'$'s module structures to a module structure on $M\otimes_K M'$

If you've got a left $A$ module $M$ and a right $A'$ module $M'$, then $M\otimes_KM'$ does have a natural left $A$ module structure, and a right $A'$ module structure, and in fact these two operations are compatible so that $M\otimes_K M'$ is rather an $A,A'$ bimodule. The operation you suggested does indeed establish this structure.

The only catch is that this describes a left $A\otimes_K (A')^{op}$ module structure on $M\otimes_K M'$ (notice the $^{op}$, denoting the opposite ring!) The $A,A'$ bimodule structure and the left $A\otimes_K (A')^{op}$ module structure are equivalent. (Let me suppress $K$ subscripts on tensors from now on.)

Making the suggested operation work

One strategy is to translate this to terms of algebra homomorphisms and exploit the universal properties of the tensor product of modules and of algebras.

The initial left $A$ module structure on $M$ is captured by an algebra homomorphism $\phi:A\to End_K(M)$, and the right $A'$ module structure on $M'$ is captured by an algebra homomorphism of $\psi:(A')^{op}\to End_K(M')$. The operations are retrieved this way: $a\cdot m :=\phi(a)(m)$, and $m'\cdot a':=\psi(a')(m')$.

You can combine these to operate on $M\otimes M'$ by starting with $(a,a')\cdot (m,m'):=(\phi(a)(m),\psi(a')(m'))$. This is bilinear in $m$ and $m'$, so the universal property of $M\otimes M'$ extends this to a linear map $(a,a'):M\otimes M'\to M\otimes M'$. If you decode the above $\phi,\psi$ in terms of module products, you'll see it is what you suggested.

This gives us a linear endomorphism of $M\otimes M'$ for every pair of elements in $A\times (A')^{op}$. In particular, verify that $a\mapsto (a, 1)$ and $a'\mapsto (1,a')$ provide algebra homomorphisms of (respectively) $A\to End(M\otimes M')$ and $(A')^{op}\to End(M\otimes M')$, and moreover $(a,1)(1,a')=(1,a')(a,1)$, so that the images of these two maps commute. This satisfies the universal property of the tensor product of algebras, so that the assignment is promoted into an algebra homomorphism of $A\otimes (A')^{op}\to End(M\otimes M')$.

share|improve this answer

This answer will only be for $M,M'$ commutative, so is in a couple ways less useful than others, but I'll leave it since I've already written it. Recall the universal property of the tensor product of commutative $\mathbb{K}$-algebras: the $\mathbb{K}$-algebra homomorphisms $M\otimes M'\to A$ are in one-to-one correspondence with the pairs of $\mathbb{K}$-algebra homomorphisms $(M\to A,M'\to A)$. What that really means is that we can define a map out of $M\otimes M'$ just by defining it on $M$ and $M'$.

A nice way to rephrase the axioms of a module that you might or might not have seen is this: an $M$-module structure on $N$ is given by a $\mathbb{K}$-algebra homomorphism from $M$ to the endomorphisms of $N$, that is, the $\mathbb{K}$-module homomorphisms $N\to N$. So the upshot is that if we want an $M\otimes M'$-module structure on $N\otimes N'$, what we need is nothing more or less than an $M$-module and an $M'$-module structure on $N\otimes N'$.

OK, let's make $N\otimes N'$ an $M$-module. So for each $m$ we need a homomorphism which we'll also call $m$ taking $N\otimes N'$ to $N \otimes N'$. How about $m(n\otimes n')=mn\otimes n'$? This is what you already suggested, and the obvious choice, when we extend by linearity to a general element $\sum n_k\otimes n'_k$ of $N\otimes N'$. We get that this describes an algebra homomorphism out of $M$ from the fact that $N$ is an $M$-module and the properties of the tensor product; for instance $$(m_1+m_2)(n\otimes n')=(m_1+m_2)n\otimes n'=m_1n\otimes n'+m_2n\otimes n'=$$ $$=m_1(n\otimes n')+m_2(n\otimes n')$$ But we've got to check that the multiplication by a given $m$ is actually an endomorphism of $N\otimes N'$. The only part of this that's not as easy as the previous calculation is showing it's well-defined. There's a special way out and a general way out, here.

The special way is to observe that $N$ and $N'$ are $\mathbb{K}$-modules, that is, vector spaces, so that $N\otimes N'$ has a basis $\{n_i\otimes n'_j\}$ where $\{n_i\}$ is a basis of $N$ and $\{n'_j\}$ is a basis of $N'$. Then each element of $N\otimes N'$ has a unique representation over this basis and there are no concerns about well-definedness at all.

If you want to be ready for working with modules over a general ring, one choice is to work explicitly with two different representations of the same element of $N\otimes N'$ (even though we've observed that in this case there never are such different representations) and show that $m$ acts the same on them. In other words, show $m$ sends all the relations $n\otimes n'-n\otimes n'$ to zero. This is inelegant and can get very confusing, so the other option is to use the universal property again and say that a map $N\otimes N'\to N\otimes N'$ is just a $\mathbb{K}$-bilinear map $N\times N'\to N\otimes N'$. Then instead of defining the action $m$ on $N\otimes N'$ we get to define it on $N\times N'$, namely as $m(n,n')=mn\otimes n'$. Bilinearity is then as quick as when we were checking properties of $m_1+m_2$ earlier on.

So, whether you used the special properties of $\mathbb{K}$-modules or worked with the universal property, you've rigorously established the $M$-module structure on $N\otimes N'$. Do the same for $N',$ and you're done!

share|improve this answer
    
1. The universal property in the first paragraph is wrong. You either need an additional assumption that $M \to A$ and $M' \to A$ commute in a certain sense, or work with commutative $M,M',A$ in the first place. 2. Your definition of the endomorphism is incredibly cumbersome. One has to work with the universal properties in order to avoid this. The universal property is not "some property" of the tensor product, but the defining property of the tensor product. I'm sure you know this, but we also have to make sure that beginners don't think that it is a good idea to define maps on elements .. –  Martin Brandenburg Jan 23 at 14:21
    
Thanks for the comments. I wrote that universal property when I thought the OP was assuming $M,M'$ commutative, then removed that assumption without fixing it-I'd better just revert. I agree that I shouldn't have tried to stick with the elementary (pun intended) definition of the action, as compared to your answer it surely doesn't clear anything up. –  Kevin Carlson Jan 23 at 15:14
    
It isn't enough to assume that $M,M'$ are commutative, you need also that $A$ is commutative, or (see rschwieb's answer) that the two homomorphisms commute. In the end you want to apply this to $A$ an endomorphism ring, highly noncommutative. –  Martin Brandenburg Jan 23 at 17:28

Your Answer

 
discard

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.