Tensor products of abelian groups
Often enough we’re going to see the following situation. There are three abelian groups — , , and — and a function that is linear in each variable. What does this mean?
Well, “linear” just means “preserves addition” as in a homomorphism, but I don’t mean that is a homomorphism from to . That would mean the following equation held:
Instead, I want to say that if we fix one variable, the remaining function of the other variable is a homomorphism. That is
we call such a function “bilinear”.
The tensor product is a construction we use to replace bilinear functions by linear functions. It is defined by yet another universal property: a tensor product of and is an abelian group and a bilinear function so that for every other bilinear function there is a unique linear function so that . Like all objects defined by universal properties, the tensor product is automatically unique up to isomorphism if it exists.
So here’s how to construct one. I claim that has a presentation by generators and relations as an abelian group. For generators take all elements of , and for relations take all the elements and for all and in and and in .
By the properties of such presentations, any function of the generators — of — defines a linear function on if and only if it satisfies the relations. That is, if we apply a function to each relation and get every time, then defines a unique function on the presented group. So what does that look like here?
So a bilinear function gives rise to a linear function , just as we want.
Usually we’ll write the tensor product of and as , and the required bilinear function as .
Now, why am I throwing this out now? Because we’ve already seen one example. It’s a bit trivial now, but catching it while it’s small will help see the relation to other things later. The distributive law for a ring says that multiplication is a bilinear function of the underlying abelian group! That is, we can view a ring as an abelian group equipped with a linear map for multiplication.
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The third condition in the definition of linearity follows from the first two… I think.
Comment by Oingo  December 8, 2007 
I think you mean the third condition in the definition of bilinearity. And I wasn’t asserting that it’s independent, but rather drawing the distinction between the previously defined condition of linearity and this condition of bilinearity.
Comment by John Armstrong  December 10, 2007 
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Pingback by The InclusionExclusion Principle « The Unapologetic Mathematician  July 24, 2008 
Hi, everyone:
This is my first time here. Great site.
I wonder what the general bilinear map t as above is ,
in the case where A=R^n and B=R^m , as Abelian groups,
i.e., what bilinear map would make the diagram:
R^n x R^m –t–> R^n (x) R^m

f
\/
etc. commute. Anyone know?
C
Comment by Bacile  March 10, 2009 
It’s pretty straightforward, Bacile. The universal bilinear map from to sends the pair to the tensor product .
Comment by John Armstrong  March 10, 2009 
O.K., say we are tensoring R^n and R^m as V.Spaces
over R
Am I wrong to assume that once we choose specific
V.Spaces to tensor , that (x) has to be a specific
map?.
Let me give an example for n=m=2, with the only
bilinear map I can think of : innerproduct in R^2 :
= xx’+yy’
This is a bilinear map from R^2xR^2 –>R
Now, we must find a linear map L: R^2(x)R^2 –>R
that makes the diagram commute, so that we must have:
L ( (x,y)(x)(x’,y’))= xx’+yy’ =
Now, in order for L to have this property, (x)
must be a specific map on (a,b)(x)(c,d) in R^2(x)R^2.
or do we just appeal to the (valid, I agree) argument
that shows the existence of this (x) making the diagram
commute.
I agree that (x) is an abstract universal bilinear map,
but I think that once we choose the V.Spaces to tensor,
(x) must become a specific map.
Maybe another way of looking at it is that we can
use the fact that R^2(x)R^2 ~ R^4 . Then it seems we
could use this isomorphism to pullback any map
R^2(x)R^2–>R into a map R^4–>R
Hope this is not too far out.
Anyway, thanks for any replies.
Comment by Bacile  March 11, 2009 
I just told you what specific map it was…
Okay, let’s try this. Say your vector space has a basis and has basis . Then you have a basis for given by . So the pair gets sent to , and we extend by bilinearity.
Say, for instance, we have the vectors and . Then the pair gets sent to . In your example, you’ve got the vectors and . As a pair, they get sent to the linear combination
Now, we can choose an isomorphism with the correspondence
Then the pair of vectors above is send to the vector
Comment by John Armstrong  March 11, 2009 
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[…] scalars and , and for any index . Equivalently, by the defining universal property of tensor products, this is equivalent to a linear function — a linear functional on . That is, the space of […]
Pingback by Multilinear Functionals « The Unapologetic Mathematician  October 22, 2009 
Hi everyone. I have a little problem… Let G be an abelian group and consider G\otimes_{\Z}\Q where \Q is the ring of rational numbers and \Z the one of integers. Take an element g\otimes q with g\in G and q\in \Q (it easily seen that any element of G\otimes_{\Z}\Q has this form).
Is it true that g\otimes q=0 if and ONLY IF g is a torsion element?
the if part is trivial but I have some truble with the other implication (in any case I’m quite sure it is true). Clearly you have to suppose q\neq 0.
Comment by Simone  October 28, 2009 
The if part isn’t even true. Let and even if is torsion the tensor isn’t zero.
Comment by John Armstrong  October 28, 2009 
of course it is! let n be the order of g, then
g\otimes 1=g\otimes n/n=ng\otimes 1/n= 0 \otimes 1/n = 0
If, instead of 1, you put a generic q in the previous line you have the proof of the if part…
Comment by Simone  October 28, 2009 
Furthermore it is not difficult to prove a more general statement:
Let G be a torsion abelian group and D a divisible abelian group, then G\otimes D=0.
Even this new statement can be generalized (it is true not only for abelian groups but for modules over a PID).
Comment by Simone  October 28, 2009 
Ah, yes, sorry. It’s been a long time since I’ve bothered with torsion elements.
Can you prove the contrapositive to handle the onlyif part?
Comment by John Armstrong  October 28, 2009 
FACT 1
Let G be a torsion free group, then G embedds into G\otimes Q. (G\otimes Q is the divisible (i.e. injective) envelope of G)
FACT 2
Let G be an abelian group, H<G, then H\otimes\Q is a subvector space of G\otimes\Q.
Using this two facts it is possible to prove the result… in fact if x is a torsion free element of G, then x generates a group isomorphic to Z. Thanks to fact 1
—>\otimes Q= Q is injective and thanks to fact 2 we have the inclusion
—>\otimes Q—>G\otimes Q.
This proves that x\otimes 1 is nonzero.
I think this can work but in any case it uses deeper results than needed… I really think that must exist a more elementary proof…
Comment by Simone  October 28, 2009 
I don’t know why but there is something missing in the message the maps were:
—> \otimes Q
and
—> \otimes Q—>G\otimes Q
Comment by Simone  October 28, 2009 
ok… I cannot write that simbol…. denot with (x) the subgroup generated by x…
(x) —> (x)\otimes Q
and
(x)—>(x) \otimes Q—>G\otimes Q
Comment by Simone  October 28, 2009 
In HTML, text contained within angle brackets is interpreted as markup. You have to write < and > to get < and >
And I don’t really see what your problem is with this proof. It gets right to the heart of the matter, that the behavior of elements under the action is entirely a matter of the subgroup they generate. Tensoring with kills torsion subgroups and preserves nontorsion subgroups.
A really nonelementary proof would use something like a structure theorem for abelian groups, which this doesn’t come close to using.
Comment by John Armstrong  October 28, 2009 
Here is the proof I was looking for! (I wrote it)
If $g$ is torsion, let $n$ be the order of $g$, then $g\otimes q=ng\otimes q/n=0$. On the other hand suppose, looking for a contradiction, that $g$ is not torsion, then $\left\langle g\right\rangle\cong \Z$ and so $G\otimes_{\Z}\Q\supseteq\left\langle g\right\rangle\otimes_{\Z}\Q\cong \Q\neq 0$. Let now $ag\otimes p$, with $p\in \Q$ and $a\in\N\setminus\{0\}$, be a nontrivial element of $\left\langle g\right\rangle\otimes_{\Z}\Q$ (we can suppose that it has this form), let $p=\frac{p_1}{p_2}$ and $q=\frac{q_1}{q_2}$ with $p_1,p_2,q_1,q_2\in\Z\setminus \{0\}$, then
$$ag\otimes p=ap_2q_1g\otimes \frac{1}{p_2q_2} \ \ \Rightarrow \ \ 0\neq ap_2q_1(g\otimes q)=a(g\otimes \frac{q_1}{q_2})=a(g\otimes q)$$
and so $g\otimes q$ is nontrivial contradicting our hypothesis. Then $g$ is torsion.
Comment by Simone  October 28, 2009 
Fact is not difficult to prove, it is essentially for definition of tensor product but fact 1 is less trivial. But now I see… Fact 1 is not really needed, in fact
< x > is isomorphic to Z whenever x is torsion free, then
< x > \otimes Q is isomorphic to Q because Z\otimes G=G for every abelian group G.
With this remark I like that proof:) thanks
Comment by Simone  October 28, 2009 
can tensor product of two abelian groups be group?
Comment by z  January 8, 2011 
The tensor product of two abelian groups is another abelian group, so yes.
Comment by John Armstrong  January 8, 2011 
Thank you, I needed a refresher !. Just curious: are we using the fact ( and can we use this fact) that , given free Abelian groups ( or free objects) C,D with respective bases {c_1,..,c_m} and {d_1,..,d_n}, that a bilinear map B: C x D –>E is then uniquelydefined once we know the values of { B(c_1,0),..,B(c_m,0), B(0,d_1,..B(0,d_m) }? , i.e., {(c_i,0), (0,d_j) ; i=1,..,m ; j=1,2,..,n} is a basis for C x D ? Then a bilinear map B in C x D is mapped to the linear map L in C(x)D so that L( c_i (x) d_j):= B( c_i, d_j)? SO it seems to come down to properties of multilinear algebra. Is there any analog of the tensor product when the objects are not free? Thanks, G.K.
Comment by G.K  May 23, 2014 
(Sorry, hit the ‘Submit’ too quickly) and, is it accurate to say that when a function f satisfies the given relations, that the function itself passes to the quotient (I think it comes down to a relation between the kernels / zero set of the spaces in a commuting triangle)? Sorry if I’m wrong, I haven’t done this stuff in a while.
Comment by G.K  May 23, 2014 
Sorry again. I meant to say the map L : C(x)D –> E factors through, so we have B((x)(m,n))= B(m(x)n)= L( C x D ) and not ‘passes to the quotient”.
Comment by G.K  May 23, 2014 
Basically yes, we are using that correspondence. And yes, if a function satisfies the bilinearity relations it factors through the quotient . That is, we can factor where .
Comment by John Armstrong  May 24, 2014 
Awesome! Thank you!!
Is the first example what’s meant by a “direct sum”?
Comment by isomorphismes  July 23, 2014 
Which example do you mean? The post is about tensor products, which are not generally the same thing. A direct sum is something like the Cartesian product of two spaces (), while a tensor product is rather different ()
Comment by John Armstrong  July 23, 2014 
Oh sorry, I meant f( a+a′ , b+b′ ) = f(a,b) + f(a′,b′).
Comment by isomorphismes  July 23, 2014 
Yes, a function that behaves like that should factor through a function defined on the direct sum.
Comment by John Armstrong  July 23, 2014 
Cool, great explanation! Thanks!
Comment by isomorphismes  July 23, 2014 
So maybe the “easiest example” would be 11×11 = ×(10+1,10+1) = ×(10,10) + ×(10,1) + ×(1,10) + ×(1,1) = 100 + 10 + 10 + 1 = 121.
Comment by isomorphismes  April 10, 2015 