Tensor products of abelian groups

Often enough we’re going to see the following situation. There are three abelian groups — $A$ , $B$ , and $C$ — and a function $f:A\times B\rightarrow C$ 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 $f$ is a homomorphism from $A\times B$ to $C$ . That would mean the following equation held:
$f(a+a',b+b')=f(a,b)+f(a',b')$

Instead, I want to say that if we fix one variable, the remaining function of the other variable is a homomorphism. That is
$f(a,b+b')=f(a,b)+f(a,b')$
$f(a+a',b)=f(a,b)+f(a',b)$
$f(a+a',b+b')=f(a,b)+f(a,b')+f(a',b)+f(a',b')$
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 $A$ and $B$ is an abelian group $T$ and a bilinear function $t:A\times B\rightarrow T$ so that for every other bilinear function $f:A\times B\rightarrow C$ there is a unique linear function $\bar{f}:T\rightarrow C$ so that $f(a,b)=\bar{f}(t(a,b))$ . 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 $T$ has a presentation by generators and relations as an abelian group. For generators take all elements of $A\times B$ , and for relations take all the elements $(a+a',b)-(a,b)-(a',b)$ and $(a,b+b')-(a,b)-(a,b')$ for all $a$ and $a'$ in $A$ and $b$ and $b'$ in $B$ .

By the properties of such presentations, any function of the generators — of $A\times B$ — defines a linear function on $T$ if and only if it satisfies the relations. That is, if we apply a function $f$ to each relation and get ${}0$ every time, then $f$ defines a unique function $\bar{f}$ on the presented group. So what does that look like here?
$f(a+a',b)-f(a,b)-f(a',b)=0$
$f(a,b+b')-f(a,b)-f(a,b')=0$
So a bilinear function $f$ gives rise to a linear function $\bar{f}$ , just as we want.

Usually we’ll write the tensor product of $A$ and $B$ as $A\otimes B$ , and the required bilinear function as $(a,b)\mapsto a\otimes b$ .

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 $R$ equipped with a linear map $m:R\otimes R\rightarrow R$ for multiplication.

April 6, 2007 - Posted by John Armstrong | Abelian Groups, Algebra, Group theory, Universal Properties

37 Comments »

[…] on tensor products and direct sums We’ve defined the tensor product and the direct sum of two abelian groups. It turns out they interact very […]

Pingback by More on tensor products and direct sums « The Unapologetic Mathematician | April 17, 2007 | Reply
[…] products of modules The notion of a tensor product also extends to modules, but the generalization is not quite as straightforward as it was for […]

Pingback by Tensor products of modules « The Unapologetic Mathematician | April 30, 2007 | Reply
The third condition in the definition of linearity follows from the first two… I think.

Comment by Oingo | December 8, 2007 | Reply
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 | Reply
[…] map defined by . Note carefully that this is linear as a function of the pair . It’s not bilinear — linear in each of and separately — which would mean bringing in the tensor […]

Pingback by The Inclusion-Exclusion Principle « The Unapologetic Mathematician | July 24, 2008 | Reply
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 | Reply
It’s pretty straightforward, Bacile. The universal bilinear map from $\mathbb{R}^n\times\mathbb{R}^m$ to $\mathbb{R}^n\otimes\mathbb{R}^m$ sends the pair $(u,v)$ to the tensor product $u\otimes v$ .

Comment by John Armstrong | March 10, 2009 | Reply
- 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 : inner-product 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 | Reply
I just told you what specific map it was…

Okay, let’s try this. Say your vector space $V$ has a basis $\left\{e_i\right\}$ and $W$ has basis $\left\{f_j\right\}$ . Then you have a basis for $V\otimes W$ given by $\left\{e_i\otimes f_j\right\}$ . So the pair $(e_i,f_j)$ gets sent to $e_i\otimes f_j$ , and we extend by bilinearity.

Say, for instance, we have the vectors $v=v^ie_i$ and $w=w^jf_j$ . Then the pair $(v,w)$ gets sent to $(v^ie_i)\otimes(w^jf_j)=v^iw^j(e_j\otimes f_j)$ . In your example, you’ve got the vectors $\begin{pmatrix}x\\y\end{pmatrix}$ and $\begin{pmatrix}x'\\y'\end{pmatrix}$ . As a pair, they get sent to the linear combination

$\displaystyle xx'\begin{pmatrix}{1}\\{0}\end{pmatrix}\otimes\begin{pmatrix}{1}\\{0}\end{pmatrix}+xy'\begin{pmatrix}{1}\\{0}\end{pmatrix}\otimes\begin{pmatrix}{0}\\{1}\end{pmatrix}+yx'\begin{pmatrix}{0}\\{1}\end{pmatrix}\otimes\begin{pmatrix}{1}\\{0}\end{pmatrix}+yy'\begin{pmatrix}{0}\\{1}\end{pmatrix}\otimes\begin{pmatrix}{0}\\{1}\end{pmatrix}$

Now, we can choose an isomorphism $\mathbb{R}^2\otimes\mathbb{R}^2\cong\mathbb{R}^4$ with the correspondence

$\displaystyle\begin{pmatrix}{1}\\{0}\end{pmatrix}\otimes\begin{pmatrix}{1}\\{0}\end{pmatrix}\leftrightarrow\begin{pmatrix}{1}\\{0}\\{0}\\{0}\end{pmatrix}$
$\displaystyle\begin{pmatrix}{1}\\{0}\end{pmatrix}\otimes\begin{pmatrix}{0}\\{1}\end{pmatrix}\leftrightarrow\begin{pmatrix}{0}\\{1}\\{0}\\{0}\end{pmatrix}$
$\displaystyle\begin{pmatrix}{0}\\{1}\end{pmatrix}\otimes\begin{pmatrix}{1}\\{0}\end{pmatrix}\leftrightarrow\begin{pmatrix}{0}\\{0}\\{1}\\{0}\end{pmatrix}$
$\displaystyle\begin{pmatrix}{0}\\{1}\end{pmatrix}\otimes\begin{pmatrix}{0}\\{1}\end{pmatrix}\leftrightarrow\begin{pmatrix}{0}\\{0}\\{0}\\{1}\end{pmatrix}$

Then the pair of vectors above is send to the vector

$\displaystyle\begin{pmatrix}xx'\\xy'\\yx'\\yy'\end{pmatrix}$

Comment by John Armstrong | March 11, 2009 | Reply
[…] the space itself. Foremost among these is the idea of a bilinear form. This is really nothing but a bilinear function to the base field: . Of course, this means that it’s equivalent to a linear function […]

Pingback by Bilinear Forms « The Unapologetic Mathematician | April 14, 2009 | Reply
[…] 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 | Reply
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 | Reply
- It’s been 11 years, but Is it too late to be pedantic and point out that the ONLY IF part fails for q = 0?
  
  Comment by Martin | March 17, 2021 | Reply
  - Simone did say at the end that “Clearly you have to suppose $q\neq 0$ “.
    
    Comment by John Armstrong | March 17, 2021 | Reply
    - I missed that part. I’ll be over there in the corner, looking ashamed 🙂
      
      Comment by Martin | March 17, 2021 | Reply
The if part isn’t even true. Let $q=1$ and even if $g$ is torsion the tensor isn’t zero.

Comment by John Armstrong | October 28, 2009 | Reply
- 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 | Reply
- 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 | Reply
Ah, yes, sorry. It’s been a long time since I’ve bothered with torsion elements.

Can you prove the contrapositive to handle the only-if part?

Comment by John Armstrong | October 28, 2009 | Reply
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 non-zero.

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 | Reply
- 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 | Reply
  - 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 | Reply
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 $\mathbb{Z}$ -action is entirely a matter of the subgroup they generate. Tensoring with $\mathbb{Q}$ kills torsion subgroups and preserves non-torsion subgroups.

A really non-elementary 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 | Reply
- 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 non-trivial 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 non-trivial contradicting our hypothesis. Then $g$ is torsion.
  
  Comment by Simone | October 28, 2009 | Reply
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
&lt x &gt is isomorphic to Z whenever x is torsion free, then
&lt x &gt \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 | Reply
can tensor product of two abelian groups be group?

Comment by z | January 8, 2011 | Reply
- The tensor product of two abelian groups is another abelian group, so yes.
  
  Comment by John Armstrong | January 8, 2011 | Reply
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 uniquely-defined 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 | Reply
(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 | Reply
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 | Reply
Basically yes, we are using that correspondence. And yes, if a function $f: C\times D\to E$ satisfies the bilinearity relations it factors through the quotient $\pi:C\times D\to C\otimes D$ . That is, we can factor $f=\bar{f}\circ\pi$ where $\bar{f}:C\otimes D\to E$ .

Comment by John Armstrong | May 24, 2014 | Reply
Awesome! Thank you!!

Is the first example what’s meant by a “direct sum”?

Comment by isomorphismes | July 23, 2014 | Reply
- 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 ( $\mathbb{R}^m\oplus\mathbb{R}^n=\mathbb{R}^{m+n}$ ), while a tensor product is rather different ( $\mathbb{R}^m\otimes\mathbb{R}^n=\mathbb{R}^{mn}$ )
  
  Comment by John Armstrong | July 23, 2014 | Reply
  - Oh sorry, I meant f( a+a′ , b+b′ ) = f(a,b) + f(a′,b′).
    
    Comment by isomorphismes | July 23, 2014 | Reply
    - Yes, a function that behaves like that should factor through a function defined on the direct sum.
      
      Comment by John Armstrong | July 23, 2014 | Reply
      - Cool, great explanation! Thanks!
        
        Comment by isomorphismes | July 23, 2014 | Reply
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 | Reply

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