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I'm reading about tensor products of modules and I don't understand the following paragraph on page 2:

Even though elements of $M \times N$ and $M \oplus N$ are written in the same way, as pairs $(m, n)$, bilinear functions $M \times N \rightarrow P$ should not be confused with linear functions $M \oplus N \rightarrow P $. For example, addition as a function $ R \oplus R \rightarrow R $ is linear, but as a function $R \times R \rightarrow R$ it is not bilinear. Multiplication as a function $R \times R \rightarrow R$ is bilinear, but as a function $R \oplus R \rightarrow R$ it is not linear. Linear functions are generalized additions and bilinear functions are generalized multiplications.

My questions are:

  1. Why the distinction between $\times$ and $\oplus$? Aren't they the same for finite sums (products)?

  2. I don't understand how addition can be linear but not bilinear. Surely, if $(a,b) \mapsto a + b$ then it doesn't matter in which argument it is linear. (Thinking of e.g. $\mathbb{Z} \times \mathbb{Z} \rightarrow \mathbb{Z}$)

  3. And if $(a,b) \mapsto ab$ then how can it be linear in both arguments (bilinear) but not be in one (linear). It's difficult to write this sentence as it makes no sense to me whatsoever.

Many thanks for your help!

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The notation M x N just means the underlying set of pairs. It has no additive structure. To define the tensor product you first start off with bilinear functions on the product set M x N but it doesn't get equipped with an additive structure. Since it's easy to imagine pairs (m,n) as elements of the direct sum, the point is not to think in that way. –  KCd Oct 13 '11 at 15:07
    
I deleted my previous stupid comment, which contained nothing useful except perhaps this link to another question about tensor products. –  Hans Lundmark Oct 13 '11 at 15:16
    
@HansLundmark: I'm glad I'm not the only one who writes stupid things. –  Matt N. Oct 13 '11 at 15:30

1 Answer 1

up vote 5 down vote accepted

Bi-linear means that fixing one argument, the resulting function of the other argument is linear. In your example $(a,b)\mapsto a+b$ (say, over $\mathbb{Z}$), fixing $a =1$, the function $b\mapsto b+1$ is not linear since $b1 + b2\mapsto b1 + b2 +1$ and not to $(b1+1)+(b2+1)=b1+b2+2$. On the other, hand the function $(a,b)\mapsto a\cdot b$ is bilinear. for example, fixing $a = 2$ we get the function $b\mapsto 2b$ which is clearly linear in $b$. I guess that once you clear this up for yourself, the rest will follow easily.

EDIT: finite sums and finite products of modules are "the same", though this should be thought of as a "theorem" and not a definition. The correct way to address those issues is using the language of category theory and universal properties. To make a long story short, I will just say that linear functions from some module $L$ to $M\times N$ are in natural correspondence to pairs of linear functions, one from $L$ to $M$ and one from $L$ to $N$ while pairs of linear functions one from $M$ to $L$ and one from $N$ to $L$ are in natural correspondence with linear functions from $N\oplus M$ to $L$. While it is easy to see that this is "the same" for every pair of modules $M,N$ this breaks for infinite families of modules (try to find the corresponding modules and see why they are different). You can also think of those notions in other "categories" like groups or topological spaces (with group homomorphisms and continues functions resp.) and see that in general the "sum" and "product" are very different. If you are interested in learning the long story (which is very interesting in my opinion), you can try reading about categorical sum and categorical product on wikipedia that will also lead you naturally to the notion of universal property.

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Thank you! Of course : / And what is your answer to my question 1? –  Matt N. Oct 13 '11 at 12:05
    
Yes, they are isomorphic for finite sums/products for abelian categories, e.g. modules over some ring. –  Julian Kuelshammer Oct 13 '11 at 13:12
    
There's even a term for this: en.wikipedia.org/wiki/Biproduct –  Dylan Moreland Oct 13 '11 at 14:32
    
yes, and there is a reason to give it a name, since it is not a mere coincidence that finite sums and finite products coincide for modules, this is characteristic to "abelian categories" (and actually, part of the definition in some formulations) so it is an important notion. –  KotelKanim Oct 13 '11 at 14:53
    
I don't see how the EDIT comment applies to the topic under discussion. A bilinear map on M x N is just defined on a raw set. In no way is M x N a "direct product of two modules" in order to understand what a bilinear map is. –  KCd Oct 13 '11 at 15:40

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