Abstract Nonsense

The Construction of the Tensor Product of Modules (Pt. I)

Point of Post: In this post we introduce the tensor product construction of modules, and discuss it as a functor, as well as define bimodules and explain their place in the theory.

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Motivation

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In this post we discuss one of the most important constructions on modules that I know of: the tensor product. The tensor product is perhaps one of the most pervasive objects in all of mathematics–it shows up everywhere. But what is it? Roughly, very roughly, the tensor product is to multiplication as the coproduct is to addition. By this, I mean that while the coproduct of modules has an additive feel (it forms disjoint unions of bases, it’s additive on dimensions) the tensor product has a multiplicative feel (it forms cartesian products of bases, it’s multiplicative on dimensions). In this way, where “multiplicative” like things show up, tensor products often follow. For example, if one takes a commutative ring $R$ and two groups $G\times H$ then $R[G\times H]$ is related to $R[G]$ and $R[H]$ by tensor product. If one takes two algebraic curves $X$ and $Y$ then their rational function field $k(X\times Y)$ is related to $k(X)$ and $k(Y)$  by tensor product. Even the polynomial ring $R[x,y]$ which could be thought of, for example, as a ring on $\{x\}\times\{y\}$,  is related to $R[x]$ and $R[y]$ by tensor product.

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So, besides acting to relate constructions based on ‘products’ to their individual factors, what else can tensor products do for us? There is a very natural way in which the tensor construction arises. Namely, suppose for a second that we have some left $R$-module $M$ and some superring $S\geqslant R$. Is there any way to naturally turn $M$ into a left $S$-module? Most likely, anyone reading this has seen a particular case of this query. Namely, one probably remembers being asked whether or not there was a natural way to take a real vector space $V$ and turn it into a complex vector space–of course there is, it’s the complexification. Thus not only is the ‘extension of ring’ problem interesting, it’s not even a new one! Tensor products allow us to answer this question affirmatively, there is a natural way to extend the scalars of a given module to a bigger ring (although, and for good reason, this extension may not be “faithful) . Moreover, the actual nitty-gritty construction of the tensor product, the actual model for it (opposed to the universal characterization) resembles precisely how one might go about trying to extend scalars–formally multiply, and then deal with the consequences later!

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There is a last, and perhaps most important (in the sense that it really comes closest to actually defining what they are) , reason to study tensor products namely if we have two modules $M$ and $N$ it’s natural to ask about bilinear maps out of $M\times N$. But, considering all of our work has been about linear maps, it would be nice if we could some way interpret bilinear maps on $M\times N$ as being linear maps on some other module, this is where the tensor product comes in. There is a fantastic expository piece written by Tim Gowers on this subject, here.

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Tensor Products

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Note: In this post all rings and all modules are assumed unital.

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Let $R$ be some ring and $M$ be a right $R$-module and $N$ a left $R$-module. Then, if $G$ is an abelian group, a map $f:M\times N\to G$ is called $R$-biadditive if

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\begin{aligned}&f(m_1+m_2,n)=f(m_1,n)+f(m_2,n)\\ &f(m,n_1+n_2)=f(m_1,n)+f(m_2,n)\\ &f(mr,n)=f(m,rn)\end{aligned}

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for all $r\in R$, $m,m_1,m_2\in M$ and $n,n_1,n_2\in N$.  We denote the set of all biadditive maps $M\times N\to G$ as $\text{BiAdd}(M,N;G)$.

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Let’s take a look at some examples of biadditive maps:

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If $r\in R$ then the multiplication map $\mu:R\times R\to R:(r,s)\mapsto rs$ is biadditive.

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Another example of a bilinear map is given to us from classic linear algebra, where if $V$ is a, say finite dimensional, $k$-space for some field $k$ then the map $\text{ev}:V\times V^\ast\to k:(v,\varphi)\mapsto\varphi(v)$ (where $V^\ast$ is the dual space).

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There is a fairly fluid way to make, given the product $M\times N$ of a fixed right $R$-module by a fixed left $R$-module, a category $\mathbf{BiAdd}(M\times N,\mathbf{Ab})$ which consists of all biadditive maps $M\times N\to G$ for some group $G$, and a morphism from $f:M\times N\to G$ to $g:M\times N\to H$ is a group homomorphism $\phi:G\to H$ such that $\phi\circ f=g$, with composition being just normal function composition. Proving that this really is a category, amounts basically to showing that if $f:M\times N\to G$ is biadditive then $\text{id}_G:G\to G$ is a homomorphism $f\to f$, and if $f_i:M\times N\to G_i$ are biadditive maps for $i=1,2,3$ and $\phi_j:f_j\to f_{j+1}$ for $j=1,2$ then $\phi_2\circ \phi_1$ really is a morphism $f_1\to f_3$. The first of these is trivial, since $\text{id}_G\circ f=f$ for all possible $f$, and the second follows by noticing that $(\phi_2\circ\phi_1)\circ f_1=\phi_2\circ(\phi_1\circ f_1)=\phi_2\circ f_2=f_3$ (and of course that the composition of group homomorphisms are group homomorphisms).  It’s clear then that $\mathbf{BiAdd}(M\times N,\mathbf{Ab})$ really is a category, and imporantly, an isomorphism $\phi$ between $f:M\times N\to G$ and $g:M\times N\to H$ is necessarily a group isomorphism $\phi:G\to H$.

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We can then define a tensor product of $M\times N$ to be an initial object in $\mathbf{BiAdd}(M\times N,\mathbf{Ab})$.

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Ok, fine, this is all nice and good. We have thus defined tensor product in a nice, compact way in which some things become immediately obvious. For example, from first principles we have that :

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Theorem: If $f:M\times N\to G$ and $g:M\times N\to H$ are tensor products of $M\times N$, then they are isomorphic up to unique isomorphism in $\mathbf{BiAdd}(M\times N;\mathbf{Ab})$. In other words, if $f:M\times N\to U$ and $g:M\times N\to V$ are two tensor product of $M$ by $N$ then there exists a unique isomorphism $\phi:U\to V$ such that $\phi\circ f=g$.

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In light of an above comment, this in particular implies that $G\cong H$ as abelian groups. But, while the above has the merit of being concise and precise, it should be (at least once) all laid out, in more apparent terms. In particular, unraveling the above definitions gives us that a tensor product of $M$ by $N$ is an abelian group $G$ with a biadditive map $f:M\times N\to G$ such that given any other abelian group $H$ and a biadditive map $g:M\times N\to H$ there exists a unique group homomorphism $\phi:G\to H$ such that $\phi\circ f=g$. In other words, all $R$-biadditive maps out of $M\times N$ factor through the tensor product.

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References:

[1] Dummit, David Steven., and Richard M. Foote. Abstract Algebra. Hoboken, NJ: Wiley, 2004. Print.

[2] Rotman, Joseph J. Advanced Modern Algebra. Providence, RI: American Mathematical Society, 2010. Print.

[3] Blyth, T. S. Module Theory. Clarendon, 1990. Print.

[4] Lang, Serge. Algebra. Reading, MA: Addison-Wesley Pub., 1965. Print.

[5] Grillet, Pierre A. Abstract Algebra. New York: Springer, 2007. Print.

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January 3, 2012 -

5 Comments »

1. […] The Construction of the Tensor Product of Modules (Pt. II) Point of Post: This is a continuation of this post. […]

Pingback by The Construction of the Tensor Product of Modules (Pt. II) « Abstract Nonsense | January 3, 2012 | Reply

2. […] that we have defined tensor products we’d like to discuss some of the more category theoretic aspects of the construction. The […]

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3. […] guessed it by now (hard to guess, yes I know) a universal element is given by where is the tensor product and is the tensor map. Why is this true? Well, probably because it’s how we (implicitly) […]

Pingback by Universal Arrows and Universal Elements (Pt. II) « Abstract Nonsense | January 12, 2012 | Reply

4. […] finally can now get around to discussing a use of tensor products that I touted in my original introduction: extension of scalars. Namely, the idea is that we are handed some -module and some superring and […]

Pingback by Extension of Scalars and Change of Ring (Pt. I) « Abstract Nonsense | January 24, 2012 | Reply

5. […] that have a commutative unital ring and two -algebras and , we can then define their usual tensor product  where, since is commutative, this naturally has the structure of an -module, such that . The […]

Pingback by Tensor Product of Algebras « Abstract Nonsense | February 7, 2012 | Reply