Abstract Nonsense

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

Point of Post: This is a continuation of this post.

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Thus, we see that despite there being, given a right $R$-module $M$ and a left $R$-module $N$ a lot of categories,multiplications, etc. etc. floating around, we see that we must only keep the (relatively) simple construction of $M\otimes_R N$ and the obvious $(A,B)$-multiplication in mind to create any type of tensor product.

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Let’s look at some particular ways in which bimodules naturally arise, and talk about how their tensor products might work.

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Let $R$ be a commutative ring, then any left $R$-module $M$ with multiplication $rm$, can be given the $R$-bimodule structure defined by $mr\overset{\text{def.}}{=}rm$–the only axiom that needs to be checked is that $r(ms)=(rm)s$ for all $r,s\in R$ and $m\in M$. But, this follows from commutativity of the $R$-multiplication: $r(ms)=r(sm)=(rs)m=(sr)m=s(rm)=(rm)s$. Of course, this is what allows us to make no distinction between left and right $R$-modules. Thus, given two $R$-modules $M$ and $N$ we see then that the $R$-biadditive $R$-bimaps $f:M\times N\to L$ (where $L$ is just any $R$-module, thought of as an $R$-bimodule in the usual way) is nothing more than the classic notion of a bilinear i.e. linear in each variable–$f(rm+m',n)=rf(m,n)+f(m',n)$ and $f(m,rn+m')=rf(m,n)+f(m,n')$. We then note that we have made $M\otimes_R N$ to be just the usual abelian group with an $R$-bimodule structure (i.e. just an $R$-module structure, I want to grill this in) defined to be such that $r(m\otimes n)=(rm)\otimes n=m\otimes (rn)$. Moreover, we see that with this $R$-module structure on $M\otimes_R N$ one has that $M\otimes_R N$ is universal with respect to bilinear maps out of $M\times N$.

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Generalizing, any left or right $R$-module $M$ can be given the structure of a $(Z(R),Z(R))$-bimodule (where, of course, $Z(R)$ is the center of $R$) defined, as usual, by $mr\overset{\text{def.}}{=}rm$–once again we note that this is practically equivalent to considering $M$ as a $Z(R)$-module. We see then that given a left $R$-module $M$ and a left $R$-module $N$, both thought of as $(Z(R),Z(R))$-bimodule (i.e. a $Z(R)$-module) , an $R$-biadditive $(Z(R),Z(R))$-bimap $f:M\times N\to L$ where $L$ is some $(Z(R),Z(R))$-bimodule (i.e. a $Z(R)$-module) is nothing more than an $R$-biadditive map which is $Z(R)$-bilinear.

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Let $M$ and $N$ be a right and left $R$-module respectively, then we can consider them as $(\mathbb{Z},R)$-modules and $(R,\mathbb{Z})$-modules respectively (in the obvious ways). We see that this line of thinking, noting that $(\mathbb{Z},\mathbb{Z})$-modules are just abelian groups, is related to our first construction of the tensor product.

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If $R$ is a unital ring and $\mathfrak{a}$ a two-sided ideal of $R$ and $S,S'$ two unital subrings of $R$. Then $\mathfrak{a}$ is an $(S,S')$-bimodule defined by just left and right multiplication–the compatibility condition between the left and right multiplication is merely just associativity.

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Remark: I want to make a point about the comments made above. One may get from my above comments that the ONLY $(R,R)$-bimodule structures are the ones that came from $R$-modules. Of course, this is not at all true. My attempt to conflate the notions of $(R,R)$-bimodules and $R$-module structures (for commutative rings) is merely that in our future encounters we shall rarely be presented, on a silver platter one may say, bimodules. More commonly we shall just be presented, for example, a right $R$-module $M$ and a left $R$-module $N$ where $R$ is commutative.  In this case, we’d like to think of $M$ and $N$ as just being $(R,R)$-bimodules in the natural way, so that if we consider $M\otimes_R N$ it will be as an $R$-module, instead of just an abelian group. More generally, given any ring $R$ we shall always consider (for the sake of tensor products) any right $R$-module $M$ and left $R$-module $N$ as $(Z(R),Z(R))$-bimodules so that we can consider $M\otimes_R N$ as a $(Z(R),Z(R))$-bimodule, or more realistically just a $Z(R)$-module.

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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 -

2 Comments »

1. […] theorem will show. Moreover, if is commutative and and we give every module in sight the usual -bimodule structure then these all collapse to the same definition we have mentioned before: […]

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2. […] and . We can then consider the set-valued functor given by on objects (where, as we have previously discussed, is the set of bilinear maps ) and given an arrow we define where (as usual) . So, what does a […]

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