Abstract Nonsense

Crushing one theorem at a time

Using Partial Exactness to Compute Things (Pt. I)


Point of Post: In this post we show how one can use partial exactness to actually compute, explicitly, the isomorphism type of certain tensor products.

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Motivation

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Up until this point I have committed a sort of embarrassing crime–I have yet to explicitly compute a tensor product! This is not good, because I strongly believe that if one doesn’t stop and compute some things, one can get a little lost in the abstraction. So, this post is devoted primarily to discussing a way in which we can use partial exactness to explicitly compute the isomorphism type of some actual examples. Roughly the idea is that if we can express a module in terms of an exact sequence of the form A\to B\to C\to0 then we know that C is the cokernel of the initial, right exactness tells us then that A\otimes D\to B\otimes D\to C\otimes D\to0 is exact and so C\otimes D will be the cokernel of the first map. After showing some examples where this is useful we shall describe a general technique for computing tensor products by using “free presentations” which ultimately amount to expressing a module in terms of generators and relations.

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Some Computations

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Let’s get warmed up by using partial exactness (as described above) to prove the following theorem. Oh, although I mentioned it already, in case one has forgotten, if we have an R-map f:M\to N we define the cokernel of f, denoted \text{coker}(f), to be N/\text{im}(f). With this in mind we get on proving:

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Theorem: Let R be a unital ring and \mathfrak{a} and \mathfrak{b} ideals of R. Then,

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(R/\mathfrak{a})\otimes_R(R/\mathfrak{b})\cong R/(\mathfrak{a}+\mathfrak{b})

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Proof: We have the following exact sequence

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\mathfrak{a}\overset{\iota}{\hookrightarrow} R\overset{\pi}{\longrightarrow} R/\mathfrak{a}\to0

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tensoring with R/\mathfrak{b} and using right exactness gives

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\mathfrak{a}\otimes_R R/\mathfrak{b}\overset{\iota\otimes 1}{\longrightarrow}R\otimes_R R/\mathfrak{b}\overset{\pi\otimes1}{\longrightarrow}(R/\mathfrak{a}))\otimes_R(R/\mathfrak{b})\to0

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Now we have that (R/\mathfrak{a})\otimes_R (R/\mathfrak{b})\cong\text{coker}(1\otimes\iota) but if we consider the isomorphism R\otimes_R R/\mathfrak{b}\xrightarrow{f}R/\mathfrak{b} we see then that

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(R/\mathfrak{a})\otimes_R (R/\mathfrak{b})\cong\text{coker}(f\circ(\iota\otimes 1))

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So, what is \text{im}(f\circ(\iota\otimes 1))? We claim that \text{im}(f\circ(\iota\otimes1))=\mathfrak{b}R/\mathfrak{a} where, as usual,

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\displaystyle\mathfrak{b}R/\mathfrak{a}=\left\{\sum_i b_i(r_i+\mathfrak{a}):b_i\in\mathfrak{b}\text{ and }r_i\in R\right\}

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To prove that \text{im}(f\otimes(\iota\otimes1))\subseteq \mathfrak{b}R/\mathfrak{a} it suffices to show that the image of simple tensors are contained, but this follows since

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b\otimes(r+\mathfrak{a})\overset{\iota\otimes1}{\mapsto}b\otimes(r+\mathfrak{a})\overset{f}{\mapsto}b(r+\mathfrak{a})

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Conversely, we see that if \displaystyle \sum_i b_i(r_i+\mathfrak{a}) is in \mathfrak{b}R/\mathfrak{a}, then this is the image of \displaystyle \sum_i b_i\otimes (r_i+\mathfrak{a}) and so the opposite inclusion follows. Thus, we have that

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(R/\mathfrak{a})\otimes_R(R/\mathfrak{b})\cong (R/\mathfrak{a})/(\mathfrak{b}R/\mathfrak{a})

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To finish we prove that the right hand side of the above is isomorphic to R/(\mathfrak{a}+\mathfrak{b}). To prove this note that since \mathfrak{a} is contained in the kernel of the canonical map \pi:R\to R/(\mathfrak{a}+\mathfrak{b}) it descends to an epimorphism \overline{\pi}:R/\mathfrak{a}\to R/(\mathfrak{a}+\mathfrak{b}). To finish we prove that \ker\overline{\pi}=\mathfrak{b}R/\mathfrak{a} from where the rest will follow from the first isomorphism theorem. Indeed, suppose that \displaystyle \sum_i b_i(r_i+\mathfrak{a}) is in \mathfrak{b}R/\mathfrak{a} then we see that

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\displaystyle \overline{\pi}\left(\sum_i b_i(r_i+\mathfrak{a})\right)=\sum_ib_ir_i+(\mathfrak{a}+\mathfrak{b})=0

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conversely, suppose that r+\mathfrak{a}\in R/\mathfrak{a} is in \ker\overline{\pi}, then r+(\mathfrak{a}+\mathfrak{b})=0 which tells us that r\in\mathfrak{a}+\mathfrak{b} so that r=a+b for a\in\mathfrak{a} and b\in\mathfrak{b} so that r+\mathfrak{a}=(a+b)+\mathfrak{a}=b+\mathfrak{a}=b(1+\mathfrak{a})\in\mathfrak{b}R/\mathfrak{a}. The conclusion follows. \blacksquare

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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 20, 2012 - Posted by | Algebra, Uncategorized | , , , , , , , , ,

4 Comments »

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