Abstract Nonsense

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.

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

January 20, 2012 -

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