# Abstract Nonsense

## Functorial Properties of the Tensor Product (Pt. II)

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

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What we’d now like to prove is that is that the tensor product functor, fixed in either variable, is right exact in the sense that right exact sequences get carried to right exact sequences. More precisely:

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Theorem: Let $M,N,L$ be left $R$-modules. If

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$M\xrightarrow{f}N\xrightarrow{g}L\to 0$

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is an exact sequence of left $R$-modules and $A$ a right $R$-module, then

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$A\otimes_R M\xrightarrow{1_A\otimes f}A\otimes_R N\xrightarrow{1_A\otimes g}A\otimes_R L\to 0$

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is an exact sequence of abelian groups.

Proof: Let’s first check that $(1_A\otimes g)\circ(1_A\otimes f)=0$, but this follows immediately since

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$(1_A\otimes g)\circ(1_A\otimes f)=1_A\otimes(g\circ f)=1_A\otimes 0=0$

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(the last equality holding since additive functors take zero maps to zero maps). Let’s now prove that $\ker(1_A\otimes g)\subseteq\text{im }(1_A\otimes f)$. This is the hard part. We have already shown that $\text{im }(1_A\otimes f)\subseteq\ker 1_A\otimes g$ and so from first principles we can factor $1_A\otimes g:A\otimes_R N\to A\otimes_R L$ through $A\otimes_R N/\text{im }(1_A\otimes f)$. In other words, we are able to get a map $p:A\otimes_R N/\text{im }(1_A\otimes f)\to A\otimes_R L$ such that if $\pi:A\otimes_R N\to A\otimes_R N/\text{im }(1_A\otimes f)$ is the natural projection, then $p\circ\pi=1\otimes g$. If $p$ is an isomorphism then we’d have that $\ker(1_A\otimes g)=\ker(p\circ\pi)=\ker\pi=\text{im }(1_A\otimes f)$. We can prove this by construction an explicit inverse $A\otimes_R L\to A\otimes_R/\text{im }(1_A\otimes f)$. To do this we note that since $g$ is surjective we can find a (set!) map $t:L\to N$ such that $t(g(n))=n$ and $g(t(\ell))=\ell$. We are then able to define

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$q:A\times L\to A\otimes_R N/\text{im }(1_A\otimes f):q(a,\ell)\mapsto a\otimes t(\ell)+\text{im }(1_A\otimes f)$

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We claim that this map is $R$-biadditive. Indeed, we need to prove that $q(ar,\ell)=ar\otimes t(\ell)=a\otimes rt(\ell)$ is equal to $q(a,r\ell)=a\otimes t(r\ell)$. To do this it suffices to prove that

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$a\otimes rt(\ell)-a\otimes t(r\ell)=a\otimes(rt(\ell)-t(r\ell))$

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is in $\text{im }(1_A\otimes g)$. To do this we note that

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$g(rt(\ell)-t(r\ell))=rg(t\ell))-g(t(r\ell))=r\ell-r\ell=0$

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and so $rt(\ell)-t(r\ell)\in\ker g=\text{im }f$ and so $a\otimes (rt(\ell)-t(r\ell))\in\text{im }(1_A\otimes f)$. Since $q$ is $R$-biadditive we are afforded a map

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$\widetilde{q}:A\otimes_R L\to A\otimes_R N/\text{im }(1_A\otimes f)$

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such that $\widetilde{q}(a\otimes \ell)=a\otimes \ell$. Thus, we find that

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$p(\widetilde{q}(a\otimes \ell))=p(a\otimes t(\ell)+\text{im }(1_A\otimes f))=a\otimes g(t(\ell))=a\otimes \ell$

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and

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\begin{aligned}\widetilde{q}(p(a\otimes n+\text{im }(1_A\otimes g))) &=\widetilde{q}(a\otimes g(n))\\ &=a\otimes t(g(n))+\text{im }(1_A\otimes f)\\ &=a\otimes n+\text{im }(1_A\otimes f)\end{aligned}

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Thus, $\widetilde{q}$ is a two-sided inverse for $p$, and so $p$ is an isomorphism and so the fact that $\ker(1_A\otimes g)\subseteq\text{im }(1_A\otimes f)$ and so exactness in the middle follows.

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It lastly suffices to prove that $1\otimes g$ is surjective. To do this, let $a\otimes \ell$ in $A\otimes_R \ell$ be arbitrary. Since $g$ is surjective there exists some $n\in N$ such that $g(n)=\ell$ and so $(1_A\otimes g)(a\otimes n)=a\otimes g(n)=a\otimes \ell$. Thus, $1_A\otimes g$ hits all the simple tensors in $A\otimes_R L$ and since the simple tensors generate $A\otimes_R L$ and $1_A\otimes g$ is a $\mathbb{Z}$-map surjectivity follows. $\blacksquare$

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Practically the same proof shows that:

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Theorem: If $M,N,L$ are right $R$-modules and we have an exact sequence

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$M\xrightarrow{f}N\xrightarrow{g}L\to0$

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is an exact sequence of $R$-maps, and if $A$ is a left $R$-module, then

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$M\otimes_R A\xrightarrow{f\otimes 1_A}N\otimes_R A\xrightarrow{g\otimes 1_A}L\otimes_R A\to 0$

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is an exact sequence of abelian groups.

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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.