Abstract Nonsense

Crushing one theorem at a time

Functorial Properties of the Tensor Product (Pt. II)


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

\text{ }

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:

\text{ }

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

\text{ }

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

\text{ }

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

\text{ }

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

\text{ }

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

\text{ }

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

\text{ }

(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

\text{ }

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)

\text{ }

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

\text{ }

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

\text{ }

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

\text{ }

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

\text{ }

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

\text{ }

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

\text{ }

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

\text{ }

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

\text{ }

and

\text{ }

\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}

\text{ }

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.

\text{ }

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

\text{ }

Practically the same proof shows that:

\text{ }

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

\text{ }

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

\text{ }

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

\text{ }

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

\text{ }

is an exact sequence of abelian groups.

\text{ }

\text{ }

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.

Advertisements

January 4, 2012 - Posted by | Algebra, Module Theory, Ring Theory | , , , , , , , , , ,

2 Comments »

  1. […] Functorial Properties of the Tensor Product (Pt. III) Point of Post: This is a continuation of this post. […]

    Pingback by Functorial Properties of the Tensor Product (Pt. III) « Abstract Nonsense | January 4, 2012 | Reply

  2. […] begin by noting how the advent of bimodules allows us to endow Hom sets with more than an abelian group structure. In […]

    Pingback by Adjoint Isomorphism for Tensor and Hom « Abstract Nonsense | January 10, 2012 | Reply


Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

%d bloggers like this: