Tensor is a right exact functor Elementary Proof

This is a relatively elementary proof (compared to others out there) of the fact that tensor is a right exact functor. Proof is taken from Hungerford, and reworded slightly. The key prerequisites needed are the universal property of quotient and of tensor product.


If A\xrightarrow{f}B\xrightarrow{g}C\to 0 is an exact sequence of left modules over a ring R and D is a right R-module, then \displaystyle D\otimes_R A\xrightarrow{1_D\otimes f}D\otimes_R B\xrightarrow{1_D\otimes g}D\otimes_R C\to 0 is an exact sequence of abelian groups. An analogous statement holds for an exact sequence in the first variable.


(Hungerford 210)

We split our proof into 3 parts: (i) \text{Im}(1_D\otimes g)=D\otimes_R C; (ii) \text{Im}(1_D\otimes f)\subseteq\text{Ker}(1_D\otimes g); and (iii) \text{Ker}(1_D\otimes g)\subseteq\text{Im}(1_D\otimes f).

(i) Since g is an epimorphism by hypothesis every generator d\otimes c of D\otimes_R C is of the form d\otimes g(b)=(1_D\otimes g)(d\otimes b) for some b\in B. Thus \text{Im}(1_D\otimes g) contains all generators of D\otimes_R C, hence \text{Im}(1_D\otimes g)=D\otimes_R C.

(ii) Since \text{Ker} g=\text{Im} f we have gf=0 and \displaystyle (1_D\otimes g)(1_D\otimes f)=1_D\otimes gf=1_D\otimes 0=0, hence \text{Im}(1_D\otimes f)\subseteq\text{Ker}(1_D\otimes g).

(iii) Let \pi:D\otimes_R B\to(D\otimes_R B)/\text{Im}(1_D\otimes f) be the canonical epimorphism. From (ii), \text{Im}(1_D\otimes f)\subseteq\text{Ker}(1_D\otimes g) so (by universal property of quotient Theorem 1.7) there is a homomorphism \alpha:(D\otimes_R B)/\text{Im}(1_D\otimes f)\to D\otimes_R C such that \displaystyle \alpha(\pi(d\otimes b))=(1_D\otimes g)(d\otimes b)=d\otimes g(b). We shall show that \alpha is an isomorphism. Then \text{Ker}(1_D\otimes g)=\text{Im}(1_D\otimes f).

We show first that the map \beta:D\times C\to(D\otimes_R B)/\text{Im}(1_D\otimes f) given by (d,c)\mapsto\pi(d\otimes b), where g(b)=c, is independent of the choice of b. Note that there is at least one such b since g is an epimorphism. If g(b')=c, then g(b-b')=0 and b-b'\in\text{Ker} g=\text{Im} f, hence b-b'=f(a) for some a\in A. Since d\otimes f(a)\in\text{Im}(1_D\otimes f) and \pi(d\otimes f(a))=0, we have
\begin{aligned}  \pi(d\otimes b)&=\pi(d\otimes(b'+f(a))\\  &=\pi(d\otimes b'+d\otimes f(a))\\  &=\pi(d\otimes b')+\pi(d\otimes f(a))\\  &=\pi(d\otimes b').  \end{aligned}

Therefore \beta is well-defined.

Verify that \beta is middle linear:
\begin{aligned}  \beta(d_1+d_2,c)&=\pi((d_1+d_2)\otimes b)\qquad\text{where }g(b)=c\\  &=\pi(d_1\otimes b+d_2\otimes b)\\  &=\pi(d_1\otimes b)+\pi(d_2\otimes b)\\  &=\beta(d_1,c)+\beta(d_2,c).  \end{aligned}

\begin{aligned}  \beta(d,c_1+c_2)&=\pi(d\otimes(b_1+b_2))\qquad\text{where }g(b_i)=c_i\\  &=\pi(d\otimes b_1+d\otimes b_2)\\  &=\pi(d\otimes b_1)+\pi(d\otimes b_2)\\  &=\beta(d,c_1)+\beta(d,c_2).  \end{aligned}

Let r\in R.
\begin{aligned}  \beta(dr,c)&=\pi(dr\otimes b)\qquad\text{where }g(b)=c\\  &=\pi(d\otimes rb)\\  &=\beta(d,rc)\qquad\text{where }g(rb)=rg(b)=rc.  \end{aligned}

By universal property of tensor product there exists a unique homomorphism \bar{\beta}:D\otimes_R C\to(D\otimes_R B)/\text{Im}(1_D\otimes f) such that \bar{\beta}(d\otimes c)=\beta(d,c)=\pi(d\otimes b), where g(b)=c.

Therefore, for any generator d\otimes c of D\otimes_R C, \displaystyle \alpha\bar{\beta}(d\otimes c)=\alpha\pi(d\otimes b)=d\otimes g(b)=d\otimes c, hence \alpha\bar{\beta} is the identity map.

\begin{aligned}  \bar{\beta}\alpha(d\otimes b+\text{Im}(1_D\otimes f))&=\bar{\beta}\alpha\pi(d\otimes b)\\  &=\bar{\beta}(d\otimes g(b))\\  &=\pi(d\otimes b)\\  &=d\otimes b+\text{Im}(1_D\otimes F)  \end{aligned}
so \bar{\beta}\alpha is the identity so that \alpha is an isomorphism.

Author: mathtuition88


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