# derive TOR long exact sequence without using right exactness of tensor

Let $R$ be a commutative ring and $M$ be an $R$ module. The only proof that I know of the right exactness of $-\otimes_R M$ uses the left exactness of $Hom_R(-,M)$. I wondered if there was a way of deriving the long exact sequence of tensor without using the right exactness of $-\otimes_R M$. This would allow me to deduce the right exactness of tensor.

Let $0 \to M \to N \to K \to 0$ be an exact sequence of $R$ modules. Let $\mathfrak M$ be a free resolution of $M$, $\mathfrak N$ a free resolution of $N$, $\mathfrak K$ a free resolution of $K$. We get as usual that there is a short exact sequence of chain complexes $0 \to \mathfrak M \to \mathfrak N \to \mathfrak K \to 0$ which stays exact after tensoring with an $R$ module $L$.

What I would appreciate assistance in is showing that the cohomology of $\mathfrak M \otimes_R L$ at the first entry is $M \otimes_R L$. This is where I see the argument would ordinarily need the right exactness of tensor.

• You are asking «how do I show that the tensor product is right exact without using that the tensor product is right exact?», essentially. Nov 25 '16 at 2:07

What I would appreciate assistance in is showing that...

Slightly more generally, suppose you have some additive functor $F$ and a projective resolution of an object $A$ $$\cdots \to P^{-2} \to P^{-1} \to P^0 \to A \to 0$$ We look at the complex $$F (P^\bullet)\colon \quad \cdots \to F (P^{-2}) \to F (P^{-1}) \to F (P^0) \to 0$$ The $0$-th cohomology is given by $$H^0 (F (P^\bullet)) = \operatorname{coker} (F (P^{-1}) \to F (P^0))$$ You want to conclude that this is naturally isomorphic to $F (A)$, but this natural isomorphism corresponds to preservation of cokernels, i.e. right exactness: $$\operatorname{coker} (F (P^{-1}) \to F (P^0)) \stackrel{???}{\cong} F (\operatorname{coker} (P^{-1} \to P^0)).$$

The only difference from right exactness is that you consider only cokernels of morphisms between projective objects; or free objects, if you take free resolutions of $R$-modules, as in your case. But checking this is not any easier than checking right exactness.

The only proof that I know of the right exactness of $-\otimes_R M$ uses the left exactness of $\operatorname{Hom}_R (-,M)$.

• This is not 100% correct: the proof uses that $-\otimes_R M$ is left adjoint to $\operatorname{Hom}_R (M,-)$, and then any left adjoint is automatically right exact and any right adjoint is left exact.

• Alternatively, you can check right exactness of $-\otimes_R M$ by hand: just take a short exact sequence $$0 \to N' \to N \to N'' \to 0$$ and look at the maps in the induced sequence $$N'\otimes_R M \to N\otimes_R M \to N''\otimes_R M \to 0$$

• $-\otimes_R M$ is not left adjoint to $Hom_R(-,M)$. I tried to come up with a proof along these lines(you deduced correctly that I never actually read the proof deducing right exactness of tensor from left exactness of Hom). Instead I got(a perfectly valid) proof by deducing the fact that if $Hom(-,M)$ applied to an exact sequence is exact for every $M$ then the sequence is exact. This part follows from a converse of the first isomorphism theorem. Nov 29 '16 at 12:17
• Would you be willing to elaborate on what you mean by using that $Hom(-,M)$ is the right adjoint of $- \otimes M$? This is not true but $Hom(M,-)$ is right adjoint to $-\otimes M$. Perhaps one can use the symmetry of Tor to conclude. Nov 29 '16 at 12:19
• @De Jong I apologize, it was a typo, I meant the covariant Hom. The contravariant Hom is adjoint to itself (for $R$ commutative, as $\operatorname{Hom}_R (L, \underline{\operatorname{Hom}}_R (M,N)) \cong \operatorname{Hom}_R (L\otimes_R M, N) \cong \operatorname{Hom}_R (M\otimes_R L, N) \cong \operatorname{Hom}_R (M, \underline{\operatorname{Hom}}_R (L,N))$.
– user144221
Nov 29 '16 at 13:30
• So all you are saying is that the left exactness of the covariant hom is completely trivial and this implies the right exactness of $\otimes M$. Ok. Not bad. Different than the proof I skimmed in Atiyah and Macdonald. Nov 29 '16 at 16:50
• @De Jong It is trivial in the sense that it follows from a more general fact: if $F$ is left adjoint to $G$, then $F$ preserves all categorical colimits (in particular cokernels, so it is right exact), and $G$ preserves all categorical limits (in particular kernels, so it is left exact). For details, see section 2.6 of Weibel's book.
– user144221
Nov 29 '16 at 16:56