My questions are about the definitions of Tor and Ext functors, only in the specific case of Abelian groups via injective and projective resolutions.
I'm sorry for the trivial questions. I have never studied Homological algebra formally, and I need these functors for a very specific reason.
I have read a little bit in the internet about these functors, and I would be happy if one could check me that I understood it correctly. Moreover, any comment would be appreciated.
The Ext functor:
Let $A$ and $B$ be Abelian groups. To define $\mathrm{Ext}_{\Bbb{Z}}^1(A,B)$ (the only one that does not vanish) there are 2 options:
$1.$ Choose a projective resolution of $A$ (which always exists): $$0\longrightarrow K\longrightarrow F\longrightarrow A\longrightarrow 0,$$ where $F$ and $K$ are projective (iff free) Abelian groups. Now, we consider:
$$0\longrightarrow \mathrm{Hom}(A,B)\longrightarrow \mathrm{Hom}(F,B)\overset{\alpha}{\longrightarrow} \mathrm{Hom}(K,B),$$ which is exact, since $\mathrm{Hom}(-,B)$ is a left exact contravariant functor (the connecting maps are the compositions with the original maps).
Now, $\mathrm{Ext}_{\Bbb{Z}}^1(A,B)$ is defined to be $\mathrm{Hom}(K,B)/Im(\alpha)$, so that we get an exact sequence:
$$0\longrightarrow \mathrm{Hom}(A,B)\longrightarrow \mathrm{Hom}(F,B)\longrightarrow \mathrm{Hom}(K,B)\longrightarrow \mathrm{Ext}_{\Bbb{Z}}^1(A,B)\longrightarrow 0.$$
Another possible definition is:
$2.$ Choose an injective resolution of $B$ (which always exists): $$0\longrightarrow B \longrightarrow D\longrightarrow Q\longrightarrow 0,$$ where $D$ and $Q$ are division Abelian groups (iff injective).Then consider $$0\longrightarrow \mathrm{Hom}(A,B)\longrightarrow \mathrm{Hom}(A,D)\overset{\beta}{\longrightarrow} \mathrm{Hom}(A,Q),$$ which is exact since $\mathrm{Hom}(A,-)$ is a covariant left-exact functor.
Then $\mathrm{Ext}_{\Bbb{Z}}^1(A,B)$ is defined to be $\mathrm{Hom}(A,Q)/Im(\beta)$ and makes the sequence: $$0\longrightarrow \mathrm{Hom}(A,B)\longrightarrow \mathrm{Hom}(A,D)\longrightarrow \mathrm{Hom}(A,Q)\longrightarrow \mathrm{Ext}_{\Bbb{Z}}^1(A,B)\longrightarrow 0$$ exact.
Similarly, we can define $\mathrm{Tor}_{\Bbb{Z}}^1(A,B)$ for Abelian groups $A$ and $B$ in two ways, using injective or projective resolutions. This time, the tensor product functor is right exact, so that we should "fix" the left side of the sequence, by considering kernels instead of co-kernels.
Did I understand correctly?
Thank you for your time!
Best Answer
We have $\mathrm{Tor}(A,B)=0$ if $A$ is projective or flat. So we need our projective objects to be the input of $A\otimes -,$ if we're viewing the projective objects as right $R-$modules. The tensor functor is right exact, so given any projective resolution of $B$
$$\cdots \overset{}{\longrightarrow}P_n\overset{d_n}{\longrightarrow} P_{n-1}\overset{}{\longrightarrow}\cdots \overset{d_1}{\longrightarrow}P_0\overset{\alpha}{\longrightarrow}B\overset{}{\longrightarrow}0$$ then the tensor functor $A\otimes-$
$$\cdots \overset{}{\longrightarrow}A\otimes P_n\overset{1\otimes d_n}{\longrightarrow} A\otimes P_{n-1}\overset{}{\longrightarrow}\cdots \overset{1\otimes d_1}{\longrightarrow}A\otimes P_0\overset{1\otimes\alpha}{\longrightarrow}A\otimes B\overset{}{\longrightarrow}0$$ is a chain complex, and $$\mathrm{Tor}(A,B)=\mathrm{ker}(1\otimes d_n)~/~\mathrm{image}(1\otimes d_{n+1}).$$
I don't know of any way to define $\mathrm{Tor}$ using injective resolution. [$\S17.1$ Dummit and Foote]