$\mathrm{Tor}_1(R/a,M)$ and $\mathrm{Ext}^1_R(R/a,M)$, $a\in R$ a non-zero divisor

In Lecture Notes in Algebraic Topology, Davis & Kirk, it is written:

Proposition $\mathbf{2.4.}\,\,$ Let $R$ be a commutative ring and $a\in R$ a non-zero divisor (i.e. $ab=0$ implies $b=0$). Let $M$ be an $R$-module. Let $M/a=M/aM$ and $_aM=\{m\in M|am=0\}$. Then

\begin{align} &1.\,\, R/a\otimes M \cong M/a, && 3.\,\,\operatorname{Hom}(R/a,M) \cong\ _aM, \\ &2. \,\,\operatorname{Tor}_1(R/a,M)\cong\ _aM, && 4.\,\,\operatorname{Ext}^1(R/a,M)\cong M/a. \\ \end{align} Proof. Since $a$ is not a divisor of zero, there is a short exact sequence $$0\to R\xrightarrow[]{\times a}R\to R/A\to 0.$$ Apply the functors -$\otimes M$ and $\operatorname{Hom}(-,M)$ to the above short exact sequence. By the axioms we have exact sequences $$0\to\operatorname{Tor}_1(R/a,M)\to R\otimes M\to R\otimes M\to R/a\otimes M\to 0\text{\,\,\, and} \\\,\\ 0\to\operatorname{Hom}(R/a,M)\to\operatorname{Hom}(R,M)\to\operatorname{Hom}(R,M)\to\operatorname{Ext}^1(R/a,M)\to 0.$$ The middle maps in the exact sequence above can be identified with $$M\xrightarrow[]{\times a}M,$$ which has kernel $_aM$ and cokernel $M/a.\color{white}{\tag{$\color{black}{\square}$}}$

Question 1: Doesn't this claim work for arbitrary $a\in R$?

Question 2: Is there any nice formula for $\mathrm{Ext}^1_R(R/I,M)$ and $\mathrm{Tor}_1^R(R/I,M)$ given any ideal $I\unlhd R$?

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1. Of course not! Try $R=\mathbb Z_4$ and $a=\hat 2$; 2. $\mathrm{Tor}_1^R(R/I,R/J)\simeq I\cap J/IJ$ (see also here). About Ext there is already a similar question. – user26857 Sep 9 '13 at 0:11
You're right! I don't know why, but strangely enough I was only checking exactness in $0\to R\to R\to R/a\to0$ at the middle and end. Thank you. – Leon Sep 9 '13 at 0:51

(Answered for future reference) YACP has already shown you why question 1 fails. For question 2, the natural thing to consider is the exact sequence $\newcommand{Hom}{\operatorname{Hom}}$ $\newcommand{Ext}{\operatorname{Ext}}$ $\newcommand{Tor}{\operatorname{Tor}}$ $\DeclareMathOperator{coker}{coker}$

$$0 \to I \to R \to R/I \to 0$$

Applying the functors $\Hom_R(\_, M)$ and $\_ \otimes_R M$ yield descriptions of $\Ext^1(R/I,M)$ and $\Tor_1(R/I,M)$ as a cokernel and kernel, respectively. To be precise,

$$\Ext^1_R(R/I,M) = \coker(M \xrightarrow{\phi} \Hom_R(I,M)),$$ $$\Tor_1^R(R/I,M) = \ker(I \otimes_R M \xrightarrow{\varphi} M)$$

where the maps $\phi, \varphi$ are defined by $\phi(m) = (i \mapsto im)$, $\varphi(i \otimes m) = im$ (incidentally, the kernel of $\phi$ and cokernel of $\varphi$ are $\Ext^0(R/I,M)$ and $\Tor_0(R/I,M)$ respectively, but these are easier to describe: they are $0 :_M I$ and $M/IM$). From this description one easily obtains that if $M = R/J$ is cyclic, then $\Tor_1(R/I,R/J) \cong I \cap J/IJ$.

In general though (and for general $M$), one is usually interested in finding when $\Ext$ and $\Tor$ vanish (vanishing typically gives the most useful information). In this regard, one can say that over a Noetherian ring $R$, if $0 :_M I = 0$ (i.e. $M$ has no $I$-torsion), then $\Ext^1(R/I,M) = 0$ iff the depth of $I$ on $M$ is at least $2$.

One last note: in certain cases, one is able to view an $\Ext$ module as a $\Hom$ in disguise, in the following sense: if $N, M$ are $R$-modules, and $\text{ann}(N)$ contains a weak $M$-sequence $a_1,...,a_n$, then $\Ext^n_R(N,M) \cong \Hom_R(N, M/(a_1,...,a_n)M)$.

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