Here is the outline of the proof I will use. The reference is Weibel's An introduction to homological algebra, section 4.5, Koszul Complexes. Each step has the corresponding place in Weibel to look. I also recommend Bruns and Herzog's Cohen-Macaulay Rings.
- (Notation 4.5.1) Define $H_{q}(x,M):=H_{q}(K(x)\otimes_{R}M)$ and $H^{q}(x,M):=H^{q}(\mathrm{Hom}(K(x),M))$ where $K(x)$ is the Koszul complex $K(x_{1})\otimes_{R} K(x_{2})\otimes_{R}\cdots\otimes_{R} K(x_{d})$ when $x=(x_{1},\ldots,x_{d})$.
- (Exercise 4.5.2) Show that there are isomorphisms $H_{p}(x,M)\cong H^{d-p}(x,M)$.
- (Corollary 4.5.5) If $x=(x_{1},\ldots,x_{d})$ is a regular sequence, then $K(x)$ is a free resolution of $k$, and $H_{p}(x,M)=\mathrm{Tor}_{p}^{R}(k,M)$ and $H^{p}(x,M)=\mathrm{Ext}_{R}^{p}(k,M)$. By 2., we have the result.
1.
Let $K(x_{i})$ be the complex $0\to R\xrightarrow{\cdot x_{i}}R\to0$. If $x=(x_{1},\ldots,x_{d})$, define $K(x):=K(x_{1})\otimes_{R}\cdots\otimes_{R}K(x_{d})$, where the tensor product of complexes $(P_{\bullet},\partial_{P})$ and $(Q_{\bullet},\partial_{Q})$ is defined to be the double complex $\{P_{p}\otimes_{R}Q_{q}\}_{p,q}$ with horizontal differentials $\partial_{P}\otimes\mathrm{id}_{Q}$ and vertical differentials $(-1)^{p}\mathrm{id}_{P}\otimes\partial_{Q}$. Note that the degree $p$ part of $K(x)$, $K_{p}(x)$, is a free $R$-module generated by the symbols
$$
e_{i_{1}}\wedge\cdots\wedge e_{i_{p}}:=1\otimes\cdots\otimes e_{x_{i_{1}}}\otimes\cdots\otimes e_{x_{i_{p}}}\otimes\cdots\otimes 1\;(i_{1}<\cdots i_{p});
$$
that is, $K_{p}(x)\cong\Lambda^{p}R^{d}$, the $p$th exterior product of $R^{n}$. Therefore $K_{p}(x)\cong R^{d\choose p}$.
Now define Koszul homology and cohomology: $H_{q}(x,M):=H_{q}(K(x)\otimes_{R}M)
$ and $H^{q}(x,M):=H^{q}(\mathrm{Hom}(K(x),M))$.
2.
Let's show $H_{p}(x,M)\cong H^{d-p}(x,M)$ for all $p\in\{0,\ldots d\}$. First, notice that $K_{d}(x)\cong R^{d\choose d}=R$ under the explicit isomorphism
\begin{align*}
\omega_{d}:K_{d}(x)&\to R\\
e_{i_{1}}\wedge\cdots\wedge e_{i_{d}}&\mapsto1.
\end{align*}
Write $(K_{\ell}(x))^{*}$ for the dual of $K_{\ell}(x)$; i.e., $(K_{\ell}(x))^{*}=\mathrm{Hom}_{R}(K_{\ell}(x),R)$. Notice that
$$
R\cong\mathrm{Hom}(R,R)=\mathrm{Hom}\left(R^{d\choose0},R\right)\cong\mathrm{Hom}(K_{0}(x),R)=(K_{0}(x))^{*}.
$$
Thus, the above map is $\omega_{d}:K_{d}(x)\to(K_{0}(x))^{*}$, and this generalizes; we may define maps
\begin{align*}
\omega_{i}:K_{p}(x)&\to(K_{d-p}(x))^{*}\\
t&\mapsto(\omega_{p}(t))(y)=\omega_{d}(t\wedge y)
\end{align*}
for $t\in K_{p}(x)$ and $y\in K_{d-p}(x)$. Now, see that for generators $e_{j_{1}}\wedge\cdots\wedge e_{j_{p}}\in K_{p}(x)$ and $e_{k_{1}}\wedge\cdots\wedge e_{k_{d-p}}\in K_{d-p}(x)$, we have
\begin{align*}
(\omega_{p}(e_{j_{1}}\wedge\cdots\wedge e_{j_{p}}))(e_{k_{1}}\wedge\cdots\wedge e_{k_{d-p}})&=\omega_{d}(e_{j_{1}}\wedge\cdots\wedge e_{j_{p}}\wedge e_{k_{1}}\wedge\cdots\wedge e_{k_{d-p}})\\
&=\begin{cases}
0&\text{if }e_{j_{\ell}}=e_{k_{m}}\text{ for some }j_{\ell}\neq k_{m};\\
\omega_{d}((-1)^{\kappa}e_{i_{1}}\wedge\cdots\wedge e_{i_{d}})=(-1)^{\kappa}&\text{else},
\end{cases}
\end{align*}
where $\kappa$ is the number of times elements $e_{j_{\ell}}$ and $e_{k_{m}}$ needed to commute in order to put
$$
e_{j_{1}}\wedge\cdots\wedge e_{j_{p}}\wedge e_{k_{1}}\wedge\cdots\wedge e_{k_{d-p}}
$$
in the order
$$
e_{i_{1}}\wedge\cdots\wedge e_{i_{d}},
$$
the basis element of $K_{d}(x)$. Hence, $\omega_{i}$ takes generators $e_{j_{1}}\wedge\cdots\wedge e_{j_{p}}$ of $K_{p}(x)$ to generators $(-1)^{\kappa}(e_{k_{1}}\wedge\cdots\wedge e_{k_{d-p}})^{*}$ on $(K_{d-p}(x))^{*}$, and thus $\omega_{i}:K_{p}(x)\to(K_{d-p}(x))^{*}$ is an isomorphism.
For our next step, consider the diagram (vertical maps are isomorphisms by work above):
$$
\newcommand{\ra}[1]{\!\!\!\!\!\!\!\!\!\!\!\!\xrightarrow{\quad#1\quad}\!\!\!\!\!\!\!\!}
\newcommand{\da}[1]{\left\downarrow{\scriptstyle#1}\vphantom{\displaystyle\int_0^1}\right.}
%
\begin{array}{llllllllllll}
K(x): & 0 & \ra{} & K_{d}(x) & \ra{\partial_{d}} & K_{d-1}(x) & \ra{\partial_{d-1}} & \cdots & \ra{\partial_{2}} & K_{1}(x) & \ra{\partial_{1}} & K_{0}(x) & \ra{} & 0\\
&&&\da{\omega_{d}}&&\da{\omega_{d-1}}&&&&\da{\omega_{1}}&&\da{\omega_{0}}&&\\
(K(x))^{*}: & 0 & \ra{} & (K_{0}(x))^{*} & \ra{\partial_{0}^{*}} & (K_{1}(x))^{*} & \ra{\partial_{1}^{*}} & \cdots & \ra{\partial_{d-1}^{*}} & (K_{d-1}(x))^{*} & \ra{\partial_{d}^{*}} & (K_{d}(x))^{*} & \ra{} & 0\\
\end{array}
$$
where $\partial_{\bullet}^{*}=\mathrm{Hom}(\partial_{\bullet},R)$. We claim the following:
(i) The squares above commute up to sign; i.e., $\omega_{p-1}\partial_{p}=(-1)^{p-1}\partial_{d-p+1}^{*}\omega_{p}$ for all $p\in\{0,\ldots d\}$.
(ii) $K(x)\cong(K(x))^{*}$ as complexes.
(iii) If $M$ is an $R$-module, then $K(x)\otimes_{R}M\cong\mathrm{Hom}(K(x),M)$.
(iv) Hence,
$$
H_{p}(x,M)=H_{p}(K(x)\otimes_{R}M)\cong H^{d-p}(\mathrm{Hom}(K(x),M))=H^{d-p}(x,M),
$$
as desired. We proceed:
(i) Fix $p$. We must show that $\omega_{p-1}\partial_{p}=(-1)^{p-1}\partial_{d-p+1}^{*}\omega_{p}$. We will do so on generators. Let $e_{j_{1}}\wedge\cdots\wedge e_{j_{p}}\in K_{p}(x)$. Let $e_{k_{1}}\wedge\cdots\wedge e_{k_{d-p}}\in K_{d-p}(x)$. Observe that
\begin{align*}
\omega_{p-1}\partial_{p}(e_{j_{1}}\wedge\cdots\wedge e_{j_{p}})&=\omega_{p-1}\left(\sum_{\ell=1}^{p}x_{j_{\ell}}e_{j_{1}}\wedge\cdots\wedge\hat{e_{j_{\ell}}}\wedge\cdots\wedge e_{j_{p}}\right)\\
&=\sum_{\ell=1}^{p}x_{j_{\ell}}\omega_{p-1}\left(e_{j_{1}}\wedge\cdots\wedge\hat{e_{j_{\ell}}}\wedge\cdots\wedge e_{j_{p}}\right)\\
&=\sum_{\ell=1}^{p}x_{j_{\ell}}(-1)^{\kappa+p-1}\left(e_{k_{1}}\wedge\cdots\wedge e_{k_{d-p}}\wedge e_{j_{\ell}}\right)^{*},
\end{align*}
since the missing $e_{j_{\ell}}$ term means there are an additional $p-1$ terms to commute past. On the other hand,
\begin{align*}
(-1)^{p-1}\partial_{d-p+1}^{*}\omega_{p}(e_{j_{1}}\wedge\cdots\wedge e_{j_{p}})&=(-1)^{p-1}\partial_{d-p+1}^{*}\left((-1)^{\kappa}(e_{k_{1}}\wedge\cdots\wedge e_{k_{d-p}})\right)^{*}\\
&=(-1)^{p-1+\kappa}\partial_{d-p+1}^{*}\left((e_{k_{1}}\wedge\cdots\wedge e_{k_{d-p}})^{*}\right)\\
&=(-1)^{p-1+\kappa}\sum_{\ell=1}^{p}x_{j_{\ell}}(e_{k_{1}}\wedge\cdots\wedge e_{k_{d-p}}\wedge e_{j_{\ell}})^{*}.
\end{align*}
Thus the squares commute up to sign, as desired.
(ii) Since $\omega_{p}:K_{p}(x)\to(K_{d-p}(x))^{*}$ is an isomorphism, letting $\tilde{\omega_{p}}=(-1)^{p(p-1)/2}\omega_{p}$ fixes the sign issue and thus gives the desired isomorphism of complexes.
(iii) Let $M$ and $N$ be $R$-modules, $N\cong R^{\alpha}$ for some $\alpha\in\mathbb{N}$. There is an isomorphism $N^{*}\otimes M\cong\mathrm{Hom}(N,M)$. (Exercise: show this!) Now, since $K_{\bullet}(x)\cong R^{\alpha}$ for $\alpha\in\mathbb{N}$, we have $(K(x))^{*}\otimes M\cong\mathrm{Hom}(K(x),M)$, and by (ii), $(K(x))^{*}\cong K(x)$. Therefore, for every $R$-module $M$,
$$
K(x)\otimes_{R}M\cong(K(x))^{*}\otimes_{R}M\cong\mathrm{Hom}(K(x),M),
$$
as desired.
(iv) Since the complexes are isomorphic by (iii), and the differentials commute with the isomorphism by (i) and (ii), $H_{p}(x,M)\cong H^{d-p}(x,M)$, as we yearned to demonstrate.
3.
Our claim that $H_{p}(x,M)=\mathrm{Tor}_{p}(k,M)$ and $H^{p}(x,M)=\mathrm{Ext}^{p}(k,M)$ certainly gives the desired conclusion. It is itself a corollary to the following claim: if $x=(x_{1},\ldots x_{d})$ is a regular sequence on $R$, then $H_{q}(x,R)=0$ for $q\neq0$ and $H_{0}(x,R)=R/xR$. This is a consequence of the Kunneth formula for Koszul complexes. It's proved in Weibel, Corollary 4.5.4.
If we take this claim, then our result follows, since it shows that the Koszul complex is an acyclic resolution of $R/\frak{m}$$=k$. Thus, definitionally, to compute Tor or Ext, we take a acyclic resolution of $k$, which we may choose to be $K(x)$, and then either tensor by $M$ or take Hom, respectively, then compute (co-)homology. And definitionally, Koszul (co-)homology is tensoring $K(x)$ by $M$ or taking Hom, and then computing (co-)homology.
Thus, finally, the result is shown.
Best Answer
The error is that $\mathfrak{a}\otimes_A N$ is not isomorphic to $\mathfrak{a}N$ in general.
As a simple example, let $A=\Bbb Z$, $\mathfrak{a}=2\Bbb Z$ and $N=\Bbb Z/2\Bbb Z$.