I can't understand the proof of Main Theorem 4.4.16 from Weibel's book "An Introduction to homological algebra".
The Theorem states
Let $R$ be a local noetherian commutative ring, then $R$ is regular iff $\operatorname{gl.dim}(R)<\infty$.
Let $\mathfrak m$ is maximal ideal of $R$ and $k=R/\mathfrak m$.
Namely I can't understand this part:
Now the image of $\mathfrak m/x\mathfrak m$ in $S=R/xR$ is $\mathfrak m/xR=\mathfrak mS$, so we get exact sequences $$ 0\ \to xR/x\mathfrak m\to \mathfrak m/x\mathfrak m\to \mathfrak m S\to0\ \operatorname{and} 0\to\mathfrak mS\to S\to k\to 0. $$Moreover, $xR/x\mathfrak m \cong \operatorname{Tor}^R_1(R/xR,k) \cong k$, and the image of $x$ in $xR/x\mathfrak m$ is nonzero.
How did he conlcude that $xR/x\mathfrak m \cong \operatorname{Tor}^R_1(R/xR,k)$?
To try to give a complete answer I use and ideal $I \subseteq R$ and ideal $\mathfrak{m}$ in our case $I=xR$ and assume $I \subseteq \mathfrak{m}$, since $x \in \mathfrak{m}$ I believe. Then we claim $\operatorname{Tor}(R/I,R/\mathfrak{m})=I/I\mathfrak{m}$
consider the free resoultion $$0\rightarrow I\rightarrow R\rightarrow R/I \rightarrow 0$$ and this gives $$0\rightarrow \operatorname{Tor}(R/I,R/\mathfrak{m})\rightarrow I\otimes R/\mathfrak{m}\rightarrow R\otimes R/\mathfrak{m}\rightarrow R/I \otimes R/\mathfrak{m}\rightarrow 0$$ so we must determine the kernel of the map $I\otimes R/\mathfrak{m}\rightarrow R\otimes R/\mathfrak{m}$ which since $I\subseteq \mathfrak{m}$ the map is zero. Thus $\operatorname{Tor}(R/I,R/\mathfrak{m})=I\otimes R/\mathfrak{m}=I/I\mathfrak{m}$.
However I suspect this follows from some previous result in the book.