Exact sequence and Tor functor.

Exact sequence and Tor functor.

Say $M$ is an $R$-module and $\operatorname{gld}(R)=n$, i.e. global
dimension of $R$ is n.
Is then $\operatorname{Tor}_i^{R}(M,N)=0$ for any $i>n$ and $M,N$ any
$R$-module?
Is it possible to choose $pd(M)\leq n$ to apply pd Lemma 4.1.6, p.93 in
Weibel's Introduction to Homological Algebra ?
I have an exact sequence \begin{align} 0 \rightarrow A \rightarrow P_2
\rightarrow P_1 \rightarrow P_0 \rightarrow
\mathbb{Z}_{(p)}/t\mathbb{Z}_{(p)}\rightarrow 0 \end{align} $P_i$ are
projective $\mathbb{Z}_{(p)}$-modules. I guess that somehow I apply
Weibel's Lemma to gain $A$ to be projective too. Applying
$\mathbb{Z}/p\mathbb{Z} \otimes_{\mathbb{Z}_{(p)}}-$ yields to
\begin{align} 0 \rightarrow \mathbb{Z}/p\mathbb{Z}
\otimes_{\mathbb{Z}_{(p)}}A \rightarrow \mathbb{Z}/p\mathbb{Z}
\otimes_{\mathbb{Z}_{(p)}}P_2 \rightarrow \mathbb{Z}/p\mathbb{Z}
\otimes_{\mathbb{Z}_{(p)}}P_1 \rightarrow \mathbb{Z}/p\mathbb{Z}
\otimes_{\mathbb{Z}_{(p)}}P_0 \rightarrow
\mathbb{Z}/p\mathbb{Z}\rightarrow 0 \end{align} Why is the first arrow
injective?