Let $R$ be a commutative ring with unity, as
$R[t]$ is a free $R-$ module it is faithfully flat. So by the property of faithfully flat modules we can say that $$ 0 \longrightarrow M \longrightarrow N \longrightarrow P \longrightarrow 0$$ is a short exact sequence of $R-$ modules iff $$ 0 \longrightarrow M\otimes R[t] \longrightarrow N\otimes R[t] \longrightarrow P\otimes R[t] \longrightarrow 0$$ exact as $R[t]-$ modules with the differentials being $*\otimes 1$.
I was wondering about this question put forward a bit differently. Let $$
0 \longrightarrow R[t]^n \longrightarrow^f R[t]^{n+m} \longrightarrow^g R[t]^m \longrightarrow 0
$$
be a short exact sequence of finitely generated free $R[t]-$ modules, from here can I construct a short exact sequence like $$0 \longrightarrow R^n \longrightarrow^{f'} R^{n+m} \longrightarrow^{g'} R^m \longrightarrow 0
$$ such that eventually $f$ and $g$ turnes out to be of the form $f' \otimes 1$ and $g' \otimes 1$ respectively.
I think if $f$ and $g$ are natural inclusion and projection then it is true. I would be greatful if you could help me out on this.Thanks in advance.
Best Answer
Let $e_1,\ldots, e_n$ be a basis for $R[t]^n$ and let $v_1,\ldots, v_m\in R[t]^{n+m}$ be such that $g(v_i)$ form a basis for $R[t]^m$. Then easy to see that $f(e_i), v_j$ form a basis for $R[t]^{n+m}$. So, take $R^n$ with basis $e_i$, $R^{n+m}$ with basis $f(e_i), v_j$ , $R^m$ with basis $g(v_i)$ with obvious maps as $f', g'$.