Let R be a ring (unital, though not necessarily commutative, and let: \begin{array}{ccccccccc} 0 & \hookrightarrow & M & \overset{f}{\hookrightarrow} & N & \overset{g}{\twoheadrightarrow} & P & \twoheadrightarrow & 0\end{array} be a short exact sequence of left R -modules.
1) Let Q be a right R -module. Consider the exact sequence:$$\begin{array}{ccccccc} Q\otimes_{R}M & \overset{\textrm{Id}\otimes f}{\longrightarrow} & Q\otimes_{R}N & \overset{\textrm{Id}\otimes g}{\rightarrow} & Q\otimes_{R}P & \twoheadrightarrow & 0\end{array}$$
Show that $\textrm{Id}\otimes f$ is injective whenever Q is a projective right R -module. Give an example where $\textrm{Id}\otimes f$ is not injective.
2) Let Q be a left R -module. Consider the exact sequence: $$0\hookrightarrow\textrm{Hom}_{R}\left(Q,M\right)\begin{array}{cccc} \overset{f_{*}}{\longrightarrow} & \textrm{Hom}_{R}\left(Q,N\right) & \overset{g_{*}}{\rightarrow} & \textrm{Hom}_{R}\left(Q,P\right)\end{array}$$
Show that $g_{*}$ is surjective whenever Q is a projective left R -module.
3) Let $\phi:R\rightarrow S$ be a homomorphism of commutative rings. Show that there is an isomorphism of S -algebras: $$R\left[x_{1},\ldots,x_{N}\right]\otimes_{R}S\cong S\left[x_{1},\ldots,x_{N}\right]$$
I've been working probably a total of ten hours on these three problems, all to no avail whatsoever. I've been given hints by my instructor, (such as for the first problem, to consider first the case where Q is a free right R-module), but I haven't been able to get anywhere with them.
As someone with a penchant for pedantry, good pedagogy, specificity concreteness, and explicit constructions, much of what we have been doing in algebra as of late has left me feeling adrift and at a loss. I spend hours trying to come up with a formula for the isomorphism in (3), only for the professor to enigmatically refer to “cokernels” or “the universal property of S -algebras”—the former of which I know not how to use; the latter of which we never even discussed in class (or, if we did, we discussed it under a different name, and no one bothered to point out that it was equivalent to a “universal property of S-algebras”.
Help (preferably in the form of solutions that I can spend time trying to wrap my head around) would be INFINITELY appreciated.
First statement
Sketch of proof.
Since a projective module is a direct summand of a free module, putting together 1, 2 and 3 gives the proof.
A counterexample is with $R=\mathbb{Z}$, $M=N=\mathbb{Z}$, $f$ the multiplication by $2$, $P=\mathbb{Z}/2\mathbb{Z}$ and $g$ the canonical projection. If you take $Q=\mathbb{Z}/2\mathbb{Z}$, the map $\mathrm{Id}\otimes f$ is the zero map.
Second statement
This is just a reformulation of the definition of projective module.
Third statement
We can be reduce to the case when $N=1$, applying induction for the general case. I'm afraid that the isomorphism $R[x]\otimes_R S\to S[x]$ is essentially the universal property of the tensor product together with the universal property of the polynomial ring in one indeterminate, but here's a sketch.
You want to show that the ring $R[x]\otimes_RS$ satisfies the required property for being $S[x]$. Suppose $f\colon S\to T$ is any ring homomorphism and $b\in T$. You want to prove that there exists a unique ring homomorphism $f_b\colon R[x]\otimes_RS\to T$ such that
Consider $f\circ\phi\colon R\to T$. Then, by the properties of $R[x]$, there exists a unique ring homomorphism $g\colon R[x]\to T$ such that $g(r)=f\circ\phi(r)$, for $r\in R$, and $g(x)=b$.
Now define $f_b\colon R[x]\otimes_RS\to T$ by $$ f_b(P\otimes s)=g(P)f(s) $$ (use the universal property of tensor products) which is a ring homomorphism (prove it). Then $f_b(x\otimes 1)=g(x)1=b$ as required. Also $$ f_b\circ j(s)=f_b(1\otimes s)=g(1)f(s)=f(s) $$ Complete the proof by showing the uniqueness of $f_b$.