Proof of $R[X] \otimes_R M \cong M[X]$ In my exercise session of commutative algebra I had the following problem: show that $R[X] \otimes_R M \cong M[X]$. There was a hint: show that $M[X]$ satisfies the universal property of tensor products. Here $R$ is a commutative ring with unity and $M$ is an $R$-module. This is my attempt:
I have defined a bilinear map
$$h: R[X] \times M \to M[X]: (\sum_{i = 0}^{n}r_ix^i, m) \mapsto \sum_{i = 0}^{n}(r_im)x^i.$$ Let us now consider a bilinear map $f: R[X] \times M \to P$, where $P$ is any $R$- module. I want to construct a linear map $\phi: M[X] \to P$ such that $\phi \circ h = f$. If a can show that this $\phi$ is unique, then I am finished, since the universal property then gives that $M[X]$ is isomorphic to $R[X] \otimes_R M$. However, I have troubles defining the map $\phi$: I want to write something like 
$$\phi= M[X] \to P: m_ix^i \to f(x^i, m_i),$$ but how to show that this map is a morphism?
 A: We know that it is possible to write
$R[x]=\{ (a_{i})\in \prod_{i\geq 0}R\ \mid \text{with the exception of a finite number of}\ a_{i},\ \text{for all}\ i\geq 0,\ a_{i}=0\}=\underset{i\geq 0}{\coprod} R$
Then we'll have 
$$\coprod_{i\geq 0} R\ \underset{R}{\otimes}\ M\cong \coprod_{i\geq 0}( R\ \underset{R}{\otimes}\ M)\cong \coprod_{i\geq 0}M= M[x]$$
A: There is indeed an obvious choice for a bilinear map
$$
h\colon R[x]\times M\to M[x]
$$
where
$$
h\biggl(\,\sum_{i=0}^k r_ix^i,m\biggr)=\sum_{i=0}^k r_imx^i
$$
Now, let $f\colon R[x]\times M\to N$ be a bilinear map. We need to find $\phi\colon M[x]\to N$ such that $f=\phi\circ h$.
Define
$$
\phi\biggl(\,\sum_{i=0}^k m_ix^i\biggr)=
\sum_{i=0}^k f(x^i,m_i)
$$
You should have no problem in proving it is $R$-linear, using bilinearity of $f$.
Next,
$$
\phi\circ h(x^i,m)=\phi(mx^i)=f(x^i,m)
$$
and use again bilinearity.
A: Say, $p = \sum m_iX^i$ and $p' = \sum m'_iX^i$. Then $p+p' = \sum (m_i+m'_i)X^i$. By definition:
$$\phi(p+p') = \sum \phi((m_i+m'_i)X^i) = \sum f(X^i,m_i+m'_i) $$
$$ \stackrel{\star}{=}\sum f(X^i, m_i) + \sum f(X^i, m'_i) = \phi(p) + \phi(p')$$
Let $r\in R$ and $p = m_nX^n+\ldots +m_1X^1 + m_0\in M[X]$. Then:
$$\phi(rp) = \sum_{i=0}^n \phi(rm_nX^n) = \sum_{i = 0}^n f(X^n,rm_n) \stackrel{\star}{=} r \sum_{i=0}^n f(X^n,m_n) = r \sum_{i=0}^n \phi(m_nX^n) = r \phi(p)$$
where $\star$ hold because $f$ is $R$-bilinear. This makes $\phi$ a homomorphism.
