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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?

3 Answers 3

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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.

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    so I can define $\phi(m_ix^i + m_jx^j)$ to be equal to $\phi(m_ix^i) + \phi(m_jx^j)$? (this was my biggest problem, to prove that $\phi$ is additive if I define it as I did). – 2017-01-15
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    Yes, you can :) – 2017-01-15
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    So way bother about proving that $\phi(rp) = r\phi(p)$? Can't we say that 'we defined $\phi$ to be R-bilinear' and that's it? – 2017-01-15
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    You defined $\phi(mX^i)$ and $\phi(rmX^i)$ separately. You have to show these definitions don't contradict each other. – 2017-01-15
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    oh, oke... I guess I'm a little confused about additivity following from the definition I gave, I mean: does it really follow from the definition I gave (I do not see this) or do I have to write that I define $\phi$ to be additive? – 2017-01-15
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    Thank you very much for the edit, Now things are really clear to me (they should have been before, since it is quite obvious now that I see it!) – 2017-01-15
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    Yes, I made an edit, but now you made me unsure myself... :D Perhaps you do need to prove at least $\phi((m+m')X^i) = \phi(mX^i) + \phi(m'X^i)$. – 2017-01-15
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    Let us [continue this discussion in chat](http://chat.stackexchange.com/rooms/51825/discussion-between-student-and-ryanblack). – 2017-01-15
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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 $$\coprod_{i\geq 0} R\ \underset{R}{\otimes}\ M\cong \coprod_{i\geq 0}( R\ \underset{R}{\otimes}\ M)\cong \coprod_{i\geq 0}M= M[x]$$

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    Oh that is indeed a nice way to prove the problem I had... Do you however have any idea on how to 'fix' the map I have tried to use? – 2017-01-15
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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.