The natural map $M \to M \otimes_R K$ is injective iff $M$ is torsion free

Question

I'm reading some lecture notes of Pete L. Clark, and there's one problem that I cannot solve. It's on page 45 of this book: Commutative Algebra. The problem reads as follow:

Exercise 3.42

Let R be a domain with fraction field K.

1. Show that K is a uniquely divisible $R-$module. (I think I get it, but I'm not sure, isn't this the way we construct a fraction field from a domain?)

2. Let $M$ be any $R-$module. Show that the natural map $M \otimes M \otimes_R K$ is injective iff $M$ is torsion-free.

I think the author made a little typo here, it should read $M \to M \otimes_R K$, instead of $M \otimes M \otimes_R K$, does my correction look good? Or should it be corrected in another way?

This is where I have no idea, I can prove $M \to M \otimes_R K$ is injective $\Rightarrow$ $M$ is torsion-free by Proof by Contradiction. But how can I prove the other way round? I would be very glad if someone can give me a push on this.

3. Show that for any $R-$module $M$, $M \otimes_R K$ is uniquely divisible.

I can show that it's divisible, but do I need to use 2. to show that it's uniquely divisible?

4. Show that $K / R$ is divisible, but not uniquely divisible.

I get this part. :)

I think I'm having some troubles with Tensor Product, it's pretty hard to visualize what its elements are, such as I have no idea what 0 looks like. I do know that $0 \otimes k = m \otimes 0 = 0$, but that's not all the 0's in $M \otimes_R K$.

So, thanks every one,

And have a good day,

Answer 1

In this case, $M\otimes_R K$ is the localization $S^{-1}M$, where $S=R\setminus\lbrace 0\rbrace$. Let $\varphi:M\rightarrow S^{-1}M$, $\varphi(x)=\frac{x}{1}$. Then:

$$ \varphi(x)=\frac{x}{1}=\frac{0}{1}\Longleftrightarrow\exists~0\ne r\in R\text{ such that }r\cdot(1\cdot x-1\cdot 0)=r\cdot x=0 $$

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To answer your question about where the equivalence comes from:

Proposition. Let $R$ be a ring, $S\subset R$ a multiplicatively closed set, and let $M$ be an $R$-module. Then $S^{-1}M\cong M\otimes_R S^{-1}R$.

Proof. Let $\varphi:M\times S^{-1}R\rightarrow S^{-1}M$, $\varphi(m,\frac{r}{s})=\frac{rm}{s}$. One can show that this map is bilinear, and so induces a well-defined map $\psi:M\otimes_R S^{-1}R\rightarrow S^{-1}M$, $\psi(m\otimes\frac{r}{s})=\frac{rm}{s}$ by the universal property. Finally, we check that $\psi$ is an isomorphism:

Surjectivity. $\psi(m\otimes\frac{1}{s})=\frac{m}{s}$.

Injectivity. First note that every element of $M\otimes_R S^{-1}R$ has the form $m\otimes\frac{1}{s}$, since: $$ (m_1\otimes\frac{r_1}{s_1})+\cdots+(m_n\otimes\frac{r_n}{s_n})=(m_1\otimes\frac{r'_1}{s})+\cdots+(m_n\otimes\frac{r'_n}{s}) $$ where $s=s_1\cdots s_n$ (just putting the fractions over a common denominator). But: $$ (m_1\otimes\frac{r'_1}{s})+\cdots+(m_n\otimes\frac{r'_n}{s})=(r'_1m_1+\cdots+r'_nm_n)\otimes\frac{1}{s} $$ from the properties of the tensor product. Now suppose that $\psi(\frac{1}{s}\otimes m)=0$. Then $\frac{m}{s}=0$, and $tm=0$, for some $t\in S$. Finally: $$ m\otimes\frac{1}{s}=m\otimes\frac{r}{ts}=tm\otimes\frac{1}{ts}=0\otimes\frac{1}{ts}=0 $$ and $\psi$ is injective. QED