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

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,

• I'm very sorry, it's my bad, I opened so many tabs, and I messed up copying the wrong link, editted. Many thanks. :) But I cannot find any typos in my enounce? Can you please tell me, so that I can fix it? Thanks, Apr 20, 2013 at 11:32
• Does anybody know the answer about the other variation with $M\otimes_k M$ i.e. If $M$ is torsion-free the natural map $$(M\otimes_k M)\to (M\otimes_k M)\otimes_k K$$ is injective ? Jan 25, 2019 at 7:17
• I finally posted this as a question there Jan 25, 2019 at 12:55

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:

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

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

• I think I get it now, thanks very much :* But can you show me, or just give me a link on how to prove: $S^{-1}M = M \otimes_R S^{-1}R$? I find this result on wikipedia, but they don't give a proof to it. Apr 20, 2013 at 12:04
• @user49685 I've edited my question with a proof of the equivalence. :) Apr 20, 2013 at 12:29
• Thank you very much Warren Moore. :) Apr 20, 2013 at 14:32
• Hi @Warren Moore, can you explain why ∃ $0≠r$ such that $r⋅(1⋅x−1⋅0)=r⋅x=0$ is necessary? Why do you write out "$r⋅(1⋅x−1⋅0)$"? Dec 3, 2013 at 4:20
• @John I do not say that there definitely does not exist such an $r$. That would depend upon the module $M$. The answer I wrote was to the question "$M\rightarrow M\otimes_R K$ is injective if and only if $M$ is torsion-free", which is a simple consequence of the equivalence relation defining the localisation. Two fractions $a/b$ and $c/d$ are equal in $S^{-1}M$ if and only if there exists $0\ne r\in R$ such that $r(da-bc)=0$ - that is how the localisation is constructed. Dec 3, 2013 at 8:43