# how to prove that $\mathbb Q$ is flat as a $\mathbb Z-$module [duplicate]

This question already has an answer here:

I know that $$Tor^{\mathbb z}_1(\mathbb Z, N) = 0$$ for any $$\mathbb Z-$$module, because free modules are flat. Then because $$Tor_1$$ is local, we have $$Tor_1^{\mathbb Q}(\mathbb Q, S^{-1}N) = 0$$, which is not useful at all. How should I use localization to prove this statement?

## marked as duplicate by Yanior Weg, Jyrki Lahtonen abstract-algebra StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); May 11 at 16:35

• I am seeking for a solution which does not use the theorem about torsion because we never did that @AlvaroMartinez – z.z May 11 at 1:10

You can think think of this as a result of the two following more general statements, which themselves should not be too difficult to prove: throughout the following discussion let $$S$$ be a multiplicative subset of a ring $$A$$.

$$1$$) Localization is exact; that is, if $$0\to M'\to M\to M''\to 0$$ is an exact sequence of $$A$$-modules, then the induced sequence $$0\to S^{-1}M'\to S^{-1}M\to S^{-1}M''\to 0$$ of $$S^{-1}A$$-modules is exact.

Just to be clear--what do we mean by "induced" sequence? If $$\varphi:M\to N$$ is a homomorphism of $$A$$-modules, then there is a natural map $$S^{-1}\varphi:S^{-1}M\to S^{-1}N$$ defined by $$(S^{-1}\varphi)(m/s):=f(m)/s$$.

$$2)$$ For any $$M$$ there is an isomorphism $$S^{-1}A\otimes_A M\cong S^{-1}M$$ of $$S^{-1}A$$-modules, which is "natural in $$M$$"; this means that if we have $$\varphi:M\to N$$, then there is a commutative diagram $$\require{AMScd} \begin{CD} S^{-1}A\otimes_A M @>{\text{id}\otimes\varphi}>> S^{-1}A\otimes_A N\\ @V{\cong}VV @VV{\cong}V\\ S^{-1}M @>>{S^{-1}\varphi}> S^{-1}N \end{CD}.$$

Combining these facts, you see that for any localization $$S^{-1}A$$, if we have an injective homomorphism $$\varphi:M\to N$$, then we will get a commutative diagram as above by fact $$(2)$$; the morphism $$S^{-1}\varphi$$ will be injective by $$(1)$$, and then commutativity will imply that $$\text{id}\otimes\varphi$$ is injective as well, proving $$S^{-1}A$$ is flat. Of course for your example we are just taking $$A=\Bbb Z$$ and $$S=\Bbb Z\smallsetminus\{0\}$$.

Any localization of a commutative ring $$R$$ is a flat $$R$$-module. For this special case, $$R= \mathbb Z$$ and the multiplicatively closed set is $$S=R\setminus \{0\}$$.

This is likely proved before proving that Tor "commutes" with localization so trying looking earlier on in whatever book/reference you are using.