# How to show fraction field is flat (without localization)

Here I asked that if one can prove the field of fraction of a domain is flat. The answers used localization, which I am not familiar with. Can anyone prove it without using localization?

• It's well worth it to learn about localization. It's a direct generalization of (tensoring with) the fraction field, and people use it constantly. t's not like it's some obscure abstract thing. – Qiaochu Yuan Apr 30 '15 at 18:50

To prove that an $R$-module $M$ is flat, it suffices to show that for every ideal $I \subset R$, the canonical map $I \otimes_R M \rightarrow M$ is injective.

When $R$ is a domain and $M$ is the field of fractions of $R$, we have that every element of $I \otimes_R M$ is expressible as a simple tensor, that is, $i \otimes m$ for some $i \in I$ and $m \in M$. This is a pleasant exercise in finding a common denominator.

With this, we must show that $i \otimes m \rightarrow im$ is an injective map. But $im = 0$ if and only if $i = 0$ or $m = 0$ in the field of fractions of $R$, and this is if and only if $i \otimes m = 0$.

Rotman in the book "An Introduction to Homological Algebra" has (Corollary 5.35 ii):

• Are we to rest assured the proof does not use localization? – rschwieb Apr 30 '15 at 18:42
• @rschwieb: The proof of prop 5.34 certainly does not use localization. However the only way I know how to express the fraction field of a domain as a direct limit is over the localizations at various nonzero elements. – RghtHndSd Apr 30 '15 at 18:46
• @RghtHndSd I guess that should have been included in my concern as well. – rschwieb Apr 30 '15 at 19:03

A long time passed, but I think the following argument deserves to be reported here as well and, in my opinion, it may be interesting in its own. It is the most elementary proof of the flatness of the field of fractions I have ever seen.

Let $$R$$ be a domain and $$Q$$ be its field of fractions. Let $$j:M \to N$$ be an injective morphism of left $$R$$-modules. Assume that there exists an element in $$\ker(Q \otimes_R j)$$. It is of the form $$\frac{1}{q} \otimes_R m$$ for some $$q\in R$$ and $$m\in M$$, because every element in $$Q\otimes_R M$$ is of that form.

Now, consider the $$R$$-linear morphism $$\lambda:R \to M, \quad r\mapsto r\cdot m,$$ and the composition $$j\circ\lambda:R \to N, \quad r\mapsto r\cdot j(m) = j(r\cdot m).$$ If $$j\circ \lambda$$ is injective, then $$R \cong R\cdot j(m)$$ as left $$R$$-modules and hence $$Q \cong Q\otimes_R R \cong Q\otimes_R R\cdot j(m)$$ via $$\frac{p}{q}\mapsto \frac{p}{q} \otimes_R j(m)$$. In particular, this would imply $$\frac{1}{q} \otimes_R j(m) \neq 0$$, which contradicts the foregoing. Therefore, $$j\circ \lambda$$ cannot be injective and, since $$j$$ is injective, we conclude that $$\lambda$$ cannot not injective: there exists $$r\in R$$ such that $$r\cdot m = 0$$. As a consequence, $$\frac{1}{q} \otimes_R m = \frac{r}{rq} \otimes_R m = \frac{1}{rq} \otimes_R r\cdot m = 0$$ and $$Q \otimes_R j$$ is injective.