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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?

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pink floyd
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    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

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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 $i \otimes m = 0$.

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

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user 1
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    Are we to rest assured the proof does not use localization? – rschwieb Apr 30 '15 at 18:42
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    @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