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What do I miss in the following?

Let $R$ be a commutative Noetherian ring with unit. A map $f:M\to N$ of $R$-modules is injective/surjective iff the associated map $f_p:M_p\to N_p$ on the localization is injective/surjective for every prime ideal $p$ of the ring $R$.

There is an isomorphism $M_p=M\otimes_R R_p$. A sequence $$0\to K\xrightarrow{k} M\xrightarrow{f} N\to 0$$ is exact iff $k$ is injective and $s$ is surjective. Hence the above sequence is exact iff for every prime ideal $p$ of $R$ the sequence $$0\to K\otimes_R R_p\xrightarrow{k'} M\otimes_R R_p\xrightarrow{f'} N\otimes_R R_p\to 0$$ is exact. But $R_p$ is not flat over $R$ in general, is it? What am I missing? Is the sequence still exact if I tensor with the residue field $R_p/m_p$?

Thank you!

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$R_p$ is flat over $R$ in general. – Robin Chapman Nov 6 '10 at 10:24
Thank you! What about the residue field $R_p/m_p$? – Zerodivisor Nov 6 '10 at 10:29
Is $\mathbb{Z}/p\mathbb{Z}$ flat over $\mathbb{Z}$? – Robin Chapman Nov 6 '10 at 10:39
It is not. Thanks. Can one test injectivity/surjectivity of the morphism $M\to N$ at the injectivity/surjectivity of all morphisms $M\otimes_R R_p/m_p\to N\otimes_R R_p/m_p$? – Zerodivisor Nov 6 '10 at 12:01
all your questions are answered in pretty much any book treating this subject. – Mariano Suárez-Alvarez Nov 6 '10 at 16:38
up vote 2 down vote accepted

Your characterisation of an exact sequence is missing exactness at $M$.

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Thanks, ok. Do you perhaps know if I can test injectivity/surjectivity with the residue fields $R_p/m_p$, i.e. is $M\to N$ injective/surjective iff $M\otimes_R R_p/m_p\to N\otimes_R R_p/m_p$ injective/surjective for every prime ideal $p$ of $R$? – Zerodivisor Nov 6 '10 at 10:31

Exact sequence of $R=\mathbb Z$ modules $0\to \mathbb Z \to \mathbb Q$ becomes not exact after choice $p=2\mathbb Z$ and tensor with $R_p/m_p =\mathbb Z /(2) $ because it becomes $0 \to \mathbb Z /(2)\to 0$

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