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given a ring $R$, a nilpotent ideal $I$ and a morphism $\phi$ of $R$-modules $M \to N$, such that $M/IM \to N/IN$ is an isomorphism.

It is easy to see that this implies $\phi$ surjective, but what about injectivity?

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    "/nilpotent/"? You mean to use *italics*? (`*italics*`)2010-12-31
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    Why/How is $I$ a submodule of $M$ and $N$?2010-12-31
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    $M/I$ denotes $M/IM$.2010-12-31

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The answer is no in general. E.g. if $M = R$ and $N = R/I$, then the natural surjection $M \to N$ will induce an isomorphism $M/I \to N/I$, which won't be an isomorphism unless $I = 0$.

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    Thank you. Now I have a question to Waterhouse, Group Schemes, Theorem 14.4, second paragraph: Why is it enough to show that $C/I_BC \to A/I_BA$ is an isomorphism?2010-12-31
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    @user5262: Dear 5262, You're welcome. Regarding Waterhous, I don't have a copy of the book at hand, so I can't tell you without more information. Regards,2010-12-31
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    I will start another thread and type the relevant pieces of Waterhouse up.2010-12-31
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    Here it is: http://math.stackexchange.com/questions/16000/question-regarding-waterhouse-affine-group-schemes2011-01-02