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Let ring A and B be a flat over k[t]/(t^{2}).

if there are morphism f from A to B then, f should be isomorphism?

if so, why?

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6 
 There is no reason that $f$ should be an isomorphism. – Kevin Buzzard May 16 2011 at 18:01
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 It's hard to answer this question because nothing like this is true. For example, $k[t]/(t^2)$ is flat over itself and $k[t, x]/(t^2)$ is also flat over $k[t]/(t^2)$, and there are plenty of maps between them. Could you give us some context about why you think this? – David Speyer May 16 2011 at 19:10
Since your question is about to get closed: I suggest that you follow David Speyer's advice, edit your question to clarify what it is you think is true, and then use the 'flag' option to contact a moderator asking if the question can be re-opened. – Yemon Choi May 16 2011 at 22:27

closed as off topic by Dan Petersen, Karl Schwede, Bugs Bunny, Yemon Choi, Andreas Blass May 16 2011 at 22:56

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