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Apr 13, 2017 at 12:58 history edited CommunityBot
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Sep 7, 2011 at 18:09 comment added user565739 @ Karl, yes and it should be finite as a $k$-vector space here to get the conclusion also. And I think that finite algebra is the common terminology for algebras which are finite as modules (over base ring), see p.30 of Atiyah & Macdonald's book "Introduction to Commutative Algebra"
Sep 7, 2011 at 17:14 comment added Karl Schwede ulrich, a dumb question when you say that ``$A$ is a finite $k$-algebra'', you mean a finite as a $k$-vector space, not finitely generated right?
Sep 7, 2011 at 14:28 comment added Karl Schwede Sorry, in my first comment I misread what was said. I got $f$ and $f'$ confused.
Sep 7, 2011 at 11:27 comment added naf You need to assume that the fibres are geometrically connected in order to get $g$ to be birational (in characteristic $0$). Over a field of characteristic $p>0$ it suffices to assume that the fibres are geometrically connected and geometrically reduced. This follows from the elementary fact that if $k$ is a field and $A$ is a finite $k$-algebra then $Spec(A)$ is geometrically connected and reduced iff $A = k$ (as a $k$-algebra).
Sep 7, 2011 at 10:28 history edited user565739 CC BY-SA 3.0
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Sep 7, 2011 at 9:38 history edited user565739 CC BY-SA 3.0
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Sep 7, 2011 at 7:43 comment added J.C. Ottem 1), 2) The fact that $g$ is birational follows from generic smoothness in characteristic zero and the reference is Hartshorne III.10.7 (and not III.10.3). I don't know if there is an analogue to the characteristic $p$ case.
Sep 6, 2011 at 20:46 history asked user565739 CC BY-SA 3.0