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Jun 1, 2010 at 19:02 comment added Karl Schwede Yes, I do mean $(a^2*c - b^2)$, thanks. Yes, the total is singular along every point of the exceptional divisor, let me think about the other question (but again, it might be better as a separate or modification of the original question as David said).
Jun 1, 2010 at 18:28 comment added David E Speyer @unkown: You'll probably get better responses if you pose you ask this question as a separate question, not a comment.
Jun 1, 2010 at 15:59 comment added unknown The example is cool, the total space is singular at every point along the exceptional divisor right? The singularities are not detected over the closed point - but they would be if we look at some neighborhood, like k[a,b]/m^3 where m = (a,b) the maximal ideal. If X \rightarrow Y, and x \in X, y = f(x), if the fiber is smooth over every infinitessimal neighborhood of y, is X smooth at x?
Jun 1, 2010 at 15:33 comment added unknown Ah, I think you are right (though you mean b^2 in your formula). Thanks, is there any other condition on the situation (besides flatness) which would guarantee that the domain is smooth or is it just hopeless?
Jun 1, 2010 at 15:15 history edited Donu Arapura CC BY-SA 2.5
added 147 characters in body; added 80 characters in body
Jun 1, 2010 at 15:11 comment added Karl Schwede Modifying Donu's example very slightly, blow up the ideal $(x^2, y^2)$. You have two charts, one is $k[x, y, (y/x)^2] = k[a,b,c]/(a^2*c - b)$ and the other is the other obvious (and symmetric) one. This chart isn't smooth, the fiber over the closed point $(x, y)$ is $k[c]$.
Jun 1, 2010 at 15:06 comment added Donu Arapura OK, sorry. I misread your question.
Jun 1, 2010 at 15:00 comment added unknown I'm asking about the domain X, not the morphism f.
Jun 1, 2010 at 14:41 comment added unknown The blowup of a smooth variety along a smooth subvariety is still smooth though?
Jun 1, 2010 at 14:35 history answered Donu Arapura CC BY-SA 2.5