# Sándor Kovács

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bio website math.washington.edu/~kovacs location Seattle, WA age member for 3 years, 5 months seen 18 hours ago profile views 6,210
I am an algebraic geometer.

# 1,948 Actions

 Mar9 revised Making Hironaka's theorem explicit for hypersurfaces deleted 1 characters in body Mar8 answered Making Hironaka's theorem explicit for hypersurfaces Feb20 comment Bertini type theorem No,of course not. I was just answering to what you wrote. Sure, you need to resolve the indeterminacies. So it is a projective closure somewhere else. I don't see the problem. I should edit what I wrote, because it maybe ambiguous. I'll do that soon. Feb20 comment Bertini type theorem Smoothness is not needed, so there is no need for a resolution. Just take a projective closure. Feb19 comment Ample divisors on $\mathbb{P}^3$ blow-up along single point Uh, did you mean that the blow up is "Fano" instead of "ample"? Feb19 comment Ample divisors on $\mathbb{P}^3$ blow-up along single point Actually, by the way, I don't understand your last sentence either. What does it mean that "the blow-up of one point in $\mathbb P^n$ is ample"? Feb19 comment Ample divisors on $\mathbb{P}^3$ blow-up along single point I think your first sentence requires a bit more explanation, because it is not true in general. Perhaps you can add that the cone of curves of the blow up has rank two and both rays are generated by actual curves, so what you claim follows from Kleiman's criterion. Feb18 answered The quotient of relative spec and proj Feb18 answered Sheaves with no cohomology Feb16 revised how does one understand GRR? (Grothendieck Riemann Roch) Updated screwed up LaTeX code. Feb13 comment Bertini type theorem In other words the intersection of an ample divisor with any effective cycle is always positive. Feb13 comment Bertini type theorem Actually for that you only need that $X\cap H$ is ample, but I wrote "very ample", because it is actually very ample. The reason is exactly what you are saying. If there were a fiber that $H$ did not intersect then the intersection cycle of $H$ and that fiber would be $0$. But then any other member of the linear system $|H|$ would have the same intersection cycle and hence all members would have to either contain or be disjoint from that fiber. If that happens, then the fiber would be mapped to a single point, so the morphism induced by $|H|$ would not be an embedding. Feb13 comment Bertini type theorem Yes, if $X$ is projective. Otherwise there may be a few hyperplanes that do not intersect $X$. I didn't want to bother with all the details... :) Feb12 comment Why is the standard flop a flop? Thanks! :)...... Feb12 comment Why is the standard flop a flop? Notice that I didn't say "$H$-flop". I said "$H$-flip" and that a "flop" is an $H$-flip for some $H$ such that the contraction os $K$-trivial. The latter is trivial on a Calabi-Yau. Very likely that anti-nef that becomes nef is his $H$. As far as contracting several curves go, you can always do the connected components separately. Otherwise, the relative Picard number one only matters for uniqueness. Also, very likely, the way the contractions are obtained, they can be decomposed into contracting one curve at a time. That gives you rel Picard number one. Feb12 comment Why is the standard flop a flop? I added more explanation to my answer to what you are asking for in your edit. Feb12 revised Why is the standard flop a flop? added explanation of uniqueness Feb12 answered Bertini type theorem Feb11 answered Why is the standard flop a flop? Feb9 revised Extend morphism between coherent sheaves in $\mathbb{P}^n$ added 1095 characters in body