Let $f:\tilde{X} \to X$ be a normalization of projective variety. Let $L$ be a very ample line bundle on $X$. Is $f^*L$ a very ample line bundle on $\tilde{X}$? If not true in general, is there any dimension restriction on $X$ for which this is true?
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1$\begingroup$ Not in general, because the sections of $f^*L$ need not separate points. E.g. take a nodal curve in the plane, and then normalize. $\endgroup$– Donu ArapuraCommented Apr 19, 2014 at 12:55
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$\begingroup$ Never, sort of definitionally, unless $X$ is normal . It does stay ample. $\endgroup$– mehCommented Apr 19, 2014 at 12:56
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$\begingroup$ I don't understand the two comments above. Take $\mathcal O(1)$ on $\mathbb P^2$ and restrict to a nodal cubic $X$. Then $f^* L$ is degree $3$ on $\tilde X$, hence very ample, no? $\endgroup$– Allen KnutsonCommented Apr 19, 2014 at 14:37
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$\begingroup$ OK, yes, my 2nd sentence was not carefully formulated. I was thinking of a scenario along the lines of abx 2). $\endgroup$– Donu ArapuraCommented Apr 19, 2014 at 14:56
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$\begingroup$ Mine too :) . I think abx took the time to think out the question more thoroughly. $\endgroup$– mehCommented Apr 20, 2014 at 12:46
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1 Answer
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1) There are certainly many cases where $f^*L$ is still very ample: e.g. if $X$ is an irreducible curve, any line bundle on $X$ of degree $\geq 2g(\tilde{X} )+1$ will have this property.
2) A typical situation where $f^*L$ is not very ample is when $H^0(\tilde{X},f^*L )\cong H^0(X,L)$. For instance, take for $X$ a curve with one ordinary double point $s$, and $L=\omega _X(-p)$, $p$ a smooth point of $X$. Then $f^*L=\omega _{\tilde{X}}(s'+s''-\tilde{p} )$, where $s',s''$ are the 2 points of $\tilde{X} $ above $s$ and $\tilde{p} $ the point above $p$. Riemann-Roch gives $h^0(f^*L)= g(\tilde{X} )=h^0(L)$ (I choose $p$ so that $h^0(s'+s''-\tilde{p} )=0$).
I think there is no simple criterion allowing to decide which situation occurs, you have to work.
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$\begingroup$ Thanks for the answer. Could you indicate how to proceed in the case of irreducible surfaces with at most rational singularities (at isolated points). $\endgroup$ Commented Apr 24, 2014 at 12:06
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$\begingroup$ But aren't these surfaces already normal? $\endgroup$– abxCommented May 5, 2014 at 5:31