Line bundles, linear systems and normalization - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-20T10:18:55Z http://mathoverflow.net/feeds/question/74742 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/74742/line-bundles-linear-systems-and-normalization Line bundles, linear systems and normalization Jodel 2011-09-07T12:48:06Z 2011-09-07T15:03:41Z <p>One example that I always have in mind is that the plane nodal (or even the plane cuspidal) cubic curve $X$ is obtained by an appropirate 2-dim linear subsystem of $|\mathcal{O} (3)|$ on $\mathbb{P}^1$. If one takes the full linear system $|\mathcal{O}_{\mathbb{P}^1} (3)|$ then we get the twisted cubic $\tilde{X}$ in $\mathbb{P}^3$, which can be seen as the normalization both of the nodal and of the cuspidal plane cubic. </p> <p>This motivates my question: is it true that, if a non-normal variety is projective, then the normalization is still projective? If the answer to this question is positive, can one always see the maps to the non-normal variety and to its normalization as given by sections of the same line bundle?</p> http://mathoverflow.net/questions/74742/line-bundles-linear-systems-and-normalization/74750#74750 Answer by Jack Huizenga for Line bundles, linear systems and normalization Jack Huizenga 2011-09-07T13:37:40Z 2011-09-07T15:03:41Z <p>It is in fact true that the normalization of a projective variety is projective, as J.C. Ottem discusses in the comments.</p> <p>It is not true that if a normal variety is mapped to a projective space by a linear series $V\subset H^0(L)$ then some larger linear series $W\supset V$ has image isomorphic to the normalization.</p> <p>For instance, let $C$ be a general smooth curve of genus $g \gg 0$, and pick a general line bundle $L$ of degree $g+2$. By Riemann-Roch, $h^0(L) = 3$, and thus the map induced by $|L|$ maps $C$ to $\mathbb{P}^2$. For large enough $g$, however, the general curve of genus $g$ is not isomorphic to a smooth plane curve, and thus the image cannot be smooth. Moreover, we're using the complete series of sections of $|L|$ already, so there aren't "more sections" to include.</p> <p>However, the following modification <em>is</em> true. Suppose $X\subset \mathbb{P}^n$ is a variety, with normalization $f:\overline{X}\to X$. Then there <em>exists</em> a line bundle $L$ on $X$ such that $f$ is given by a collection of sections of $L$ and the complete series $|L|$ gives an embedding of $\overline{X}$ into some big projective space.</p> <p>Roughly, if $L = f^* \mathcal{O}(1)$, we can modify $L$ by adding a sufficiently ample divisor $nH$ so that $L+nH$ gives an embedding. But if $V \subset H^0(L)$ corresponds to $f$, then multiplying by a fixed section $D$ of $nH$ gives us an inclusion $D + V \subset H^0 (L+nH)$; note that this series has $D$ as a base locus. The map corresponding to this series is just $f$, realized as a projection from the big projective space which $L+nH$ maps $\overline{X}$ to. </p>