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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.

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?

Jodel
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  • By a normalization I take it you mean a birational finite birational morphism from a normal variety to $X$? In that case I think following works: Take a projective embedding $X\to \mathbb{P}^n$ and let $A$ be the homogenous coordinate ring of $X$. The integral closure $B$ has a veronese subring $B^{(d)}$ which is generated in degree 1 and which is the integral closure of $A^{(d)}$. Then a normalization map is given by $Proj(B)=Proj(B^{(d)})\to Proj(A^{(d)})=Proj(A^{(d)})$. – J.C. Ottem Sep 07 '11 at 13:25
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    In fact, I guess you can show projectivity of the normalization by finding an ample line bundle on $\bar{X}$. If $f:\bar{X}\to X$ is the normalization, it is finite and surjective (by going up/down) and so $f^*L$ is ample on $\bar{X}$ whenever $L$ is ample on $X$. – J.C. Ottem Sep 07 '11 at 13:28
  • I suppose properness of $\overline{X}$ is also needed, but that also follows from finiteness of $f$. – Jack Huizenga Sep 07 '11 at 14:35

1 Answers1

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It is in fact true that the normalization of a projective variety is projective, as J.C. Ottem discusses in the comments.

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.

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.

However, the following modification is true. Suppose $X\subset \mathbb{P}^n$ is a variety, with normalization $f:\overline{X}\to X$. Then there exists 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.

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.

Jack Huizenga
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