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May 21, 2015 at 22:14 comment added Ricardo Andrade You can create bold text using the markdown syntax **bold text**. See here for more information on markdown. Please do not use mathjax to format text. Thanks.
May 21, 2015 at 21:59 review Close votes
May 22, 2015 at 8:26
May 21, 2015 at 7:07 comment added user5117 @Ritwik: yes, that's correct. Only those three are complete intersections.
May 20, 2015 at 14:14 comment added Ritwik @Rene: If I understand your (and Smith's) remark correctly, the answer to my question is no in general; in particular if $k=1,2,3,4$, $7$ or $8$? The answer is only yes if $k=0,5$ or $6$.
May 20, 2015 at 11:23 comment added user5117 @René: good point (about quadrics)!
May 20, 2015 at 11:20 comment added R.P. To add to Artie Prendergast-Smith's comment: this means that $X_k$ is a complete intersection iff $k = 0,5,$ or $6$ (the case $n=3,d=2$ he mentioned does correspond to a del Pezzo, but not to a blow-up of $\mathbb{P}^2$).
May 20, 2015 at 11:08 comment added user5117 "I think": yes, any projective variety can be embedded in some $\mathbf P^n$. "Can $X_k...$ ?" Assume wlog that all the polynomials have degree $\geq 2$. Use adjunction to calculate the anticanonical bundle of a complete intersection of $n-2$ things in $\mathbf P^n$. Observe that the answer is positive only for $n=3,d=2,3$ or $n=4,(d_1,d_2)=(2,2).$
May 20, 2015 at 11:08 comment added Lev Borisov Of course, $X_6$ is a cubic in $\mathbb P^3$. If you allow weighted projective spaces, then $X_7$ is a degree $4$ hypersurface in $\mathbb P(1,1,1,2)$. I can't think of other constructions off the top of my head.
May 20, 2015 at 10:39 history asked Ritwik CC BY-SA 3.0