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Given $\mathbb{P}^k$ and $\mathbb{P}^{\ell}$, the Segre variety is projective subvariety of $\mathbb{P}^{(k+1)(\ell+1)-1}$ defined as the common vanishing set of quadratic polynomials which can be defined as the $2 \times 2$ minors of a $(k+1)\times (\ell +1)$ matrix.

For geometric reasons, I would expect this projective variety to have dimension $k + \ell$, and thus to be defined by $[(k+1)(\ell+1)-1]-k + \ell=k\ell$ polynomials.

However, combinatorially I would expect the number of $2 \times 2$ minors of a $(k+1)\times (\ell +1)$ matrix to be $\binom{(k+1)}{2}\cdot \binom{(\ell +1)}{2} = \frac{1}{4}(k^2+k)(\ell^2+\ell)$, thus the Segre variety is defined by that many polynomials.

But $\frac{1}{4}(k^2+k)(\ell^2+\ell) \not= k\ell $. (Take, e.g., $k=4, \ell=8$.)

Question: Where are the errors in my reasoning?

If there are no errors in my reasoning, then does this imply that some of the defining polynomials for the Segre variety are redundant? Why or why not? If so, which ones? ("Soft" answers would be sufficient.)

(Would this be a case where, for degeneracy reasons, we can't use the Implicit function theorem to easily use the minimum number of defining polynomials to compute the dimension of the variety at each point?)

There are almost certainly major mistakes in my reasoning, so I apologize in advance for the embarrassingly stupid question.

Arkady
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Chill2Macht
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    That should be dimension $k+\ell$, no? Take $k = \ell = 1$; then $\mathbb{P}^1 \times \mathbb{P}^1$ is a surface, cut out in $\mathbb{P}^3$ by the equation $xw - yz = 0$. Moreover, this agrees with how things usually work: $\mathbb{R}^m \times \mathbb{R}^n \cong \mathbb{R}^{m+n}$. – Viktor Vaughn Aug 18 '17 at 05:36
  • @Quasicoherent You are 100% correct, the dimension should be $k + \ell$, I am stupid and don't know why I made that mistake -- I will edit. Thank you for pointing this out. – Chill2Macht Aug 19 '17 at 16:08

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What you are noticing is that it is not, in general, true that the codimension of a projective (or affine) variety is equal to the number of equations used to define it (or even the minimum number of generators of it's ideal). When the codimension is equal to the minimum number of generators of the ideal of the variety, it is called a complete intersection and is quite special. See examples and non-examples here on the wikipedia page.

Arkady
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  • As a sanity check, the dimension of the Segre variety is $k \ell$, right? Also the non-example given in Wikipedia is locally defined by only codimension many polynomials, i.e. is a local complete intersection: in general projective varieties aren't even local complete intersections, correct? The stacks project talks about "local complete intersections" in terms of commutative algebra, but doesn't translate it into geometric language -- it says a Noetherian local ring is a local complete intersection, so unless (for Noetherian coefficients) the projective variety corresponds to – Chill2Macht Aug 18 '17 at 02:58
  • a local ring, it cannot be a local complete intersection? http://stacks.math.columbia.edu/tag/00S8 This webpage (also by a Ravi) seems to be discussing a similar topic: https://math216.wordpress.com/2012/09/05/twenty-third-post-local-complete-intersections/ EDIT: here is the translation to geometric language it seems: https://mathoverflow.net/questions/131454/local-complete-intersection EDIT: if the Segre variety is smooth, apparently it is a local complete intersection https://math.stackexchange.com/questions/2002667/smooth-varieties-are-local-complete-intersection – Chill2Macht Aug 18 '17 at 02:58
  • This book says that the Segre variety is smooth (also dually degenerate): https://books.google.com/books?id=ThrvBwAAQBAJ&pg=PA71&lpg=PA71&dq=dually+degenerate+smooth+variety+Segre+variety&source=bl&ots=oye6mzToBb&sig=5GDnBryjjBIX8B5qU5DyEwDPeF4&hl=en&sa=X&ved=0ahUKEwiJgsaM59_VAhXHjFQKHULoCroQ6AEINzAD#v=onepage&q=dually%20degenerate%20smooth%20variety%20Segre%20variety&f=false so it would seem that the Segre variety is a local complete intersection, but not itself a complete intersection. This would be an interesting exercise for me to prove one day - I don't think I'm quite there yet though – Chill2Macht Aug 18 '17 at 03:11
  • @Chill2Macht Cheers! Glad you did some searching by yourself and figured it out. – Arkady Aug 18 '17 at 14:45