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It is known (see the MO question " Varieties cut by quadrics") that every projective variety can be realized as a scheme-theoretic intersection of quadrics. Is there a way to determine if the intersection of irreducible quadric hypersurfaces is irreducible? For example, consider equations of the following form: $$ (x_2-x_1)(y_{2i}-y_{1j})-(x_3-x_1)(y_{3k}-y_{1j})=0, $$ where $i=1, 2, \dots, M$, $j=1,2,\dots, N$ and $k=1, 2, \dots, R$, and $M$, $N$ and $R$ are fixed natural numbers. Is there a way to determine if the intersection is irreducible? More generally, what about equations of the following form:

$$ (x_n-x_m)(y_{ni}-y_{mj})-(x_r-x_m)(y_{rk}-y_{mj})=0, $$ where $m:n:r\neq 1:1:1$.

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What exactly are you looking for? The obvious (although sarcastic-sounding) answer is "Sometimes," and it's hard to improve on that answer without further information. –  Jack Huizenga Nov 30 '11 at 15:01
    
For example, take the two quadrics; $f = x^2 + y^2 + z^2 +w^2$ and $g = x^2 + y^2 - z^2 -w^2$ . Then the ideal $I =(f,g) = (x^2 + y^2,z^2 +w^2)$ which is 4 lines. Hard to find two nicer quadrics. –  aginensky Dec 6 '11 at 20:57
    
Hey, thanks for the comments. I have edited my question to reflect the concern. Thanks again! –  Fei YE Dec 8 '11 at 14:38
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What space does the intersection take place in, and how do $i,j,k$ vary? Also, where does this come from? Typically studying the geometric properties of a variety by studying its defining equations is a difficult thing to do. If these came up more naturally as determinantal varieties or something, that could perhaps be useful. –  Jack Huizenga Dec 8 '11 at 18:13
    
Hi Jack, You can assume that $i$, $j$, $k$ vary in the natural number set and intersection is taken in the complex affine space of those variables. The equations are determinants of $2\times 2$ matrices [\begin{vmatrix} x_2-x_1& y_{2i}-y_{1j}\\ x_3-x_1& y_{3k}-y_{1j} \end{vamtraix}.] –  Fei YE Dec 9 '11 at 15:36

2 Answers 2

It seems that you are taking 2x2 minors of a matrix. This gives you an irreducible determinantal variety. For more on the subject see the book of Harris, Algebraic Geometry or a paper of Eisenbud, Linear sections of determinantal varieties.

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Hey cdm80, you are absolutely right, the equations are determinants of $2\times 2$ matrices. Thank you for the references! –  Fei YE Dec 9 '11 at 15:46

This is not quite an answer to your question, but you may find it useful. Miles Reid's thesis has a chapter on intersections of two quadrics, which is already an interesting case. Associated to an intersection of two quadrics is a polynomial: express your two quadrics as symmetric matrices $A$ and $B$, and then look at $\textrm{det}(A + \lambda B)$. One of the results you'll find in Miles' thesis (Proposition 2.1) states that the intersection is non-singular if that polynomial has distinct roots (over an algebraically closed field).

There is also a section on intersections of two quadrics in Colliot-Thélène, Sansuc & Swinnerton-Dyer, "Intersections of two quadrics and Châtelet surfaces. I", Crelle vol. 373. One result which you might find useful is Lemma 1.11: let $\Phi_1, \Phi_2$ be the two quadric forms; suppose that they have no common factor; that there exists a form of rank at least 5 in the pencil $\lambda \Phi_1 + \mu \Phi_2$; and there is no non-zero form of rank strictly less than 3 in that pencil. Then the variety is geometrically integral. (They point out that this criterion is definitely not necessary, since it's impossible to satisfy it for intersections of two quadrics in $\mathbb{P}^3$, yet there certainly are integral intersections of two quadrics in $\mathbb{P}^3$.)

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Hey Martin, Thanks! Miles Reid's result is quite interesting. Do you know that is there a way to generalize it to more quadrics? –  Fei YE Dec 9 '11 at 15:45
    
I don't know. As soon as you move to three quadrics, the singular ones in the family are cut out by a plane curve rather than a few points, which sounds distinctly more complicated. As the comments above point out, it might be unreasonable to expect a very general answer. –  Martin Bright Dec 12 '11 at 10:08

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