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Let $k$ be an algebraically closed field of characteristic $0$. For a homogeneous ideal $I=(q_1,\dots, q_k)\subset k[x_0,\dots,x_n]$ generated by quadrics, is there a method to decide whether the generators of $I$ admit relations "of degree $1$" i.e. if there are linear polynomials $L_i\in k[x_0,\dots,x_n]_1$ such that $\sum_iL_ig_i=0$? Can one compute (in explicit cases) the dimension of such $k$-uples of linear polynomial?

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These relations are called "linear syzygies". A lot are known about them, and interesting questions are abound too. You can compute them with Macaulay 2 and read the number off the Betti table.

In this particular case, there is a convenient formula using Hilbert function. The number of linear syzygies is $(n+1)r - \dim_k I_3$, where $r$ is the number of generators. The reason is that the space of cubics contains $(n+1)r$ cubics from the generators, but each linear relation translates to a $k$-linear relation among these cubics and make the dimension of $I_3$ goes down $1$.

For instance, if $I=(x^2,y^2,x^2)$, the Hilbert series of $R/I$ is $1+3t+3t^2+1$. It follows that $\dim_k I_3=10-1=9$, so there are $3\times 3-9=0$ linear syzygies. If $I= (x^2,y^2,x^2,xy+yz+zx)$, then the Hilbert series of $R/I$ is $1+3t+2t^2$, so $\dim I_3=10$ and there are $4\times 3-10=2$ linear syzygies.

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  • $\begingroup$ Thank very much for your detailed answer. $\endgroup$ – pi_1 May 17 at 21:31

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