# Decide whether there are “linear” relations between quadrics

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?

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.