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Suppose that $X$ is a complete intersection in projective space, with $X = \cap Z(f_i)$. If the Hilbert polynomial is known, what relations must the $f_i$ satisfy? Can we get $n$ independent equations?

I know that the Hilbert series determines these degrees, but we lose information when passing to the Hilbert polynomial.

I am trying to argue by Grobner degeneration to an initial ideal: the hilbert polynomial doesn't change, but I'm not sure if the degeneration is still a complete intersection and moreover if we can get a generating set for the initial ideal from the leading terms of a generating set - though if it is the case then one knows the degrees of the hypersurfaces from their initial ideals. If it is, then one just needs to argue for the case of monomial ideals. This case is also not obvious, so maybe one can find a counter example among monomial ideals.

I know that curves in projective space, if we have $Z(f) \cap Z(g) = X$, $\deg f= s$ and $\deg g = t$, then $\deg X = st$, and $\mathrm{genus}(X) = \frac{1}{2} (st) (s + t - 4) + 1$. So we get two equations, and we can solve using the quadratic formula.In some situations, there is only one solution when $s$ and $t$ are both positive.

So I would hope that this generalizes, and there are $\operatorname{codim}X$ independent equations in terms of the degrees of the $f_i$.

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    I hope I'm not missing something here, but we know the leading term of the Hilbert polynomial $h_X(m)$ is $(d/k!)m^k$, where $k$ is the dimension and $d$ the degree of $X$. If $X$ is a complete intersection then $d=d_1\cdots d_r$, where each $f_i$ is of degree $d_i$. This will give you your finite list in terms of all the possible decompositions of $d$ as a product of $r$ numbers (remember that if $d_i=1$ for some $i$ you may as well consider $X$ to live in a projective space of one dimension less and dismiss this polynomial). Even if you don't know the number $r$ you get a finite list. – 2017-01-03
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    @user347489 You are right, I'm not asking exactly what I want to. Let me edit. – 2017-01-03
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    If I understand correctly you want to know how much information about the syzygies (In particular the second syzygy) associated to a complete intersection you can recover from the Hilbert polynomial, right? I don't know the answer to this, but Harris' discussion about Hilbert polynomials in 'Algebraic Geometry: A First Course' may be useful... – 2017-01-03
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    I'm not sure about the set up in algebraic geometry. But in terms of algebra, the Koszul complex determined by the $f_i$ is in fact a minimal graded free resolution from which you can read everything. If they coincide, then the Koszul relations are the ones they satisfy. – 2017-01-03
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    @Youngsu Do you mean that the Hilbert polynomial is completely unnecessary as long as you know the generators? What if you know the polynomial but not the generators? – 2017-01-04
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    I do not follow what you mean by "completely unnecessary". A free resolution contains finer information that a Hilbert function. I am not sure of what your question is. Are you asking if the Hilbert function for a complete intersection graded ring determines the degrees of $f_i$? – 2017-01-05

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