User:LkNsngth/hilbert polynomial

But the binomial coefficients are all polynomials in $$z$$ so the expanded form is the

associated Hilbert polynomial. Thus $$P_M(z)=

\frac{1}{n!}(z+n)(z+n-1)...(z+1)-\frac{1}{n!}(z-d+n)(z-d+n-1)...(z-d+1)$$

Now the arithmetic genus is just $$(-1)^{n-1}(P_M(0)-1)$$ which is

just $$(-1)^{n-1}(1-\frac{1}{n!}(n-d)(n-d-1)...(-d+1))-1$$. The 1s

cancel, so the only thing left is the long product. However, because

we multiply by $$(-1)^{n-1}$$ we effectively switch all the negative

signs, so the product looks like $$\frac{1}{n!}(d-1)(d-2)...(d-n)$$

which is $$(^{d-1}_n)$$.