ag.algebraic geometry - Minimal number of generators of a homogeneous ideal (exercise in Harsthorne)

Dear Andrea: Hartshorne was right, but we need to do some work. Let $mu(I)$ be the minimal number of generators of $I$, and $mu_h(I)$ be the minimal number of a homogenous system of generators of $I$. Let $R=k[x_1,cdots,x_n]$ and $m=(x_1,cdots,x_n)$. Suppose $mu_h(I)=n$ and $(f_1,cdots, f_n)$ is a minimal homogenous set of generators. At this point we switch to the local ring $A=R_m$ (the reason: it is easier to do linear algebra over local rings, as anything not in $m$ is now invertible). It will not affect anything since $Isubset m$.

Construct a surjective map $F_0 = oplus_1^n A(-deg f_i) to I to 0$ and let $K$ be the kernel. We claim that $K subset mF_0$. If not, then one can find an element $(a_1,...,a_n) in K$ such that $sum a_if_i=0$ and $a_1$, say, has a degree $0$ term $u_1neq 0$. By considering terms of same degree in the sum one sees that there are $b_i$s such that:

$$u_1f_1 = sum_{2}^n b_if_i$$
so the system is not minimal, as $u_1 in k$, contradiction.

Now tensoring the sequence

$$0 to K to F_0 to I to 0$$ with $k=A/m$. By the claim $Ksubset mF_0$, so $Fotimes k cong Iotimes k$. It follows that $n= rank F_0 = dim_k(Iotimes k)$. But over a local ring, the last term is exactly $mu(I)$, and we are done.

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