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 $f_1,\cdots, f_n$ (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 $I\subset m$.
Construct a surjective map $F_0 = \oplus_1^n A(-deg \ f_i) \to I \to 0$ and let $K$ be the kernel. By grading consideration, one can show We claim that $K \subset mF_0$. If not, otherwise 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 $u_1\neq 0$. 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 $K\subset mF_0$, so $F\otimes k \cong I\otimes k$. It follows that $n= rank\ F_0 = dim_k(I\otimes k)$. But over a local ring, the fact that the first syzygy last term is inside $mF_0$ characterize minimal system of generators, so exactly $n=\mu(I)$, \mu(I)$, and we are done.
