# Minimal number of generators of a homogeneous ideal (exercise in Harsthorne)

In the very first chapter Hartshorne proposes the following seemingly trivial exercise (ex. I.2.17(ii) ):

Show that a strict complete intersection is a set theoretic complete intersection

Here are Hartshorne's definitions:

A variety $Y$ of dimension r in $\mathbb{P}^n$ is a (strict) complete intersection if $I(Y)$ can be generated by $n-r$ elements. $Y$ is a set-theoretic complete intersection if $Y$ can be written as the intersection of $n-r$ hypersurfaces.

Here $I(Y)$ is the homogeneous ideal of $Y$. The point is that the first definition seems wrong, since one would naturally require that $I(Y)$ can be generated by $n-r$ homogeneous elements (with this definition the exercise becomes trivial).

I have never made my mind if this is a misprint by Hartshorne. So the question is

Is it true that in a polynomial ring any homogeneous ideal generated by k elements is also generated by k homogeneous elements?

If I recall well it is not difficult to find counterexamples in graduate rings which are not polynomial rings, so the point of the exercise may be to show that polynomial rings have this special property.

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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 $I\subset m$.
Construct a surjective map $F_0 = \oplus_1^n A(-deg \ f_i) \to I \to 0$ and let $K$ be the kernel. 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 $u_1\neq 0$. 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 last term is exactly $\mu(I)$, and we are done.
I used $n$ for the number of variables too, which I should not. I did not want to make minor changes and bump this to the top, so I put this correction as a comment. –  Hailong Dao Feb 18 '10 at 0:24