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



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> Show that a strict complete intersection is a set theoretic complete intersection.
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Here are Hartshorne's definitions:



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> 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.
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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



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> Is it true that in a polynomial ring any homogeneous ideal generated by $k$ elements is also generated by $k$ homogeneous elements?
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If I recall well it is not difficult to find counterexamples in graded rings which are not polynomial rings, so the point of the exercise may be to show that polynomial rings have this special property.



Minimal number of generators of a homogeneous ideal (exercise in Hartshorne)
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 $R=k[x\_1,\dots,x\_n]$ and $\mathfrak m=(x\_1,\dots,x\_n)$. Suppose $\mu\_h(I)=m$ and $f\_1,\dots, f\_m$ is a minimal homogenous set of generators. At this point we switch to the local ring $A=R\_{\mathfrak m}$ (the reason: it is easier to do linear algebra over local rings, as anything not in $\mathfrak m$ is now invertible). It will not affect anything since $I\subset\mathfrak m$.


Construct a surjective map $F\_0 = \bigoplus\_{i=1}^m A(-\deg \ f\_i) \to I \to 0$ and let $K$ be the kernel. We claim that $K \subset\mathfrak mF\_0$. If not, then one can find an element $(a\_1,...,a\_m) \in K$ 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:$$u\_1f\_1 = \sum\_{i=2}^m b\_if\_i$$
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 $k=A/\mathfrak m$. By the claim $K\subset\mathfrak mF\_0$, so $F\otimes k \cong I\otimes k$. It follows that $m= \operatorname{rank} F\_0 = \dim\_k(I\otimes k)$. But over a local ring, the last term is exactly $\mu(I)$, and we are done.


