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Define a projective variety to be a subspace $V \subset \mathbb P^n$ such that $V$ is the zero set of some set $T$ of homogenous polynomials in $k[x_0, \ldots , x_n]$. My book claims that "as with affine varieties, we can assume $T$ is finite.

I'm having trouble seeing why this is true. $Z(T) = Z(\langle T \rangle)$, and $\langle T \rangle $ is a homogeneous ideal since it's generated by homogenous polynomials. Since $k[x_0, \ldots , x_n]$ is Noetherian, $\langle T \rangle $ is finitely generated, but why can we take those generators to be homogeneous?

Thanks

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Hint: $<T>$ homogenous Ideal (generated by homogeneous Polynomials) $\Rightarrow$ for all $f\in <T>$ every homogeneous component $f_i$ of $f=f_0+f_1+f_2+...+f_n$ is also in $<T>$ –  Blah Mar 12 '12 at 19:20

2 Answers 2

If you take a homogeneous ideal $I$ in $k[x_0,\cdots,x_n]$, then $I=\langle f_1,\cdots,f_m\rangle$, for some polynomials $f_i$. Now take the homogeneous components of each of the $f_i$ and you get a finite set of homogeneous polynomials generating $I$.

(note that $I$ is homogeneous if and only if the homogeneous components of all polynomials in $I$ are still in $I$)

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Also, you can just as well use the equivalent characterization of Noetherian rings: there are no infinite ascending chains of ideals. This holds, in particular, for ideals generated by homogenous polynomials. Hence, they are all finitely generated (just add generators until the chain becomes stationary).

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