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Exercise 5.3.22 of Algebraic Geometry: A Problem Solving Approach asks:

Show that $\mathbb{P}^n(k)$ is a projective variety when $k$ is an infinite field.

This is easy -- $\mathbb{P}^n(k)$ corresponds to the zero set of $\langle 0 \rangle$, which is always prime since fields are always integral domains, and thus polynomial rings over fields are always integral domains.

Usually the cardinality of a field is important because, for finite fields, there exist non-zero polynomials which evaluate to $0$ at every point of the space (e.g. see here). But I don't see why that matters here -- even if $\mathbb{P}^n(k)$ could possibly be generated as the zero set of an even larger ideal than just $\langle 0 \rangle$ for finite fields, that still doesn't mean that $\mathbb{P}^n(k)$ isn't a projective variety, right?

Question: Do I actually need to use the hypothesis that $k$ is infinite to show this?

Chill2Macht
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    What is your definition of projective variety ? For me a projective variety is an algebraic closed subset of $\Bbb P^n(k)$, and then it's true that $\Bbb P^n(k)$ is a projective variety for any field $k$. –  Jun 24 '17 at 12:20
  • @N.H. Good point. The definition of projective variety in my book is an irreducible algebraic subset of $\mathbb{P}^n(k)$. When you say closed, do you mean in the Zariski topology, i.e. that the ideal generated by the set is a prime ideal? Because by a previous exercise in the section, which shows that an algebraic set is irreducible if and only if its ideal is prime, then our definitions coincide. So I guess my question is not so much about whether $\mathbb{P}^n(k)$ is a projective variety for any field $k$, but whether the proof I propose needs to be modified for finite fields. – Chill2Macht Jun 24 '17 at 15:03
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    Ok, then yes your proof works exactly the same. –  Jun 24 '17 at 15:14
  • @N.H. I appreciate the sanity check -- enjoy the rest of your weekend! – Chill2Macht Jun 24 '17 at 15:56
  • Sure, have a good week end too ! (I have a terrible exam to prepare haha) –  Jun 24 '17 at 16:16
  • if your projective varieties are subsets of $\Bbb P^n(k)$ without any algebraic structure attached, then over finite fields this is not particularly interesting as every subset is an algebraic subset – mercio Jul 03 '17 at 08:51
  • @mercio A projective variety is the zero locus of a prime homogeneous ideal in $k[x_0, \dots, x_n]$. – Chill2Macht Jul 03 '17 at 08:54

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