# Punctured plane algebraic over a finite field?

I've been asked to prove, by my algebraic geometry teacher, that the punctured affine plane $\mathbb{A}_k^2 \backslash \{0,0\}$ is not an algebraic set, i.e. is not the zero set of any set of polynomials, even in the case of a non-algebraically closed field. I was going to proceed by showing that it has the same ring of regular functions as $\mathbb{A}_k^2$, then use the fact that we have a bijection between maximal ideals of the ring of regular functions and points of the algebraic set, then say that there is no maximal ideal in $\mathbb{A}_k^2 \backslash \{0,0\}$ corresponding to the maximal ideal in $\mathbb{A}_k^2$ corresponding to the origin i.e. the bijection between the sets misses the origin. However, let $k$ be finite and let $a_1,...,a_l$ be its nonzero elements. Define $p(x, y) = \prod(x-a_i)(y-a_i)$. Clearly $p(a,b) = 0$ if at most one of $a, b$ are zero, and $p(0, 0) \neq 0$ as $k$ is in specific a domain. Then we have that $V(p) = \mathbb{A}_k^2 \backslash \{0,0\}$. What am I missing here? Do any of the results I wanted to use break down in the case of finite fields?

The definitions I'm using: a set is algebraic if it is the zero locus of an ideal $I$ of A = $k[x_1,...,x_n]$, in which case the ring of regular functions is $A/I$ (or equivalently a function is regular on an algebraic set if it is the restriction of a polynomial on $\mathbb{A}_k^n$ to the set). I do not think I'm assuming that the field is algebraically closed; there are other problems in the same assignment that expressly say to separate finite and infinite cases.

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Over a finite field every set in affine space is algebraic. Perhaps your teacher meant for you to assume $k$ was infinite. –  JSchlather Feb 17 '13 at 6:06
The affine plane over a finite field as a scheme is not a finite set, though. The problem with your approach is over finite fields the ring of regular functions has to be defined with more care... –  Mariano Suárez-Alvarez Feb 17 '13 at 6:14
@MarianoSuárez-Alvarez Can you point me towards such a definition? –  Julien Clancy Feb 17 '13 at 6:24
@Julien: Perhaps you should better tell us the definitions which you have learned and should use here. But probably these are not correct if $k$ is not algebraically closed, and this should be implicitly assumed in a course on classical algebraic geometry. –  Martin Brandenburg Feb 17 '13 at 13:33
They don't have the same points when $k$ is not algebraically closed. For example, $\mathbb A^2_k$ consists in maximal ideals of $k[X,Y]$, but $k^2$ correspond to those maximal ideals of the form $(X-a, Y-b)$ for some $(a,b)\in k^2$. This difference really counts over finite fields. –  user18119 Feb 19 '13 at 22:57