Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.