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I've read the wikipedia article but I don't know what an affine plane is and the definition/example did not seem clear. What I know is that in the 1880s mathematicians like Hilbert, Kronecker, Lasky and Macauley were responsible for the development of the algebraic variety concept. This is for a history of mathematics paper that I am working on.

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As far as I know, Algebraic varieties are (irreducible) set of solutions of a polynomial equation. Affine spaces are very similar to a linear space but with no real attention being paid to the existence of an origin, zero element (in a vector space, one of the axioms required the zero element to be a part of the space, in an affine space, that is no matter of concern really) –  Bidit Acharya Mar 27 '12 at 5:14
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Why isn't this sufficient: en.wikipedia.org/wiki/Algebraic_variety ? –  William Mar 27 '12 at 5:17
    
I don't have the math background to understand that article. –  organgina Mar 27 '12 at 5:32
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You've told us what you don't understand, but you haven't told us what you do understand, so it's hard to know where to begin. –  Gerry Myerson Mar 27 '12 at 5:48
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Are you trying to write a paper about the historical development of 19th century algebra and geometry without understanding the related mathematical ideas? If so, I think you have not chosen a topic well. Getting stuck on the meaning of the affine plane does not put you in a good position to see the point of the work of Hilbert, Kronecker, et al., since it appears that you can't give an example of the objects that interested them. –  KCd Mar 28 '12 at 1:51

2 Answers 2

An affine plane is the plane you learned about in childhood, distinguished in this context from a projective plane, which is a plane with a line at infinity, and it doesn't matter which of its lines is considered the one at infinity.

In the ordinary ("affine") plane, two lines may be parallel, so they never meet, or they may meet at one point. Now extend those two parallel lines to infinity and say there's a point at infinity where they meet, and it doesn't matter which direction you go to get there. If the lines go east and west, going infinitely far east or infinitely far west brings you to the same point at infinity either way. But if a line runs northeast and southwest, then the point at infinity on that line is a different point at infinity. The set of points at infinity is the line at infinity. But it doesn't matter which line in the projective plane is considered the one at infinity. When you work with the projective plane, you don't know or care which line is the one at infinity. But if you pick any line---it doesn't matter which one---and remove it from the projective plane, then what you're left with is the affine plane.

Say you have an ellipse in the projective plane. You delete one line from it, which does not touch or cross the ellipse, and then you have an ellipse in the projective plane. But now suppose the line you delete touches the ellipse at one point but does not cross the ellipse. Then what's left is a parabola. The point where the line you deleted touches the ellipse became the point at infinity on the parabola. And then suppose the line you deleted crosses the ellipse twice. Then what you're left with is a hyperbola in the affine plane. Those two points became the two points at infinity on the hyperbola.

The equation of the ellipse in the affine plane may be $x^2/a^2 + y^2/b^2 = 1$. That ellipse is an algebraic variety since it's the graph of an algebraic equation and it's not a union of more than one such graph.

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I'm not sure why you mention the word "affine plane" since it is not used in the Wikipedia article about algebraic varieties.

In any case, it's not as hard as it might seem. I have not studied algebraic geometry yet but I have come across one type of variety in one of my other courses. As you can see in the Wikipedia article, there are two types of algebraic varieties, affine and projective. Let's ignore projective varieties for now and first understand what an affine variety is:

You start with a ring of polynomials in $n$ variables $\mathbb F [x_1, \dots, x_n]$ where $\mathbb F$ is any algebraically closed field. Let $S \subset \mathbb F [x_1, \dots, x_n]$ be any set of polynomials. Now you can define the affine variety associated to $S$ as $$ V(S) = \{ \vec{x} \in \mathbb F^n \mid p(\vec{x}) = 0 \text{ for all } p \in S \}$$

So, an affine algebraic variety is a special subset of $\mathbb F^n$, namely, the set of all points that make all polynomials in $S$ vanish.

If $F= \mathbb C$ and $S = \{ p_n(z) = z^n + 1 \mid n \in \mathbb N \}$ then $V(S)$ would be all the roots of unit on the circle in the complex plane.

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It might be better to demand that $\mathbb{F}$ be algebraically closed when using this definition. Otherwise some strange things can happen... for example, $V(S)$ can be empty while $\operatorname{Spec} \mathbb{F}[x_1, \ldots, x_n] / S$ is not... –  Zhen Lin Aug 28 '12 at 15:26
    
@ZhenLin I did require it to be closed. Stupidly, I used a non-closed field for my example (which now is a non-example) because I thought $\mathbb R^3$ is oh-so-nice to picture. : / –  Rudy the Reindeer Aug 28 '12 at 15:29
    
@ZhenLin I tried to fix the non-example into an example. –  Rudy the Reindeer Aug 28 '12 at 15:32

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