# “The notion of an affine algebraic set is still not satisfactory”

In their book "Algebraic Geometry I" Gortz and Wedhorn at page 16, after defining the category of affine algebraic sets (Zariski topology in $\mathbb A^n_k$, corrispondence between radical ideals and closed subsets, morphisms of affine algebraic sets ecc...), say:

The notion of an affine algebraic set is still not satisfactory. We list three problems:

1. Open subsets of affine algebraic sets do not carry the structure of an affine algebraic set in a natural way. In particular we cannot glue affine algebraic sets along open subsets (although this is a “natural operation” for geometric object

2. Intersections of affine algebraic sets in $\mathbb A^n_k$ are closed and hence again affine algebraic sets. But we cannot distinguish between $V (X) \cap V (Y ) ⊂ \mathbb A^2_k$ and $V (Y ) \cap V (X^2 −Y ) \subset \mathbb A^2_k$ although the geometric situation seems to be different (we will see similar phenomena later when we study fibers of morphisms)

3. Affine algebraic sets seem not to help in studying solutions of polynomial equations in more general rings than algebraically closed fields.

The first problem is due to the fact that affine algebraic sets are necessarily embedded in an affine space.

Now, for me the third problem is obviously clear. But what is the meaning of the first two problems? I need some further explanations. Moreover, why they say that the problem 1. is due to the embedding in $\mathbb A^n_k$?

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A comment because it's a vague, partial answer: the problem in question 2 is that in the case of $V(Y)\cap V(X^2-Y)$ there are "supposed" to be two intersection points, because you're intersecting a quadratic curve with a line, but the geometry says there is one, exactly like the intersection of two lines. You want some notion of multiplicity of intersection, so that you can rigorously say that the intersection point is really two points on top of each other. – Matthew Pressland Mar 14 '13 at 11:05

1) Everything manifold can be embedded in one $\mathbb R^n$ ; this is false in algebraic geometry. For example, if I take $\rm N$ copies of $\mathbb R^2$ and that I tell you how I decide to glue them, I get a manifold (if I glue well). This manifold can be embedded in $\mathbb R^4$, so it is affine.

In Algebraic Geometry, if I give you affine schemes in $\mathbb A^2$ and that I tell you how I glue them nicely, then the result need not be embeddable !

2) For this problem, there is a very nice way to see it. Consider $\mathrm{V(X( X} - a))$ in $\mathbb A^1$. It has two points if $a \not = 0$. Then let $a \to 0$, you will get $\mathrm{V(X^2)}$ which has only one points, but which has to be thought as two points infinitessimaly close.

In other words, the theory of schemes will allow you to make a difference between $\rm V(X)$ which represent only one point and $\rm V(X^2)$ which represent the infinitesimal process of two points coliding and becomming infinitessimaly close.

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About 1. The Point is that, for example, in differential geometry a manifold $M$ is obtained gluing the charts that are open subset with a structure of manifold. For an affine algebraic set $X$, this is wrong because its open sets are not affine algebraic sets. One can only take the irreducible components of $X$ that are a finite number. Is this right? – Dubious Mar 14 '13 at 12:47
An open set of an affine set is affine. But the gluing of many affine sets may not be affine. – Damien L Mar 14 '13 at 12:57
@Damien L It's not true that an open set of an affine set is affine! – Piotr Pstrągowski Mar 14 '13 at 16:03
@Galoisfan The problem lies even deeper. For example, if you work with $k = \mathbb{C}$, then it's not that hard to show that any algebraic subset of positive dimension is non-compact in it's complex topology. But you actually can take two affine curves (ie. algebraic, irreducible subsets of dimension 1) and glue them along an open subset (which is again affine!) to obtain a curve that is projective, in particular it is compact in the complex topology. Such a curve cannot be affine! (This can be also established rigorously using only algebra, see the notion of properness.) – Piotr Pstrągowski Mar 14 '13 at 16:06

(1) It can be shown for instance that $\mathbb A^2\setminus \{(0,0)\}$ is not an affine algebraic set in any affine space $\mathbb A^n$.

(2) Both intersections are reduced to the origin $(0,0)$. But clearly, in the first case, the curves $V(X), V(Y)$ meets transversally, while in the second case, both curves share the same tangent line.

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"reduced" might not be the best word for (2). – zyx Mar 14 '13 at 19:01