Algebraic sets

by ayushkhaitan3437

This is a kind of longish post on algebraic sets. I bought Hartshorne, then ended up studying from Fulton’s “Algebraic curves”. I will focus on the geometric aspects of algebraic sets.

So what are algebraic sets? They could be points, curves, surfaces. But if they are curves, they have to be infinitely long curves on both ends, and not one of those finitely long segments. If they are surfaces, they have to stretch infinitely in all directions. In short, some geometric entities qualify as algebraic sets, whilst others don’t. Confused?

Algebraic sets V(S) in affine space k^n are those sets of points which satisfy all polynomials in S\subset k[x_1,x_2,\dots,x_n]. More on algebraic sets can be found in Hartshorne and Fulton amongst other books on introductory Algebraic Geometry.

As an example, take all points satisfying y-x=0. This is an algebraic set, as it satisfies all polynomials in \langle y-x\rangle.

Now we move on to discuss the nature of k. It is quite clearly a field. But does it matter if it is algebraically closed or not? It does. Take \Bbb{R} and \Bbb{C} for instance. In affine spaces k^n where n\geq 2, all polynomials in x,y have infinite zeroes if k=\Bbb{C}, but may have a finite number of zeroes if k=\Bbb{R}. For example, take the curve x^2+y^2=0 in k[x,y]. Clearly, it has only one zero in \Bbb{R}^2, but infinitely many in \Bbb{C}^2. Now you might wonder why does n have to be greater than 2? Why does it not work for n=1? Figure that out for yourself [hint: something to do with the algebraically closed nature of \Bbb{C} and the fact that you can substitute any value for y in every equation and get values for x in \Bbb{C}, while there being only one variable in k^1.

Now let us move on to more important things. Say you have k^n. Can you construct a set of polynomials which have only one point (say point a) as its zero set? What about a set of polynomials which have a curve as a zero set? What about a surface? What kinds of lines and surfaces are possible? I will try to answer all these questions in the subsequent paragraphs. But first a note about polynomials in n variables. These polynomials are not the usual polynomials that you can draw on a plane or a sheet of paper, like y=x^2. These polynomials can be planes, surfaces, cylinders, 25-dimensional quantum surfaces, whatever you will. Please spend some time thinking about these concepts.

What is a zero-dimensional dot in n-dimensional affine space A? What is a one dimensional curve in A? What is a two dimensional surface in A? How are these things defined? In n-dimensional space, all polynomials of lower dimensions are defined through intersections. For example, in k^3, all curves are defined by the intersection of surfaces. It is possible to define the curve as the intersection of two surfaces. A zero-dimensional point can be defined as the intersection of 3 surfaces. Generalizing this idea, in n-dimensional space, n-a dimensional polynomials can be defined as intersections of a n-1-dimensional polynomials. This is probably the most important line in the whole article. Read it again. Draw it. Gulp it down with some lemon sikanjee.

What are algebraic sets geometrically speaking? In an n-dimensional affine space, they are a finite union of polynomials of dimensions \leq n. They are polynomials. Polynomials they are (says Yoda). Why do they have to be polynomials? Because they are in essence either n-dimensional polynomials, or the finite intersection of n-dimensional polynomials. Who says that? The definition. An algebraic set is the set of points satisfied by every polynomial in an assorted bunch of polynomials in k[x_1,x_2,\dots,x_n]. In other words, it is the intersection of all polynomials in that set of polynomials. Can it be an infinite intersection? Yes. But you will always end up with a polynomial with a degree \leq 2. That is the thing with polynomials. They intersect to give other polynomials.

Why finite union? Because you need to multiply sets of polynomials with each other to get the union of algebraic sets, and the multiplication of an infinite number of polynomials is not defined as every polynomial needs to have a finite degree.

Hence, an infinite bunch of disconnected points or a finitely long line segment cannot be algebraic sets.

Let S be an algebraic set. Then I(S) is quite clearly an ideal. I will try to motivate why I(S) needs to be finitely generated. First let us get a lemma out of the way. If an ideal P is generated by polynomials p_1,p_2,\dots,p_n, then Z(P) is exactly equal to Z(p_1)\cap Z(p_2)\cap\dots Z(p_n). OK. Now let S be a dimensional, where a\leq n. Then we can surely find n-a n-1-dimensional polynomials which intersect to give S. These polynomials have to exist (this becomes clear on thinking a little about the concepts involved). Do these n-a polynomials generate I(S)? This is a question that I am yet to answer. Hopefully I will continue this post tomorrow.

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