Mathematicians, of course, are known to be liars. They tend to be treacherous little beasts and ought not to be trusted without supervision.
@crypticdesign031297 disagrees, but she can't do anything about it until after she hears the recording.
And I have proof that they are liars. For example, they name a subject "algebraic geometry" and expect you to think it has something to do with those nice ordinary polygons they taught you about previously when they said the word "geometry", but they do, in fact, want to just do more algebra, and they snuck the word geometry in (figuring they already had the algebra crowd at the word 'algebraic') to make it seem slightly more comforting to the geometry crowd. Because all mathematicians are secretly algebraists, except the ones who aren't.
Oh, they pretend it has something to do with geometry. They sketch some nice little circles at first and proceed to do some suspiciously analytical manipulations.
They sketch cusps, curves, surfaces, and higher-dimensional objects no sane artist would attempt, all in an effort to lull you into a (very false) sense of understanding, and then they turn around and replace the whole business with rings, ideals, sheaves, and other apparatuses designed to scare off the faint of heart before they can reach the proper and polite mathematics society.
This may seem somewhat severe of them, but it's well deserved. They have a lot of strange objects that look quite harmless and are quite dangerous, you have to prove your mettle.
The much, much, bigger proof that mathematicians are liars is that one of them lured me here with promises of a large audience who would be very, very, amused by me, and has instead tricked me into giving an explanation of algebraic geometry.
I will assume you are familiar with the basics of algebra itself.
Much of our study will be about varieties. So let us consider, first, what we will call the vanishing set of a polynomial. Let k be an algebraically closed field, and consider the affine space πΈβΏβ, representing all the possible inputs to polynomials in k[xβ...xβ]. We will call F the ring of polynomials k[xβ...xβ], and A the affine space πΈβΏβ, because I do not want to try to convey unicode out loud again.
Let fβF, then the 'vanishing set' of f is the set of all points a in A such that f(a) is 0.
And spending some time thinking about it, you realise you can take unions by just multiplying functions together.
This is because we are working in a field k, and fields are domains, and in domains, 0 is prime, so therefore if the result of fg is 0, then either the result of f is 0, or the result of g is 0, thus
But in the other direction, if either are 0, then the product must be 0. Hence
The natural consequence that also occurs to you is that you can factor a vanishing set into distinct curves. Since k is a field and thus a UFD, so then k[xβ] is a UFD, and thus... and also that repeated factors don't "add" anything.
But what about intersections. You faff around for a bit and quickly realise that there isn't a good way to do this. There can be finite amounts of intersection points, but functions of more than one variable of algebraically closed fields must have a number of solutions equal to the degree for every input of the first n-1 variables.
(That is, for f(xβ...xβ) of degree d, each input xβ...xβββ provides d witnesses xβ such that xβ...xβ is in the vanishing set.)
Naturally, at this point, you give up and decide to never do math again and... ow! Really? Okay. Fine.
It occurs to you to modify the set. Instead of taking in functions, you take in a set of functions, and you find all the zeros they have in common.
V(F)={x|βfβF, f(x)=0}
V({f}) happens to be exactly like Vn(f).
We will call the function V the "variety".
Now you can do intersections by taking the union of the sets!
And then from there, you also notice that intersecting two sets that have two varieties returns the elements that have all zeroes in both sets. This MIGHT give you the union, as certainly all points in the union must be in the new variety, but it can give you extra objects. For example:
V({f}β©{g})= V(β¦°} which equals everything, not just the points on f and g.
Maybe products might be more useful. The products of elemenrs in F and G? But it takes some working out.
We noticed previously that Vn(ff)=Vn(f), but now, we can consider the set of ALL functions that have 0s in the variety of some F.
I(C), where C is some set of points, is the set of all polynomials functions which are 0 at C.
{fβk[xβ...xβ]| βcβC, f(c)=0}
And my new claim is that for any curve, I(C) is an ideal.
Proof; let f, g be in I(c) and cβC arbitrary. Then (f+g)(c) f(c)+g(c)=0+0=0, so f+g is in I(c). Similarly, for any g in k[xβ...xβ], gf at c is g(c)f(c)=g(c)0=0, so fg is in I(c), and it is maximal. No external elements are 0 everywhere there.
Now the setup is nice, as you get.
But applying V, V(F) βV(I(V(F))), by I, I(C)βI(V(I(C)))
And in the other direction, you know that I(C)βI(V(I(F))), since a bigger set will make I smaller, so.
Therefore, F always has a square free ideal π¦=I(V(F)) such that V(π¦)=V(F).
Notably, this must be an intersection of maximal ideals. Proof;
Assume it is not. Then consider the intersection of all maximal ideals that contain I(V(F)), and consider a polynomial g in that intersection not in I(V(F)), then this g(a)β 0 for some a in V(F). Now considee the maximal ideal generated by a. (xβ-aβ, xβ-aβ...xβ-aβ). This element cannot contain g, since g is not 0 at a, but it must contain all of I(V(F)), since all of those points are 0 in f, so by intersection with this maximal ideal, we filter it out.
So therefore, we really only need to consider curves generated by ideals that are intersections of maximal ideals.
This is part of something called nullstelenstatz. Which is a horrible name for me to try and spell.
And now I have picked the lock and am fleeing so I can stop! Goodbye.
