Subvarieties in algebraic geometry

1. Apr 24, 2007

AlphaNumeric

I'm slowly working my way through 'Ideals, Varieties and Algorithms' by Cox, Little and O'Shea and the bit about subvarieties has confused me. If those in the know have access to the book it's page 236 I'm talking about.

Definition : Given an affine variety $$V$$ in $$k^{n}$$, let $$k[V]$$ represent the collection of polynomial functions $$\phi : V \to k$$

No problem there, you define a variety somehow (often as the common zeros of a set of functions) and then you have a set of functions defined on the points of V such that they take a point and map it to the field k, so points in n dimensional space to some number (typically k is the complex numbers).

Definition : Given $$V$$ in $$k^{n}$$,
1) For any ideal $$J = \langle \phi_{1},\ldots,\phi_{m}\rangle \subset k^{n}$$ we can define $$\mathbf{V}_{V}(J) = \{ (a_{1},\ldots,a_{n}) \in V \, : \, \phi(a_{1},\ldots,a_{n}) = 0 \, \forall \phi \in J\}$$. This is a subvariety of V.
2) For each subset $$W \subset V$$, we define $$\mathbf{I}_{V}(W) = \{ \phi \in k[V] \, : \, \phi(a_{1},\ldots,a_{n}) =0 \, \forall (a_{1},\ldots,a_{n}) \in W\}$$

Definition one is picking out the zeros of the ideal J, fine. Definition two pick out the "functions everywhere zero" on a subset of the variety. Up to this point I feel I've got a decent grip of things. However the book has an example which kinda demolishes that :

"Let $$V = \mathbf{V}(z-x^{2}-y^{2}) \subset \mathbb{R}^{3}$$. Let $$J = \langle [x]\rangle \in \mathbb{R}[V]$$. Then we have $$W = \mathbf{V}_{V}(J) = \{(0,y,y^{2}) : y \in \mathbb{R}\} \subset V$$"

Fine, no problem. J basically says "Let x be zero, what's the part of V with this true? It's W, parameterised by y. Happens to be, as pointed out by the book, the same as $$\mathbf{V}(z-x^{2}-y^{2},x) \subset \mathbb{R}^{3}$$. The book then continues :

"Similarly, if we let $$W = \{(1,1,2)\} \subset V$$, then we leave it as an exercise to show that $$\mathbf{I}_{V}(W) = \langle [x-1],[y-1]\rangle$$."

This I don't get. If $$\mathbf{I}_{V}(W)$$ is picking out the "functions everywhere zero on W in V", shouldn't it be $$\mathbf{I}_{V}(W) = \langle [x-1],[y-1],[z-2]\rangle$$ ? Otherwise you've not specificed in your new ideal that z=2 leads to the function being zero. Or are you supposed to 'glean' this information from the fact the ideal is defined from W in V[/tex] and since V is defined by $$\mathbf{V}(z-x^{2}-y^{2})$$, you have this additional bit of information?

I'm just generally confused entirely by that second part of the example. I'd be really grateful if someone could spell out what exactly is going on. Thanks very much.

2. Apr 24, 2007

StatusX

You know that everywhere in V, z=x^2+y^2. So if x=1 and y=1, it follows z=2.

3. Apr 24, 2007

mathwonk

you have it. in k[V], the functions [z] and [x^2+y^2] are equal. so knowing [x] and [y] are equal to 1, forces [z] to equal 2.

i.e. in the ring k[R^3], the ideal <x-1, y-1, z-2> is the same as the ideal <x-1,y-1,z-x^2-y^2>.

since k[V] is the ring k[R^3]/<z-x^2-y^2>, modding out k[V] by <x-1,y-1> is the same as modding it out by <x-1,y-1,z-x^2-y^2> = <x-1,y-1,z-2>.

by the way you are reading the an earlier editiion of the book and in the new third edition this page is numbered 239.

you also have a typo above where you say the ideal consists of functions in k^n instead of in k[V].

Last edited: Apr 24, 2007
4. Apr 24, 2007

AlphaNumeric

Thanks guys, I was sort of stumbling in the right direction. That explains to V subscript in $$I_{V}(W)$$, it gives you that exact bit of information to use V.

Mathwonk, I've got two copies of the book (from different libraries and a mix up!) and both are the second edition. I think the latest one is the 3rd edition.

5. Apr 24, 2007

mathwonk

yes the publisher just sent it to me. i also have the prepublication notes version.

6. Apr 24, 2007

AlphaNumeric

It's a pretty nice book and the fact they wrote a few Mathematica files which impliments things like the S polynomials makes it easier to follow some of the worked examples or to do the questions.

Though I've not got that far, they cover the projective plane in quite a lot of detail which is something I remember you answering a question of mine about a few months ago. I warn you, I doubt this will be my last algebraic geometry question :tongue:

7. Apr 29, 2007

AlphaNumeric

As predicted, I've more questions.....

I've got the whole "How to decompose a variety into it's irreducible component varieties" thing. Each i.v. can be parameterised by as many variables as the dimension of the variety (or more, but you know what I mean). If you work out the dimension of a variety, you're basically computing the dimensionality of it's largest irred. variety.

ie V = union of line + a plane, dim(V) = dim(plane) = 2. W = union of line and a line, dim(W) = 1 etc etc.

The main emphasis of what I want to use algebraic geometry for involves finding zero dimensional varities. I have a bunch of polynomials, compute the ideal and work out the dimensionality of the variety it relates to. If this is just the sum of points, spiffy, easy to find (ala section 5.3 in the book I mentioned). However, if it's the union of finitely many points and an extended variety (or varieties) then it's not immediately obvious and the problem is that decomposing such a variety into it's component irred. varieties involves both splitting the extended varieties from the isolated ones but also all the isolated ones up!

That is more than I'm interested in. Is there a way to see, if given I, wether V(I) contains zero dimensional component varieties without actually solving all the equations involved? Some algorithm or method which basically says

Dimensions of subvarities of V = $$\{0^{n_{0}},1^{n_{1}},\ldots\}$$

where there's $$n_{0}$$ zero dimensional varieties etc.

I just want to know if a given variety defined by bunch of polynomials has any zero dimensional subvarities, but with the minimal amount of computation. Some of these things manage to hammer even a 3Ghz 2Gb RAM computer as it is.

The paper which suggested this methodology says that the only way is to do a prime decomposition and work from there. That's a lot of computing effort. Is it the only way?

Perhaps I'm way off the plot (which is true more often than not) but if you've a Groebner basis $$G = \{g_{1},\ldots, g_{n} \}$$ over $$\mathbb{C}[x_{1},\ldots,x_{m}]$$ and (by relabelling if needed) the first m of G have the right leading orders so that you've a leading order term for each $$x_{i}$$, such a subset of G would represent a bunch zero dimensional varieties, wouldn't it? Would that be enough? Can I then ignore the other elements of G and still get the same zero dim. varieties, all the remaining n-m are not important? Am I making any sense or am I just making myself look like I don't have a clue about this stuff? Be nice, I'm an applied mathematician who didn't know anything about algebraic geometry this time last month.

Thanks guys

Last edited: Apr 29, 2007
8. May 1, 2007

mathwonk

"Each i.v. can be parameterised by as many variables as the dimension of the variety (or more, but you know what I mean)."

actually i do not know. avariety in general cannot be the image of a dense map from a coordinate space, which to me is the usual meaning of parametrize.

"If you work out the dimension of a variety, you're basically computing the dimensionality of it's largest irred. variety."

yes, this is the definitionof dimension of a reducible variety.

"ie V = union of line + a plane, dim(V) = dim(plane) = 2. W = union of line and a line, dim(W) = 1 etc etc." yes.

"The main emphasis of what I want to use algebraic geometry for involves finding zero dimensional varities. I have a bunch of polynomials, compute the ideal and work out the dimensionality of the variety it relates to. If this is just the sum of points, spiffy, easy to find (ala section 5.3 in the book I mentioned). "

"However, if it's the union of finitely many points and an extended variety (or varieties) then it's not immediately obvious and the problem is that decomposing such a variety into it's component irred. varieties involves both splitting the extended varieties from the isolated ones but also all the isolated ones up!"

huh?

"That is more than I'm interested in. Is there a way to see, if given I, wether V(I) contains zero dimensional component varieties without actually solving all the equations involved? "

"I just want to know if a given variety defined by bunch of polynomials has any zero dimensional subvarities, but with the minimal amount of computation. Some of these things manage to hammer even a 3Ghz 2Gb RAM computer as it is."

"The paper which suggested this methodology says that the only way is to do a prime decomposition and work from there. That's a lot of computing effort. Is it the only way?"

whats wrong with that? its almost the exact question you asked. well but youwant to igniore the positive dimensional pieces. maybe you can modify the decomposition to ignore non maximal ideals.

Perhaps I'm way off the plot (which is true more often than not) but if you've a Groebner basis over and (by relabelling if needed) the first m of G have the right leading orders so that you've a leading order term for each , such a subset of G would represent a bunch zero dimensional varieties, wouldn't it? Would that be enough? Can I then ignore the other elements of G and still get the same zero dim. varieties, all the remaining n-m are not important? Am I making any sense or am I just making myself look like I don't have a clue about this stuff?

this is alittle hard to follow. especially since i reallydon'tknow anything ABOUT GROBNER bases. maybe if you give some page refernces in the book i willlook at them.

9. May 4, 2007

AlphaNumeric

It's best to ignore my last post, I was trying to nail down a few concepts I didn't get but now that they've been rolling around in my head (and I managed to reproduce the results of a paper using Singular while before I was chugging along with Mathematica so compute resources aren't an issue now), I'm happy about them.

A more straight forward question I have is wether the following is true :

$$((I:f^{\infty}):g^{\infty}) = (I:(fg)^{\infty})$$

where I is a polynomial ideal in $$\mathbb{C}[x_{1},\ldots,x_{n}]$$ and f,g are just two polynomials in $$\mathbb{C}[x_{1},\ldots,x_{n}]$$. My vaguely informed guess is yes.

$$((I:f):g) = I:(fg)$$ is true and $$(I:f^{\infty}) = \bigcup_{n=1}^{\infty}(I:f^{n})$$ and these lead to

$$((I:f^{\infty}):g^{\infty}) = \bigcup_{m=1} \left( \bigcup_{n=1}^{\infty}(I:f^{n}) : g^{m}\right) = \bigcup_{m=1}^{\infty} \bigcup_{n=1}^{\infty}\left( I : f^{n}g^{m}\right)$$

The variety of $$(I:f^{\infty})$$ is the space of points where I's functions are zero but f is certainly non-zero. That would mean that $$((I:f^{\infty}):g^{\infty})$$ is the space of solutions to I's functions but certainly not to f and g. Since polynomials over C are an integral domain, fg = 0 means f=0 or g = 0 (or both) and so $$(I:(fg)^{\infty})$$ is the space of zeros for I's functions but certainly not f.g and so by the integral domain property of polynomials, neither f nor g can be zero. Thus $$((I:f^{\infty}):g^{\infty}) = (I:(fg)^{\infty})$$.

Sound right or am I missing something subtle?

Last edited: May 4, 2007