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I Affine Algebraic Sets - General Question

  1. Jun 18, 2016 #1
    I am trying to gain an understanding of the basics of elementary algebraic geometry and am reading Dummit and Foote Chapter 15: Commutative Rings and Algebraic Geometry ...

    At present I am focused on Section 15.1 Noetherian Rings and Affine Algebraic Sets ... ...

    I am trying to gain a full understanding of the nature of affine algebraic sets ...

    If we take an arbitrary subset A of affine space ##\mathbb{A}^n## ... how can we determine whether A is an affine algebraic set ... ?

    Are they any methodical approaches ... ?

    Do we just have to creatively come up with a polynomial or set of polynomials whose set of zeros equals A?

    Any clarifying comments are welcome ...

    Peter
     
  2. jcsd
  3. Jun 18, 2016 #2

    fresh_42

    Staff: Mentor

    An arbitrary subset of ##\mathbb{A}^n## isn't much of information. This could be literally everything. So to start with you have to have some definition of it. In general it is the other way round: Given such a set, i.e. defined by any algebraic equations, it's an algebraic variety, an affine or projective algebraic set. This leaves really many examples: all curves, hyperspaces and so on. Even such things like for instance linear algorithms are algebraic varieties because they are defined by algebraic equations.

    The ring of polynomials which don't vanish on such a set is also called coordinate ring, i.e. the quotient ##k[V] = k[X_1, \dots ,X_n] / \mathfrak{I}(V)##. It helps to do calculations on it.
    For an overview of the matter you might read these Wiki articles:

    https://en.wikipedia.org/wiki/Algebraic_variety
    https://en.wikipedia.org/wiki/Sheaf_(mathematics)

    So to answer your question one has to ask: How do you describe an arbitrary subset of an affine space?
    (This is absolutely necessary in order to decide which properties it has or has not.)

    By the way: if you have inequalities, e.g. ##xy \neq 0##, you get dense open subsets (which are not affine algebraic sets of course).
     
    Last edited: Jun 18, 2016
  4. Jun 18, 2016 #3

    mathwonk

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    Science Advisor
    Homework Helper

    algebraic subsets are very special, so almost any "arbitrary" subset is not algebraic. for one thing, a proper algebraic subset of affine n space has a finite number of irreducible components, and each of them has dimension less than n. so an open ball is not algebraic, nor is any proper subset that contains a non empty open ball. also they have finiteness properties, so an infinite sequence of distinct points is not algebraic, and neither is an infinite sequence of distinct lines. assuming we are over say the complex numbers. Moreover each individual component of an algebraic set has a fixed dimension that is visible at every point of it. thus in affine 3 space, an algebraic set consists of a finite number of points, a finite number of irreducible curves, and a finite number of irreducible surfaces. oh yes, and over the complex numbers say, a positive dimensional affine algebraic set is unbounded, it cannot be say a finite line segment. this boundedness of course is not true over the reals, since the unit circle and the unit sphere are algebraic over R.

    Over any field k, the coefficient ring R of an irreducible affine algebraic set V is a finite extension of a polynomial ring, i.e. R is a module finite extension of a ring of form k{X1,...,Xr], where r is the dimension of the algebraic set, (see noether normalization, DF, p.699). When k is algebraically closed, this implies that the algebraic set V is a finite cover of affine r space via projection. So every irreducible curve in this setting is a finite cover of the affine line, every irreducible surface is a finite cover of the affine plane, etc.... In particular this shows the unboundedness in that case.

    In simple cases one can also try to list all algebraic sets. In the plane, over an algebraically closed field k say, to list curves we can proceed by degree. all degree one sets are lines. all degree two sets are classified up to isomorphism by rank, either a double line (rank 1), two distinct lines (rank 2), or a smooth conic (rank 3). In degree three we can prove, up to isomorphism, they are either three lines, or a line and a conic, or if irreducible they have the very special form y^2 = x(x-1)(x-c), in some choice of coordinates. after that it gets harder.

    edit: oops, those last special ones are only the "non singular" cubics, (and some "nodal" cubics when c= 0 or 1). there are also "cuspidal" cubics such as y^2 = x^3. that may be all.

    it is traditional , after riemann, to try rather to classify algebraic varieties up to abstract isomorphism, and then for each abstract variety to try to classify all ways it can be embedded in (not affine but) projective space. for irreducible curves the basic classification is by the genus, and for each curve of a given genus, one studies its projective models by means of the "Jacobian" variety of line bundles, which correspond to projective mappings of the curve to projective space. riemann knew the space of all irreducible non singular curves of fixed genus g > 1, can itself be given the structure of a variety of dimension 3g-3. In genus one, all curves have the special form of the irreducible plane cubics given above, so they form a one dimensional family, with the constant c as (almost) a parameter.
     
    Last edited: Jun 23, 2016
  5. Jun 23, 2016 #4
    Hi fresh_42, mathwonk,

    Firstly ... so sorry for late reply ... had a bad case of the flu ... indeed still not fully recovered ...

    However, I managed to read both of your posts at least twice ... and as a result have a better understanding and appreciation of the nature and character of affine algebraic sets ...

    Thanks for your help in this ... really appreciate your help in my getting an understanding of the basics of elementary algebraic geometry ... from skimming some books, it looks like a fascinating subject ...

    Thanks again,

    Peter
     
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