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Galois in a nutshell

  1. May 28, 2012 #1
    Hi I struggled with Galois theory as a senior, and would like to improve my ability to jump in and out of the subject with a sort of big picture to details map.

    So as I see it, there are four things to be proficient at:

    Knowledge and understanding of:
    1. Theory,
    2. Particular examples,
    3. Relevant group theory
    4. General examples,
    and how they all relate.

    I conjecture that to feel comfortable in the subject at the undergraduate level, you have to cover these four bases.

    Now I will go into more detail about what I mean by these each of these four.

    By THEORY, I mean understand all the theorems and what they say, and given any question on the theory, be able to dig through the chain of arguments to answer questions about the inner working of the theory. Also be able to recognize which facts tend to be use for which parts of examples.

    By PARTICULAR EXAMPLES, I mean taking a specific polynomial, or a field extension, and finding the splitting field, or irreducible polynomial respectively, and understand the group structure and the structure of the field extension.

    By RELEVANT GROUP THEORY, I mean it would be nice if so late in an algebra course, they did not fly past all the facts they need from group theory, it would be nice to have a focused presentation of group theory and for those facts that are used over and over in computing galois groups at this level.

    By GENERAL EXAMPLES, I mean along the idea that at some point they go through the arguments for general quadratics, general cubics, and some of the facts can be used for general nth degree polynomials, like the notion of discriminant.


    So it is a project of mine right now to find good sources to get a better feel for Galois so that I will not feel so disoriented in the subject. I get stuck way too often on reading people's arguments in the subject, because for instance Artin has very abbreviated reasoning.

    Please let me know if you think I'm missing something in the big picture.
     
    Last edited: May 29, 2012
  2. jcsd
  3. May 29, 2012 #2
    "The Galois group of a polynomial extension can be computed readily if the
    zeros of the polynomial are known and can sometimes be obtained indirectly when
    they are not." -http://web.science.mq.edu.au/~chris/galois/chap04.pdf

    So if we know the roots, we might be asked to find the group and the field structure.

    If the roots are not known, we may still be able to find the group.
     
  4. May 30, 2012 #3

    Stephen Tashi

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    One of my professors once remarked "The half life of Galois Theory is about 10 seconds". He didn't elaborate on that statement, but we understood it to mean that you forget half of what you know in that time.

    I think an important thing to establish about mathematical topics where students have heard a popularized version is how well the subject will live up to the popularized version. The popularized version of Galois theory is that it was developed to 1) Show the impossibility of the general solution of certain polynomial equations and 2) Solve certain special polynomial equations.

    From what I remember about the subject, using it to solve polynomial equations isn't a step-by-step process (like an algorithm that either produces an answer or returns "not solvable by Galois Theory"). It seems to me to require a miscellaneous set of deductions that don't follow any general pattern. I find it like the group theory problems where you are given an integer N and asked "How many groups are there of order N?" or asked to prove some special property about the group(s) of that order.

    On the one hand, it's nice for people to exercise deductive powers in mathematics, but on the other hand the "power" of mathematics seems to be most evident when it can applied without much thinking! Machinery that can grind out an answer is more impressive that a collection of hints.

    Anyway, that's my impression of the applicability of Galois theory (and the theory of group "structure"). I'm not an expert in either one and it would be interesting to know if my impression is wrong.
     
  5. May 30, 2012 #4
    Stephen, I love that comment by your professor! I can never remember what the Galois theory details are. With that disclaimer, I would like to submit my 2 cents on what the subject is about in a larger context.

    It is a general phenomenon in math that we can find explicit solutions/formulas for problems that have enough inherent symmetry. The symmetry is characterized by a group action that has certain properties the formulation of which depend on the problem. But if you know the group of symmetries and the explicit group action, then you can use that to explicitly construct solutions. For polynomial equations, this is called Galois theory.

    The same idea works in differential equations. Examples:
    1. y'(x) = f(y(x)). This is symmetric with respect to translations in t, and an explicit solution happens to exist (see separable equations)

    2. y' = f(x). This is symmetric with respect to translations in the x direction. Explicit solutions also exist (antiderivatives of f)

    3. dy/dx = f(y/x) . This problem symmetric under a scale change of both variables.

    4. y'(x) + p(x) y(x) = 0. This problem is scale invariant in the y variable. Note that means that if we set y = e^u, then scale invariance with respect to y becomes translation invariance with respect to u. And you can see how that is related to the solution to the original equation using integrating factors.

    5. Eulers' equation: x^2y'' + a xy' + by=0. Scale invariant with respect to x. Hence the substitution x=e^t to switch the symmetry to translation invariance in t.

    Anyways, in Diffy Q there is an explicit method to construct solutions if you know enough symmetries of your original equation. Mechanical systems are special cases of this. Ever wonder why the 2 body gravitation problem has an explicit solution but not the general 3 body problem. Just enough symmetry in the former, not enough in the latter.

    So while the details of Galois theory sound quite different to all that, the spirit of analyzing solution through groups of symmetries is very much the same.
     
  6. May 30, 2012 #5
    Thanks for the notes Stephen and Vargo!

    Stephen, I would like to expend some energy trying to make Galois theory and practice at the introductory level feel less "miscellaneous". And while I may not succeed, at least through these efforts, it's half life will be increased.

    Vargo, yes, symmetry to reduce problems is a huge idea in math, and I actually like to absorb it into the bigger umbrella of choosing coordinates in math. I will edit this later if I remember more, but I think of the choice of coordinates, whther motivated by groups of symmetries or other ways, can be seen as the main idea. Whether we decide to "coordinatize" a polynomial as a sum of terms, or a product of terms, depending on what question we want to answer. The sum of terms is one choice of "coordinates", good for taking derivatives. The product of terms is a good choice of "coordinates" if we want to recognize the roots, or maybe even the local behavior. Expressing spaces such as groups or manifolds as a quotient or product of spaces is another good way to "coordinatize", or break the problem down into pieces.
     
  7. May 30, 2012 #6

    mathwonk

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    The idea of Galois theory is to analyze the complexity of solving a polynomial by measuring how many times you have to add in more roots to the coefficient field until you have them all.

    E.g. an irreducible quadratic polynomial is not complicated at all, because when you add in one root you automatically get the other one as well. So you made one degree 2 extension and the Galois group is Z/2.

    Irreducible Cubics come in two types, first those for which you only have to add in one root to get both the others also. In this case when you add in one root r, the cubic already factors into three linear factors.

    For the more complicated ones, when you you add in one root r, the cubic factors into one linear factor and one irreducible quadratic factor. Then you have to add in another root of that quadratic factor before getting the third root for free.

    In the first case the "root field" has extension degree 3, so the Galois group has order three, and must equal Z/3, while in the second case the dimension of the root field extension is 3.2 = 6, and the group has order 6. Since the group is always a subgroup of permutations of the three roots, in that case the group must be S(3) the permutation group on 3 elements.

    In general, the Galois group is determined by how the original irreducible polynomial factors in each successive extension field, as you add in more and more roots until you have them all.

    E.g. if as you add in more roots, you never get any for free until the last quadratic step, then for an irreducible polynomial of degree n, you have to make n-1 extensions, of degrees n, n-1, n-2,.....,2. Thus the full root field extension has degree n!, and must equal the full permutation group S(n).

    If p is prime then X^p-1 = (X-1)(X^p-1 +.....+1) where the second factor is irreducible with roots equal to the "primitive" pth roots of 1. Any of these roots gives all the others as powers, so you only have to add in one of them, a degree p-1 extension, to get them all.

    Thus the Galois group has degree p-1, and it is just Z/(p-1) ≈ the multiplicative group of units in the ring Z/p.

    I once wrote a free book on this when I taught the course long ago.

    http://www.math.uga.edu/%7Eroy/843-2.pdf [Broken]

    Reading it I find the following proposition: an irreducible polynomial over Q of prime degree p, with exactly two non real roots, has Galois group S(p).

    E.g. X^5 - 80X + 2 has group S(5) (hence is not solvable by radicals).

    proof: Since it has prime degree p. the Galois group has order divisible by p. I.e. adding in the first root gives an extension of degree p, and degrees of successive extensions multiply. Since the Galois group is a transitive subgroup of S(5) it contains a cycle of length p, such as ( 1 2 3.....p). Now by definition, the Galois group is the automorphism group of the root field fixing the coefficient field Q. But complex conjugation must restrict to an automorphism of the root field that interchanges precisely the two non real roots. Thus the group G is a transitive subgroup of the group S(5) containing a p cycle and a transposition. Hence (exercise) it is the whole group S(5).
     
    Last edited by a moderator: May 6, 2017
  8. May 30, 2012 #7

    mathwonk

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    In a nutshell, the Galois group is the automorphism group of the root field, fixing the coefficient field, let's say Q for simplicity.

    The order of the root field as a vector space over the coefficient field Q, is the product of the degrees of the successive extensions used to obtain it, hence is always ≤ n!, where n = degree of the polynomial.

    The Galois group is a subgroup of the full permutation group of the distinct roots, and the order of the Galois group is equal to the degree of the root field (dimension as a vector space over the coefficient field).

    To "determine" the galois group, say by writing down a multiplication table, one needs much more than just a list of the roots {a1,....,an}.

    One needs to know exactly how the polynomial f factors in each of the successive extensions toward obtaining the full root field,
    i.e. one needs the irreducible factorizations of f in all the partial root fields:
    Q, Q(a1), Q(a1,a2),.....,Q(a1,....an).

    Exercise: Suppose f has degree n and f = gh, where g and h are distinct irreducible polynomials of degree > 1. Can the Galois group of f equal S(n)?
     
    Last edited: May 31, 2012
  9. Jun 7, 2012 #8
    PARTICULAR EXAMPLES

    (expanding on subtopic of first post)

    Thank you Mathwonk. As I've been studying, I've made it to "knowing" half of what you've said, so I've more to do before I might claim I've paraphrased the techniques you've mentioned.

    I've been lookng through examples trying to figure out the methods (for analyzing the extensions associated with a polyomial) generally used, and so far, here are the commonalities I've found.

    1. Find the roots
    2. Find algebraic generators of extension
    3. Determine the group actions on roots
    4. Determine the group, and recall the structure
    5. Find intermediate fields (for instance, using linear basis)

    So this is my attempt to cover methods commonly used for "particular examples", as I mentioned above.
     
  10. Jun 8, 2012 #9

    Hurkyl

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    Are you comfortable at all with finite fields? Galois theory there is a basic fact of life, but fortunately turns out to be somewhat simplified too. e.g.

    (q is always a power of the prime p)

    • The Galois group of GF(qn) / GF(q) is cyclic of order n, generated by the Frobenius action x -> xq. (The same is true of infinite algebraic extensions, if you think in terms of procyclic groups)
    • There is a unique field of each size. e.g. GF(q3) is the splitting field of every irreducible cubic polynomial over GF(q), and every irreducible degree 6 polynomial factors into two quadratics over GF(q3)
    • Polynomials like xp2+axp + bx turn out to be linear functions on GF(q), which entails a lot of weirdness both good and bad in how polynomials behave.

    Knowing finite fields thoroughly can help with more advanced topics in algebraic number theory (which I'm just beginning to learn), in terms of the relationships between a field extension and how it behaves on the residue fields of the base field: the residue fields are finite fields in the most interesting cases like Q or Fq(x).
     
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