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What I find after reading books is that they explain only the conceptual definition and no one mentions the explanation behind it; I have been reading the Galois theory as many people told me to read, so I am wondering:

  1. Why are permutation of roots important, and how do they influence the solvability of a polynomial?

    Can anyone briefly explain the direct intuition behind why did Galois consider the permuations?

  2. Please tell me: what does the profinite group $\mathrm{Gal}(\mathbb{\bar{Q}}/\mathbb{Q})$ contain?

    Is there any logical intuition behind that, like I want the answer in the form, that if someone asks what do homology sequence account for, its intuition can be given as "it counts the holes that map to another surface".

After reading about Galois theory, I find that I understood everything, but there are large blocks that stumble me.

Please help me in understanding intuition behind the theory.

Thanks a lot every one,

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"i find that i understood everything,but there are large blocks that stumble me" - if the latter is true, the former is false, so... –  J. M. Sep 5 '11 at 16:32
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its just my notion,i feel as if i understood something but i have doubts,when i go to microscopic level @J.M –  Iyengar Sep 5 '11 at 16:35
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About Ramanujan: I'd sure like to think he did mathematics because he enjoyed it, and that any acclaim was secondary. –  J. M. Sep 5 '11 at 16:46
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@iyengar For sharing your wisdom with us, please go here: philosophy.stackexchange.com. The present site is for mathematics. As for Ramanujan, you got it slightly wrong there. He was invited to Cambridge and spent the best part of his mathematical career being mentored and working together with one of the most eminent number theorists of his time. He published lots of papers, and nobody told him that they didn't make any sense. I don't know who told you that no one wanted to help him, but whoever did, he lied. –  Alex B. Sep 6 '11 at 0:42
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@Alex: That might not be a very charitable interpretation. It's true that Ramanujan sent letters to several mathematicians (among them Hardy), and that nobody but Hardy replied; it is unlikely that this was because "nobody wanted to help him", but more likely a simple result of this being but one more of numerous unsolicited messages they would get claiming great discoveries (it is worth noting that the discovery Ramanujan claimed that caught Hardy's attention, a formula to generate primes, turned out to be incorrect). So it sounds like mangling of the story and/or iyengar's interpretation. –  Arturo Magidin Sep 6 '11 at 3:26
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1 Answer 1

up vote 32 down vote accepted

First, remember that "solvability of a polynomial" had a very specific meaning: it meant finding the roots as expressions of the coefficients, and specifically expressions involving only the basic operations of additions (including subtraction), multiplication (including division) and extraction of radicals. For example, the solutions to the linear and quadratic equation are of this sort, since the roots of $p(x)=ax+b$ can be expressed as $-b/a$; the roots of $p(x)=ax^2+bx+c$ can be expressed as $\frac{-b+\sqrt{b^2-4ac}}{2a}$ and $\frac{-b-\sqrt{b^2-4ac}}{2a}$, etc. Similar with cubics and biquadratics.

Why are permutations of the roots interesting? You want to remember that if you have a monic polynomial, then the coefficients are symmetric functions of the roots: $$(x-r_1)(x-r_2)\cdots(x-r_n) = x^n -(r_1+\cdots+r_n)x^{n-1} + \cdots + (-1)^n(r_1\cdots r_n).$$ These coefficients are such that if you permute the roots, the coefficients don't change.

If we define the elementary symmetric functions on $r_1,\ldots,r_n$ as follows: $$\begin{align*} s_0(r_1,\ldots,r_n) &= 1\\ s_1(r_1,\ldots,r_n) &= r_1+\cdots + r_n\\ s_2(r_1,\ldots,r_n) &= r_1r_2 + r_1r_3 + \cdots + r_1r_n + r_2r_3 + \cdots + r_{n-1}r_n\\ &\vdots\\ s_n(r_1,\ldots,r_n) &= r_1\cdots r_n; \end{align*}$$ that is, $s_i(r_1,\ldots,r_n)$ is the sum of all possible products of $i$ distinct roots; then we have $$(x-r_1)\cdots(x-r_n) = x^n + (-1)s_1(r_1,\ldots,r_n)x^{n-1} + \cdots+ (-1)^n s_n(r_1,\ldots,r_n).$$

Suppose that $\sigma$ is a permutation of $\{1,\ldots,n\}$. If $\mathbb{Q}[x_1,\ldots,x_n]$ is the set of all rational polynomials in $n$ variables, then $\sigma$ acts on $\mathbb{Q}[x_1,\ldots,x_n]$ by mapping $p(x_1,\ldots,x_n)$ to $p(x_{\sigma(1)},\ldots,x_{\sigma(n)})$. We can then ask: what is the subset of $\mathbb{Q}[x_1,\ldots,x_n]$ that is fixed pointwise by the action of $S_n$? Clearly, the elementary symmetric functions are fixed pointwise; so are others. For instance, $x_1^2+\cdots + x_n^2$ is fixed pointwise.

The polynomials that are invariant under the action of $S_n$ are the "symmetric functions" on $x_1,\ldots,x_n$. Newton proved that the elementary symmetric functions generate the symmetric functions: every symmetric function can be expressed as combination of the elementary symmetric functions.

So the coefficients of a polynomial are intimately related to the symmetric functions of the roots, which are in turn intimately connected with the action of $S_n$ on the roots.

For example, let's consider the quadratic equation in this light. We have $$x^2 +bx + c = (x-r_1)(x-r_2),$$ so $b=-(r_1+r_2)$, $c=r_1r_2$. To express $r_1$ and $r_2$, separately, using $b$ and $c$, consider the symmetric polynomials $(r_1+r_2)^2$ and $(r_1-r_2)^2$ on the roots. Since these are symmetric, they can be expressed in terms of $b$ and $c$ (which generate the symmetric polynomials). Indeed, $$\begin{align*} (r_1+r_2)^2 &= (-(r_1+r_2))^2 = b^2;\\ (r_1-r_2)^2 &= (r_1+r_2)^2 - 4r_1r_2 = b^2 - 4c. \end{align*}$$ Thus, $|r_1-r_2| = \sqrt{b^2 - 4c} $ , so $r_1-r_2 = \sqrt{b^2-4c}$ or $r_1-r_2=-\sqrt{b^2-4c}$. Since $r_1+r_2 = -b$, we have $$\begin{align*} r_1 &= \frac{1}{2}\Bigl( (r_1+r_2) + (r_1-r_2)\Bigr) =\left\{\begin{array}{l} \frac{1}{2}\Bigl( -b +\sqrt{b^2-4c}\Bigr)\\ \text{or}\\ \frac{1}{2}\Bigl( -b -\sqrt{b^2-4c}\Bigr) \end{array}\right.\\ r_2 &=\frac{1}{2}\Bigl( (r_1+r_2) - (r_1-r_2)\Bigr) = \left\{\begin{array}{l} \frac{1}{2}\Bigl( -b - \sqrt{b^2-4c}\Bigr)\\ \text{or}\\ \frac{1}{2}\Bigl( -b+\sqrt{b^2-4c}\Bigr) \end{array}\right. \end{align*}$$ which gives the usual quadratic formula: one root of $x^2+bx+c$ equals $\frac{-b+\sqrt{b^2-4c}}{2}$, the other equals $\frac{-b-\sqrt{b^2-4c}}{2}$.

A similar approach can be used for the cubic and the biquadratic. The question is whether something similar can be done with the quintic and higher. This particular straightforward approach (originally due to Lagrange) runs into a problem: to solve a cubic, you end up having to solve a quadratic equation on the elementary symmetric polynomials. To solve a biquadratic, you end up having to solve a cubic. But to solve a general quintic, you end up having to solve a polynomial of degree six! So you run into a roadblock.

Galois Theory studies the roots by studying the "symmetries" among the roots, by considering their permutations (which necessarily leave the coefficients fixed), and considering how certain subgroups of $S_n$ leave (or not) the coefficients or other functions of the roots fixed.

As to your second question: the group $\mathrm{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})$ has, as a quotient, every single Galois group over $\mathbb{Q}$. This is known to include at least all the solvable groups (a theorem of Shafarevich), as well as many of the nonabelian simple groups. It is still an open question whether every group is in the Galois group of a polynomial over $\mathbb{Q}$. The group itself can be described abstractly, but we still don't have a very good "feel" for it. In fact, a lot of the work on Galois representations (which was key to the proof of the Taniyama-Shimura Conjecture) has to do with understanding "just" the image of this group in suitable matrix groups (i.e., trying to understand the representation theory for this group, in order to gain some insight into the group itself).

As for "intuition": any infinite Galois extension is completely determined by its finite Galois sub-extensions; this is why $\mathrm{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})$ is a profinite group: it is determined by the Galois groups of the finite sub-extensions it has as quotients. The possible images of an element $a\in\overline{\mathbb{Q}}$ under any homomorphism $\overline{\mathbb{Q}}\to\overline{\mathbb{Q}}$ that fixes $\mathbb{Q}$ must be another root of the minimal polynomial of $\alpha$, and so the homomorphism will restrict to an automorphism of the Galois closure of $\mathbb{Q}(\alpha)$. Any element of $\mathrm{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})$ is thus determined by its action on these finite Galois extensions, and the inverse limit is a way of "gluing" all this information together in a coherent way. But the group is far from understood.

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:most elegant answer,i thought of giving $+\infty$ but no way of doing it,you helped me a lot sir,even though you dont know me personally,i am in debt with you,and never forget your help,touching your feet –  Iyengar Sep 6 '11 at 14:47
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i can only give +1 –  Iyengar Sep 6 '11 at 14:47
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:and you even took risks to edit my stupid english,and correct my sentences,just infinitely many thanks to you sir –  Iyengar Sep 6 '11 at 14:55
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you dont know how happy i felt after reading the answer,i understood it perfectly,and its after long search i got the answer,someone feels much happy when they find the answer they are looking for ,i never forget you,but a small advice ,why cant you author some good books on the theories sir,like everyone looks for the exact way of explanation you gave ,so it is the great service you render this world ,and your explanation cant be found in any of the books,its just by intuition one gets such a realization –  Iyengar Sep 6 '11 at 15:23
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