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I get the basic idea of the Galois group and Galois correspondence but I am missing the connection between them and symmetric polynomials (in terms of intuition).

What I understand about Galois group and correspondence

  1. Let $\mathbb{Q}$ be the base field and $F(r)$ be an extension field of $\mathbb{Q}$ where $r$ is a root of an irreducible polynomial $f$ over $\mathbb{Q}$. Then the set of all automorphisms on $F$ that fixes $\mathbb{Q}$ is essentially a subgroup of the symmetric group $S_n$ and is known as the Galois group denoted by $Gal(F/\mathbb{Q})$.
  2. But not all subgroups of $S_n$ will induce automorphisms and the central idea behind Galois theory is about those that do.
  3. If $F$ happens to be the splitting field of $f$, then the subgroup will turn out to be a normal subgroup of $S_n$. And for any intermediate field extension $E$ such that $\mathbb{Q} \subseteq E \subseteq F$, $E$ is a splitting field of an irreducible polynomial $g$ (over $\mathbb{Q}$) if and only if $Gal(E/\mathbb{Q})$ is a normal subgroup of $Gal(F/\mathbb{Q})$.

So far so good. Now let me tell something I understand about symmetric polynomials and group actions.

What I understand about polynomials and group actions

Consider this set consisting of three polynomials:

$p_1 = x_1^2+x_2^2+x_3^2+x_4^2$, $p_2=x_1x_2+x_3+x_4$, and $p_3=x_1x_2^2-x_3^3x_4^4+5x_1x_2x_3x_4$

Now, if we compute the stabilizers of this set with respect to $S_4$ acting on these polynomials, we can see $Stab(p_1)$ will be all of $S_4$ (since any permutation of $S_4$ will leave $p_1$ unchanged), $Stab(p_2)$ will be $(12)$, $(34)$, $(12)(34)$ and the identity $\{e\}$ whereas $Stab(p_3)$ will only be the identity $\{e\}$.

If we compute the orbits of this group action, we will have:

$Orbit(p_1) = \{p_1\}$, $Orbit(p_2) = \{x_1x_2+x_3+x_4, x_1x_3+x_2+x_4,x_1x_4+x_2+x_3,x_2x_3+x_1+x_4,x_2x_4+x_1+x_3,x_3x_4+x_1+x_2 \}$ and the $Orbit(p_3)$ will be all $24$ polynomials corresponding to $S_4$.

Thus,

  1. The stabilizer of a polynomial is a measure of its symmetry, the larger the stabilizer, more symmetric the polynomial is.
  2. The orbits are also a measure of its symmetry but in the reverse direction. The smaller the orbit, more symmetric the polynomial is.

What I don't understand

I know these two ideas are somehow intimately connected since I have read Galois theory is about the permutations of the roots of the polynomials. That is if we construct a set of polynomials with the roots, some may be symmetric and others may not. But I am failing to understand how this symmetry/asymmetry of polynomials of the roots are tied to the notion of Galois group and correspondence, especially in the light of field extensions and irreducible polynomials.

I understand it is a very broad question and I am looking for an intuitive explanation and not theorems and proofs. Any help will be much appreciated. Thanks for your patience with this novice.

Update From this article I see that the Galois group consists of such permutations where solutions of polynomial expressions involving the roots remain invariant under those permutations. Not sure how they are equivalent in an intuitive way.

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  • $\begingroup$ "3. ...will turn out to be a normal subgroup..." That's wrong. For $n\geq 5$ $S_n$ has only two normal subgroups: $1$ and $A_n$. $\endgroup$
    – TomTom314
    Commented Sep 29, 2019 at 9:28

1 Answer 1

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For $f\in K[X]$, irreducible and separable with roots $a_1,\ldots a_n\in L$ in a splitting field there is a homomorphism $$\varphi: K[X_1,\ldots,X_n]\to L: X_i\mapsto a_i$$ For $\mu \in Aut_K(K[X_1,\ldots,X_n])$ you can show: $\exists \sigma \in Aut_K(L/K):\sigma\circ\varphi =\varphi\circ\mu \iff \mu(\ker\varphi)\subset \ker(\varphi)$. Now you can compare polynomials $g\in K[X_1,\ldots,X_n]$ with permutations of $a_1,\ldots a_n$.

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