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When talking about field extensions of degree two I understand the automorphisms in the Galois group intuitively as analogous to complex conjugation. I lose my understanding when going to field extensions generated by cubics.

Suppose we have roots $a_1, a_2, a_3$ so that $K=F(a_1, a_2, a_3)$. It's possible for $G(K/F)$ to have order 3 even though there are six possible permutations of these roots. So where did the other 3 go? I see two possibilities:

  1. Some permutations of the roots are not automorphisms. This seems unlikely because I think $F(x,y)\cong F(y,x)$ always, correct?
  2. Some permutations are equivalent. I don't understand how this could be.

Is there a third option I didn't think of? Am I missing the point entirely?

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Basically, what you are missing is that the roots need not be algebraically independent: there could be some $F$-relation between $a_1$, $a_2$, and $a_3$, and the automorphism has to respect that relation; this is what is happening in Jack's example, where every root is a power of any other root. –  Arturo Magidin Sep 8 '11 at 21:45
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3 Answers

up vote 8 down vote accepted

The first option is what happens. While $F(x,y) \cong F(y,x)$ the isomorphism may not take $x$ to $y$ and $y$ to $x$.

I think an easy example (with more roots) is the polynomial $x^4+x^3+x^2+x+1 = (x^5-1)/(x-1)$. The roots are the four non-identity powers of $\zeta_5 = \exp(2\pi/5)$.

If a permutation sends $\zeta_5$ to $\zeta_5^3$, where does it send $\zeta_5^2$? Well, as a permutation it could send it to anything except $\zeta_5^3$, but as a field homomorphism it has to respect multiplication, and so it has to send it to $(\zeta_5^3)^2 = \zeta_5^6 = \zeta_5$. In other words, out of the three possibilities of where to send it as a permutation, only one of the three possibilities worked out to be a homomorphism.

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Let me give an example, which I learned from Toby Gee.

Let $f(X) = X^3 - 3X + 1$, a polynomial in $\mathbf Q[X]$. By the rational root test from high school algebra, this has no roots in $\mathbf Q$ and is hence irreducible. It is easy to check that if $\alpha_1$ is a root of $f$ in some extension of $\mathbf Q$, then so is $\alpha_2 = 1 - 1/\alpha_1$. Repeating this operation, we find the last root $\alpha_3 = 1 - 1/\alpha_2 = 1/(1 - \alpha_1)$.

We see that the splitting field $K$ of $f$ is equal to $\mathbf Q(\alpha_1)$, and that while the Galois group acts transitively on the $\alpha_i$, an automorphism of the extension is completely determined once we have chosen the image of an $\alpha_1$, as the other two roots are rational expressions in $\alpha_1$. So not all permutations of the roots appear and the Galois group is isomorphic to $A_3$.

For more on the Galois groups of cubics — a criterion on the discriminant and many examples — read Keith Conrad's handout.

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Your alpha_3 should be a function of alpha_2, shouldn't it? I think you've got a misprint there. –  Marty Green Sep 8 '11 at 21:41
    
@Marty I was trying to emphasize that $\alpha_3$ has a rational expression in terms of $\alpha_1$. Maybe it's just confusing, though? I do think the expression is correct. I'll spell it out. –  Dylan Moreland Sep 8 '11 at 21:44
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Not a complete answer but this might help: In a more general sense we can view Galois groups of irreducible polynomials as transitive subgroups of $S_n$ where $n$ is the degree of the polynomial. This different definition will probably give you more intuitive feel for that: Let $f$ be an irreducible polynomial of degree n with coefficients in some field $K$ whose roots are $a_1,..,a_n$ then the Galois group $G$ consists of those permutations which preserves the relations between them i.e. G is the set of all those permutations σ of the symbols $1,..,n$ such that $\phi(a_{σ(1)},..,a_{σ(n)})=0$ for every $n$ variable polynomial $\phi$ for which $\phi(a_1,..a_n)=0$.

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