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154 CHAPTER 13
We call the three roots of this cubic u, v, and w. They are all
irrational real numbers:
u =−4.786 ...
v =−0.446 ...
w = 4.230 ...
In this case, it turns out that G(f ) has only three elements: ι,
σ , and τ . As usual, ι is the neutral permutation, while σ
causes the permutation u v w u on the roots of f , and
τ causes the permutation u w v u.
You may think that G(f ) should have six elements, because
there are six possible permutations of the three numbers u, v,
and w. However, each of the other three permutations (e.g.,
u v u, w w) fails to belong to the Galois club: They do
not extend to permutations of Q(f ) that preserve addition and
multiplication.
That is enough examples. This is already looking too much like a
textbook on Galois theory. You can look at some actual texts (Artin,
1998; Edwards, 1984; Fenrick, 1998; Gaal, 1998; Garling, 1986;
Rotman, 1998; Stewart, 1989) if you want to see more complicated
examples.
Digression: The Inverse Galois Problem
We say that two groups H and K are “isomorphic” if there is a
one-to-one correspondence between them that is a morphism, that
is, that preserves the group laws. For instance, in the examples
above,
{
5,
5}
is isomorphic to
{1,2}
,
{r,s,t}
is isomorphic to
{1,2,3}
, and
G(f ) in the last example is isomorphic to F
3
with the group

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