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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