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Fearless Symmetry by Robert Gross, Avner Ash

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THE GALOIS GROUP OF A POLYNOMIAL 155
law of addition, where
ι 0
σ 1
τ 2.
If H and K are isomorphic, then everything “grouplike” about
them is the same. If H is infinite, K is infinite, and vice versa. Or if
H has 43 elements, so does K and vice versa. If H is commutative,
so is K and vice versa. And on and on, for all the fine structure you
can think of, as long as you can state it in terms of the group law.
The inverse Galois problem is the following problem: Given
any finite group H, can you find a Z-polynomial f (x) with G(f )
isomorphic to H? If you can prove that this is always possible, give
us a call! It is an important unsolved problem. It is not easy to see
why it is so difficult, but you can take this as an indication of just
how complicated these fields of the form Q(f ) can be.
This problem may also give you a little insight into what mathe-
maticians do. They do not just keep proving the same old theorems
over and over, or add up large columns of numbers. Some of them
work on the inverse Galois problem. This means trying to come up
with a general proof where, given any H, you can always find an f (x)
as above. If that is too difficult, they try to do it for certain H’s or
classes of H’s. Here there has been some success, and some very
complicated and large finite groups are known to be isomorphic
to Galois groups of the form G(f ). One of them is a huge group
known as the “Monster.” You will have to find a book on group
theory to find out more about what that is. Suffice it to say that
it has 2
46
· 3
20
· 5
9
· 7
6
· 11
2
· 13
3
· 17 · 19 · 23 · 29 · 31 · 41 · 47 · 59 · 71
elements!
Two More Things
Here is a difficult and lengthy exercise:
EXERCISE: Let f (x) = (x
2
2)(x
2
3). The roots of f are
2,
2,
3, and
3.
156 CHAPTER 13
Show that G(f ) has exactly four elements, ι, σ , τ , and ρ,
where ι is the identity permutation on the roots and the other
permutations are the following:
3
σ (±
2) =∓
2, σ (±
3)
3;
τ (±
2)
2, τ (±
3) =∓
3;
ρ(±
2) =∓
2, ρ(±
3) =∓
3.
SOLUTION: Here is a sketch: Let γ be an element of G(f ). We
must show that γ is one of ι, σ , τ ,orρ.Nowγ (
2) must be a
root of x
2
2, so γ (
2) =
2orγ (
2) =−
2. Similarly,
γ (
3) =
3orγ (
3) =−
3. Therefore, ι, σ , τ, and ρ are the
only possibilities for γ .
The harder part is to show that all four possibilities work.
First you have to identify Q(f ), which you can check is the set
of all numbers of the form a + b
2 + c
3 + d
6, where
a, b, c, and d are rational numbers. Then check that each of
ι, σ , τ , and ρ preserves addition and multiplication in Q(f ).
To get you started, this is how you show that σ preserves
addition: If s = a + b
2 + c
3 + d
6 and s
= a
+ b
2
+ c
3 + d
6, then σ(s) = a b
2 + c
3 d
6 and
σ (s
) = a
b
2 + c
3 d
6.
Now you check that σ(s + s
) = (a + a
) (b + b
)
2 +
(c + c
)
3 (d + d
)
6 = (a b
2 + c
3 d
6) +
(a
b
2 + c
3 d
6) = σ(s) + σ (s
).
By the way, you may have noticed that in all our examples, the
number of elements in G(f ) is the same as the number of rational
parameters it takes to describe Q(f ). This is always true, and is one
of the main theorems of Galois Theory.
3
The , ± notation means that in any single equation, you can resolve ± as + or ,in
which case you must resolve as the opposite sign. For example, σ (±
2) =∓
2isa
concise way of writing two equations: σ (+
2) =−
2 and σ (
2) =+
2.

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