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Fearless Symmetry by Robert Gross, Avner Ash

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THE GALOIS GROUP OF A POLYNOMIAL 151
means it must be a lot simpler than the big Galois group G.In
the next chapter, we will see some connections between all of these
groups. (It may be a good idea to review chapter 3 at this time.)
Why is G(f ) finite? The reason is that any permutation in G(f )
must permute the roots of f , and there are at most d different roots.
After you know how the roots are permuted, this fixes how all the
other numbers in Q(f ) must be permuted, because they are cooked,
using the operations of basic arithmetic, only out of these roots and
Q—no other ingredients. And elements in G(f ), just like elements
of G, must fix every rational number, as we saw on page 97. So there
are at most d! elements in G(f ).
We summarize with the following theorem:
THEOREM 13.1: Let f be a Z-polynomial and let A be the set
of all its roots in Q
alg
.Letr be the function from G(f )to
A
which sends any element of G(f ) to the permutation it causes
on A. Then r is a faithful morphism and hence a faithful
permutation representation of G(f ).
In fact, it is useful at this point to redefine G(f ) as the set of
permutations of the roots of f that, when extended to Q(f ), preserve
addition and multiplication. Thus:
DEFINITION: Let f be a Z-polynomial and let A be the set of
all its roots. Then:
Q(f ) is the smallest field containing Q and all of the
elements of A.
G(f ) is the set of permutations σ of A with the following
property: There is some permutation of Q(f ) that
preserves addition and multiplication and that is the
same as σ on the elements of A.
Examples
Phew! Maybe some examples will help.
EXAMPLE: f (x) = x 3. This is a really easy example. At the
beginning, only the root(s) of f (x) and the rational numbers Q
152 CHAPTER 13
are in the pot. The only root of f (x) is 3, so our pot contains
only all of Q. If we start adding, subtracting, multiplying, and
dividing elements of Q (where we, of course, remember not to
divide by 0), we only get other rational numbers. So Q(f )is
just Q. Any permutation of Q that preserves arithmetic must
preserve 0 and 1 and hence all integers and hence all rational
numbers. Such a permutation can only be the identity
permutation. So G(f ) is the “trivial group” that only has one
element in it, the neutral element. We generally call the
neutral element in a group e, but from now on, we will call
the neutral element in a Galois group ι, the Greek letter
iota. It stands for “identity.
EXAMPLE: f (x) = x
2
5. This one is not too difficult. The two
roots of f(x) are
5 and
5. If you multiply by various
rational numbers and add to other rational numbers you find
that Q(f ) contains every number of the form a + b
5, where
a and b are arbitrary rational numbers. Some more work will
convince you that this is all there is in Q(f ).
Now we try to figure out what G(f ) is. Of course, G(f )
always possesses the neutral element ι.Ifσ is another
element of G(f ), what could it do to
5? It must take it to one
of the roots of f (x), so either to
5orto
5. Remember that
any element of any Galois group fixes every rational number.
If σ takes
5to
5, then because it respects all arithmetic
operations, it will have to take a + b
5toa + b
5. That is, σ
would be the neutral element ι. Because we are assuming it is
not, the only choice is for σ to take
5to
5. Then, it must
take a + b
5toa b
5, no matter what rational numbers a
and b may be.
If you want a good algebra exercise, you can prove that if σ
is the permutation of Q(f ) defined by σ(a + b
5) = a b
5,
then σ does preserve the four arithmetic operations.
(Remember, this means you have to check that
σ (u + v) = σ (u) + σ (v) and σ (uv) = σ (u)σ (v) for all elements u
and v in Q(f ). The other two requirements,

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