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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