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Fearless Symmetry by Robert Gross, Avner Ash

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186 CHAPTER 16
which says that 5|N(Frob
5
(i) i). If we try Frob
5
(i) =−i,we
get 5|N(2i), which is not true, so the only possibility is that
Frob
5
(i) = i.
EXERCISE: Let p be an odd prime. Show that
Frob
p
(i) =
i if p 1(mod4)
i if p 3(mod4).
Notice that this formula looks an awful lot like equation (7.4) on
page 79. This is not a coincidence. See chapter 19.
Unfortunately, this example is misleading, for in the general case
there can be more than one root of the minimal polynomial that
satisfies the divisibility relationship. For the complete definition of
Frob
p
(θ), see the second appendix to this chapter.
We have been discussing ramification with respect to θ ,and
at the beginning of this chapter we spoke of ramification with
respect to a Galois representation r. What is the connection? Well,
we would have to spell out the connection between number fields
and Galois representations, which we do in the first appendix to
this chapter.
Frob
p
and Factoring Polynomials modulo p
Now that we have told you what Frob
p
looks like, at least some of
the time, we should tell you something about it. A lot of the amazing
things will wait for future chapters, but here is one thing that we
can use to finish off this chapter.
Pick an irreducible Z-polynomial f (x)ofdegreen with leading
coefficient 1. (Irreducible just means that it does not factor into
Z-polynomials of smaller degree. A consequence of irreducibility
is that f (x) is the minimal polynomial of each of its roots.) We
can take the n roots α
1
, α
2
, ... , α
n
of this polynomial. Then we
know that any element of the Galois group G permutes these roots.
In particular, pick any prime p so that
f
is not evenly divisible
by p.Thenp is unramified with respect to any of the roots of f (x).
FROBENIUS 187
We know that Frob
p
permutes the roots. What can we say about this
permutation?
As we have seen, permutations can be broken up into cycles.We
can start with a root α
1
, apply the permutation Frob
p
to it to get
another root, apply Frob
p
to that root, and keep going. Eventually,
we have to get back to α
1
, because there are only finitely many roots.
The number of different roots that we visit on our trip is called the
length of a cycle. For example, if Frob
3
(α
1
) = α
2
,andFrob
3
(α
2
) = α
11
and Frob
3
(α
11
) = α
8
,andFrob
3
(α
8
) = α
1
, then this cycle has length 4,
because it visits the four different roots α
1
, α
2
, α
11
,andα
8
.
There will also be another cycle, starting at α
3
,becauseα
3
was
not part of the previous cycle. We can keep going, putting each root
of the polynomial into a cycle, and counting the lengths of the cycles .
The lengths have to add up to n, because each root is in exactly one
cycle. (If you are worried about the possibility that Frob
3
(α
5
) = α
5
,
we call that a cycle of length 1.) So we have a bunch of positive
integers n
1
, n
2
, ..., n
k
,sothatn
1
+ n
2
+···+n
k
= n.Theseintegers
are the lengths of the cycles produced by the permutation on the
roots. We say that Frob
p
“has cycle type n
1
+ n
2
+···+n
k
.”
4
We can also try factoring the polynomial f (x)inF
p
. For exam-
ple, x
2
11 = (x + 10)(x 10) in F
89
. Here is an amazing theorem
connecting cycle types and factorizations:
THEOREM 16.1: If f (x) is an irreducible Z-polynomial and p
is a prime not dividing
f
,andiff (x) factors in F
p
into k
factors, and the degrees of those k factors are n
1
, n
2
, ..., n
k
,
then the cycle type of Frob
p
is n
1
+ n
2
+···+n
k
.
In general, it is easy to factor polynomials with elements in F
p
(there are fast computer algorithms for this), but it is difficult to
figure out Frob
p
. So we usually use the factorization to tell us about
the cycle type of Frob
p
.
4
You may object that we are ignoring the ambiguity in the notation Frob
p
here. However,
the cycle type will turn out to be the same, whatever choice of Frob
p
you make. Also,
remember that the + sign in the cycle type does not denote addition, but is the
traditional symbol for separating the lengths of the cycles from each other.

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