| >> Let F be a finite field with characteristic p (p is prime) and
>> |F| = p^n where n is an integer > 0.
The map taking an element b of F into b^p is a field
isomorphism (from F to F, so it is also called an
automorphism) leaving 0,1,...,p-1 fixed and moving
everything else.
If f(x) is a polynomial in F and a,b are in F with
a = f(b), then as b -> b^p is a field isomorphism you
will have a^p = g(b^p), where the polynomial g is what
you get by replacing each coefficient c of f(x) with c^p.
If the coefficients of f(x) are all in 0,1,...,p-1 and if
a = 0, then f = g and a^p = 0 so 0 = f(b^p) as well.
The claim in .0 is that f(b) = 0 implies f(b^p) = 0 for
arbitrary polynomials f(x) in F[x], and is false. For
example, consider p = 2, F = {0,1,a,a+1} where
a^2 + a + 1 = 0. Let f(x) be x + a. Then f(a) = 0 but
f(a^2) = f(a + 1) = 1.
So you do need to restrict f(x) to be in (Z/p)[x], not
F[x]. Given that, proving .0 reduces to proving the
statement made in the first paragraph.
So let h(b) = b^p. To prove h is an automorphism you must
show h(0) = 0, h(1) = 1, h(a + b) = h(a) + h(b), and
h(ab) = h(a)h(b). The three easy cases are
h(0) = 0^p = 0
h(1) = 1^p = 1
h(ab) = (ab)^p = (a^p)(b^p) = h(a)h(b)
The almost easy case is
h(a+b) = (a+b)^p
= a^p + terms with coefficients dividible by p + b^p
(using known facts about the binomial expansion)
= a^p + b^p
= h(a) + h(b)
So it's an automorphism. As F is a field the polynomial
x^p - x has at most p zeroes, and by Fermat's little
theorem a^p - a = 0 (mod p) for a = 0,1,...,p-1. That
establishes the rest of what was claimed in the first
paragraph.
Dan
|
| re .1,
>> So let h(b) = b^p. To prove h is an automorphism you must
>> show h(0) = 0, h(1) = 1, h(a + b) = h(a) + h(b), and
>> h(ab) = h(a)h(b).
Actually, that just proves h is a homomorphism of F into
itself. To show it is an isomorphism you must also show
that it is a bijection. However, that result isn't used
in the proof that f(c^p) = 0. The fact that F is finite
isn't used, either. So if the root c of f(x) isn't
assumed to be in F, then let K be the unique (up to
isomorphism) algebraic completion of F and let c be any
zero of f(x) in K. Then h(a) = a^p is a field
homomorphism of K into itself that leaves each of
0,1,...,p-1 fixed, h maps F into F, and for any
polynomial f(x) in (Z/p)[x] and f(c) = 0 then h(f(c)) =
f(h(c)) = f(c^p) = h(0) = 0.
Dan
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