Frobenius endomorphism

In commutative algebra and field theory, which are branches of mathematics, the Frobenius endomorphism (after Ferdinand Georg Frobenius) is a special endomorphism of rings with prime characteristic "p", a class importantly including fields. The endomorphism maps every element to its "p"th power. In certain contexts it is an automorphism, but this is not true in general.

Definition

Let "R" be a commutative ring of positive and prime characteristic "p" (the characteristic is always prime when "R" is an integral domain, for example). The Frobenius endomorphism "F" is defined by

:"F"("r") = "r"^"p"

for all "r" in "R". Clearly this respects the multiplication of "R": "F"("rs")=("rs")^"p" = "r"^"p""s"^"p", and "F"(1) is clearly 1 also. What is interesting, however, is that it also respects the addition of "R". The expression ("r" + "s")^"p" can be expanded using the binomial theorem. Because "p" is prime, it divides "p"! but not any "q"! for q < p; it therefore will divide the numerator, but not the denominator, of the explicit formula of the binomial coefficients

$frac\; \{p!\}\{k!\; (p-k)!\}$

for 1 &le; "k" &le; "p" − 1. Therefore the coefficients of all the terms except "r"^"p" and "s"^"p" are divisible by "p", the characteristic, and hence they vanish. This shows that "F" is a ring homomorphism.

In general, "F" is not an automorphism. For example, let "K" be the field F_"p"("t"), that is, the finite field with "p" elements together with a single transcendental element. We claim that the image of "F" does not contain "t". We will prove this by contradiction: Suppose that there is an element of "K" whose image under "F" is "t". This element is a rational function "q"("t")/"r"("t") whose "p"'th power ("q"("t")/"r"("t"))^"p" equals "t". This makes p(deg "q" - deg "r") = 1, which is impossible. So "F" is not surjective and hence not an automorphism. It is also possible for "F" to be non-injective. This occurs if and only if "R" has a nilpotent of order less than or equal to "p".

Fixed points of the Frobenius endomorphism

Say "R" is an integral domain. The Frobenius map fixes all the elements of "R" which satisfy the equation "x"^"p" = "x". These are all the roots of the equation "x"^"p" - "x", and since this equation has degree "p", there are at most "p" roots. These are exactly the elements 0, 1, 2, ..., "p" - 1, so the fixed point set of "F" is the prime field.

Iterating the Frobenius map gives us a sequence of elements in R::$x,\; x^p,\; x^\{p^2\},\; x^\{p^3\},\; ldots$Applying the "e"'th iterate of "F" to a ring which contains a field "K" of "p"^"e" elements gives us a fixed point set equal to "K", similar to the example above. The iterates of the Frobenius map are also used in defining the Frobenius closure and tight closure of an ideal.

Frobenius for finite fields

Let F_"q" be the finite field of "q" elements, where "q"="p"^e. "F" fixes F_"p" by the argument above. If "e"=2, then "F"², the second iterate of Frobenius, fixes "p"² elements, so it will fix F_p². In general, "F"^"e" fixes F_p^e. Furthermore, "F" will generate the Galois group of any extension of finite fields.

= Frobenius for schemes =

Using the setup above, it is easy to extend the Frobenius map to the context of schemes. Let "X" be a scheme over a field "k" of characteristic "p". Choose an open affine subset "U"=Spec "R". Since "X" is a "k"-scheme, we get an inclusion of "k" in "R". This forces "R" to be a characteristic "p" ring, so we can define the Frobenius endomorphism "F" for "R" as we did above. It is clear that "F" commutes with localization, so "F" glues to give an endomorphism of "X".

However, "F" is not necessarily an endomorphism of "k"-schemes. If "k" is not F_"p", then "F" will not fix "k", and consequently "F" will not be a "k"-algebra map. A partial resolution of this problem is to look at the inclusion of "F"("k") = "k"^"p" in "k": Since "X" is a "k"-scheme, it is also a "k"^"p"-scheme. "F" is then a map of "k"^"p"-schemes.

Frobenius for local fields

The definition of "F" for schemes automatically defines "F" for local and global fields, but we will treat these cases separately for clarity.

The definition of the Frobenius for finite fields can be extended to other sorts of field extensions. Given an unramified finite extension "L/K" of local fields, there is a concept of Frobenius endomorphism which induces the Frobenius endomorphism in the corresponding extension of residue fields.

Suppose "L/K" is an unramified extension of local fields, with ring of integers "O_K" of "K" such that the residue field, the integers of "K" modulo their unique maximal ideal &phi;, is a finite field of order "q". If &Phi; is a prime of "L" lying over &phi;, that "L/K" is unramified means by definition that the integers of "L" modulo &Phi;, the residue field of "L", will be a finite field of order "q^f" extending the residue field of "K" where "f" is the degree of "L/K". We may define the Frobenius map for elements of the ring of integers "O_L" of "L" by

:$s\_Phi(x)\; equiv\; x^q\; mod\; Phi$.

Frobenius for global fields

In algebraic number theory, Frobenius elements are defined for extensions "L/K" of global fields that are finite Galois extensions for prime ideals &Phi; of "L" that are unramified in "L/K". Since the extension is unramified the decomposition group of &Phi; is the Galois group of the extension of residue fields. The Frobenius then can be defined for elements of the ring of integers of "L" as in the local case, by

:$s\_Phi(x)\; equiv\; x^q\; mod\; Phi$,

where "q" is the order of the residue field "O_K" mod &phi;.

Examples

The polynomial

:"x"⁵ − "x" − 1

has discriminant

:19 &times; 151,

and so is unramified at the prime 3; it is also irreducible mod 3. Hence adjoining a root &rho; of it to the field of 3-adic numbers $Bbb\{Q\}\_3$ gives an unramified extension $Bbb\{Q\}\_3(\; ho)$ of $Bbb\{Q\}\_3$. We may find the image of &rho; under the Frobenius map by locating the root nearest to &rho;³, which we may do by Newton's method. We obtain an element of the ring of integers $Bbb\{Z\}\_3\; [\; ho]$ in this way; this is a polynomial of degree four in &rho; with coefficients in the 3-adic integers $Bbb\{Z\}\_3$. Modulo 3⁸ this polynomial is

:$ho^3\; +\; 3(460+183\; ho-354\; ho^2-979\; ho^3-575\; ho^4)$.

This is algebraic over $Bbb\{Q\}$ and is the correct global Frobenius image in terms of the embedding of $Bbb\{Q\}$ into $Bbb\{Q\}\_3$; moreover, the coefficients are algebraic and the result can be expressed algebraically. However, they are of degree 120, the order of the Galois group, illustrating the fact that explicit computations are much more easily accomplished if p-adic results will suffice.

If "L/K" is an abelian extension of global fields, we get a much stronger congruence since it depends only on the prime &phi; in the base field "K". For an example, consider the extension of $Bbb\{Q\}$ obtained by adjoining a root &beta; satisfying

:

to $Bbb\{Q\}$. This extension is cyclic of order five, with roots

:$2\; cos\; frac\{2\; pi\; n\}\{11\}$

for integer "n". It has roots which are Chebyshev polynomials of &beta;:

:&beta;² - 2, &beta;³ - 3&beta;, &beta;⁵-5&beta;³+5&beta;

give the result of the Frobenius map for the primes 2, 3 and 5, and so on for larger primes not equal to 11 or of the form 22n+1 (which split). It is immediately apparent how the Frobenius map gives a result equal mod "p" to the "p"-th power of the root &beta;.

References