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Frobenius morphism : ウィキペディア英語版
Frobenius endomorphism

In commutative algebra and field theory, the Frobenius endomorphism (after Ferdinand Georg Frobenius) is a special endomorphism of commutative rings with prime characteristic , an important class which includes finite fields. The endomorphism maps every element to its -th power. In certain contexts it is an automorphism, but this is not true in general.
==Definition==
Let be a commutative ring with prime characteristic (an integral domain of positive characteristic always has prime characteristic, 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^ps^p = F(r)F(s)\ ,
and is clearly 1 also. What is interesting, however, is that it also respects the addition of . The expression can be expanded using the binomial theorem. Because is prime, it divides but not any for ; it therefore will divide the numerator, but not the denominator, of the explicit formula of the binomial coefficients
:\frac,
if . Therefore the coefficients of all the terms except and are divisible by , the characteristic, and hence they vanish.〔This is known as the Freshman's dream.〕 Thus
:F(r + s) = (r + s)^p = r^p + s^p = F(r) + F(s)\ .
This shows that ''F'' is a ring homomorphism.
If is a homomorphism of rings of characteristic , then:
:\phi(x^p) = \phi(x)^p.
If and are the Frobenius endomorphisms of and , then this can be rewritten as:
:\phi \circ F_R = F_S \circ \phi.
This means that the Frobenius endomorphism is a natural transformation from the identity functor on the category of characteristic rings to itself.
If the ring is a ring with no nilpotent elements, then the Frobenius endomorphism is injective: means , which by definition means that is nilpotent of order at most . In fact, this is an if and only if, because if is any nilpotent, then one of its powers will be nilpotent of order at most . In particular, if is a field then the Frobenius endomorphism is injective.
The Frobenius morphism is not necessarily surjective, even when is a field. For example let be the finite field of elements together with a single transcendental element; equivalently, is the field of rational functions with coefficients in . Then the image of does not contain . If it did, then there would be a rational function whose -th power would equal . But the degree of this -th power is , which is a multiple of . In particular, it can't be 1, which is the degree of . This is a contradiction, so is not in the image of .
A field is called ''perfect'' if either it is of characteristic zero or if it is of positive characteristic and its Frobenius endomorphism is an automorphism. For example, all finite fields are perfect.

抄文引用元・出典: フリー百科事典『 ウィキペディア(Wikipedia)
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