Re: every finite field is perfect
warren065@xxxxxxx wrote:
> A field F of prime characteristic p is said to be perfect
> if the mapping a -> a^p isa surjection on F.
>
> 1) Show that every finite field is perfect
>
> 2) Let F be an arbitrary field of characteristic p not
> equal to 0. Show that the field F(x) is not perfect.
>
> Proof: 1) [...]
> Can I assume the surjection or do I need to prove that?
> What would be a proper proof?
You need to show that the map f(a) = a^p is a surjection. So, given
some b in F, find an a such that f(a) = b.
> Proof: 2) No clue whatsoever how to approach this.
Suppose the map f(a) is a surjection. Since F has characteristic p,
then p * 1 = 0. Derive a contradiction.
--- Christopher Heckman
.
Relevant Pages
- Re: Cantor Confusion
... I give a surjection on all existing paths. ... Each part of an edge may map to a single path, ... The crucial thing in a surjection (and indeed for a mapping) from A to B ...
(sci.math) - Re: CANTORs theorem
... >> Since that estabishes that there cannot be any surjection from N to ... This is *not* the set M as defined in Cantor's proof for this mapping. ... in the image of the map. ... And that shows that the cardinality ...
(sci.math) - Re: question on compact spaces
... Every quasicompact space is the surjective image under a continuous map of a compact Hausdorff space. ... I.e. quasicompact just means what most books call compact, though Bourbaki includes the Hausdorff property in the word compact. ... There is a continuous surjection from 2^X to X, ...
(sci.math) - Re: Continuous maps between boolean/stone spaces
... http://planetmath.org/encyclopedia/MHStonesRepresentationTheorem.html) we can assume that the surjection of Stone spaces comes from a homomorphism phi: A -> B of Boolean algebras. ... Let's assume in the first step that A and B are *free* Boolean algebras with basis S and T, resp., with S contained in T. Then the Stone spaces A* and B* are homeomorphic to ^S and ^T, resp., and the map B* -> A* induced by the restriction from T to S is clearly open. ... Denoting the free Boolean algebra over S by Fr, the homomorphism Fr-> A is a surjection inducing the map A* -> Fr* which is injective and the topology of A* coincides with the trace topology induced by Fr. ...
(sci.math) - Re: Continuous maps between boolean/stone spaces
... http://planetmath.org/encyclopedia/MHStonesRepresentationTheorem.html) we can assume that the surjection of Stone spaces comes from a homomorphism phi: A -> B of Boolean algebras. ... Let's assume in the first step that A and B are *free* Boolean algebras with basis S and T, resp., with S contained in T. Then the Stone spaces A* and B* are homeomorphic to ^S and ^T, resp., and the map B* -> A* induced by the restriction from T to S is clearly open. ... Denoting the free Boolean algebra over S by Fr, the homomorphism Fr-> A is a surjection inducing the map A* -> Fr* which is injective and the topology of A* coincides with the trace topology induced by Fr. ...
(sci.math) |
|
