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Extensions fields

  • Thread starter ehrenfest
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  • #26
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Take a nonzero element a in the algebraic closure. If pa != 0, then p*1 = pa*a^-1 != 0.
I wish you would use the dot notation because I do not think that multiplication by the element p in the algebraic closure of of Z/pZ is necessarily the same as [itex] p \cdot a = (a +...+a) = \sum_{i=1}^p a[/itex]. I would use the asterick only for multiplication in the field.

So, you are saying that if [itex] p \cdot a \neq 0 [/itex], then [itex] p \cdot 1 = (p \cdot (a*a^{-1}) ) = (p \cdot a)* a^{-1} [/itex] where in the last step I used the distributive property. That is obviously impossible because 1 is in Z/pZ and we thus know its characteristic is p. So I guess what you did checks outs, but I still think you should always use the dot notation when dealing with characteristics of fields.

But really, the easiest way to see this is to think about characteristic in terms of prime subfields. Since the prime subfield of the algebraic closure of Z/pZ is Z/pZ, we're done.
There is a theorem in my book that says: "A field is either of prime characteristic p and contains a subfield isomorphic to Z_p or of characteristic 0 and contains a subfield isomorphic to Q."

Simply containing a subfield isomorphic of Z_p is not enough to conclude that the field has prime characteristic p (at least from this theorem).
 
  • #27
Hurkyl
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I wish you would use the dot notation because I do not think that multiplication by the element p in the algebraic closure of of Z/pZ is necessarily the same as [itex] p \cdot a = (a +...+a) = \sum_{i=1}^p a[/itex]. I would use the asterick only for multiplication in the field.
Aren't fields distributive?
 
  • #28
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Aren't fields distributive?
Yes. So what? You could have a field that does not even contain an element called p and then p*a does not even make sense unless it means [itex]p \cdot a [/itex].
 
  • #29
morphism
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I wish you would use the dot notation because I do not think that multiplication by the element p in the algebraic closure of of Z/pZ is necessarily the same as [itex] p \cdot a = (a +...+a) = \sum_{i=1}^p a[/itex].
It is. p is simply shorthand for 1+1+...+1 (p times).

So I guess what you did checks outs, but I still think you should always use the dot notation when dealing with characteristics of fields.
Exercise: Prove that the dot notation is as compatible as we want it to be with field multiplication. Conclude that it's alright if we don't bother to use it anymore.
 
  • #30
Hurkyl
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Yes. So what? You could have a field that does not even contain an element called p
How? Can you come up with an example?
 
  • #31
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How? Can you come up with an example?
The dot notation is definitely not always compatible with field notation, as I will show. Consider the field Z/3Z. Writing 5*2 is absurd because 5 is not an element of the field Z/3Z. You need to write [itex]5 \cdot 2 [/itex] if you want to represent the sum [itex]2+2+2+2+2 [/itex].

morphism, if you are saying that if F is a field that contains f, then [itex]n \cdot f = nf[/itex] WHEN the characteristic of F is greater than n or equal to 0, then I think that is true.
 
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  • #32
Hurkyl
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Writing 5*2 is absurd because 5 is not an element of the field Z/3Z.[/itex]
Er, yes it is. 5 denotes the same element that 2 does.
 
  • #33
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OK. This is getting really technical. I would rather move on.
 

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