Solved: Field Theory Problem: Z_p in Algebraic Closure of F

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Homework Help Overview

The discussion revolves around the properties of fields of characteristic p, specifically whether a field F must be contained within the algebraic closure of Z_p. The original poster questions an assumption made in their textbook regarding this relationship.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Some participants question the validity of the assumption that F is an arbitrary field of characteristic p, suggesting that it may not hold true in all cases. They provide examples to illustrate their points, particularly focusing on finite fields and rational function fields. Others explore the implications of F being a finite extension of Z_p and its relationship to algebraic closures.

Discussion Status

The discussion is ongoing, with participants exploring different interpretations of the problem. Some have provided insights into the nature of finite extensions and their properties, while others are still clarifying their understanding of splitting fields and algebraic extensions.

Contextual Notes

There is a mention of specific sections in the textbook that participants are referencing, indicating varying levels of familiarity with the concepts involved, such as splitting fields and algebraic extensions.

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[SOLVED] field theory problem

Homework Statement


If F is a field that has characteristic p, it must contain a copy of Z_p. Is it true that F must sit inside of the algebraic closure of Z_p? My book assumes that it does and I do not understand why?


Homework Equations





The Attempt at a Solution

 
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Are you sure it says F is an arbitrary field of characteristic p? Because this is obviously false. For instance take the rational function field of the alg closure of Z_p - this still has characteristic p and properly contains the alg closure of Z_p.

On the other hand, this is true if F is a finite field of characteristic p.
 
morphism said:
Are you sure it says F is an arbitrary field of characteristic p? Because this is obviously false. For instance take the rational function field of the alg closure of Z_p - this still has characteristic p and properly contains the alg closure of Z_p.

On the other hand, this is true if F is a finite field of characteristic p.

Sorry. F has p^r elements.
 
Then F is the splitting field of x^(p^r) - x over Z_p.
 
morphism said:
Then F is the splitting field of x^(p^r) - x over Z_p.

I am in section 33 and splitting fields are in section 50. Thus, I do not even know what they are.

I figured it out though. F must be a finite extension of Z_p and finite extensions are always algebraic extensions and thus F must sit inside the algebraic closure of Z_p.
 

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