Splitting Fields: Anderson and Feil, Theorem 45.6 ....

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SUMMARY

The discussion centers on Theorem 45.6 from "A First Course in Abstract Algebra" by Anderson and Feil, specifically regarding the proof that states if \( K \) is a normal extension of \( F \) (a field with characteristic zero), then \( K = F(\alpha) \) for some algebraic element \( \alpha \) over \( F \). The user initially struggled to connect the conditions of Theorem 45.5 with the proof of Theorem 45.6 but later recognized that the conditions were indeed satisfied. The focus is on ensuring that all elements of \( K \) are algebraic over \( F \) before applying Theorem 45.5.

PREREQUISITES
  • Understanding of normal extensions in field theory
  • Familiarity with algebraic elements and their properties
  • Knowledge of Theorem 45.5 from Anderson and Feil's text
  • Basic concepts of field characteristics, specifically characteristic zero
NEXT STEPS
  • Review the proof of Theorem 45.5 in "A First Course in Abstract Algebra"
  • Study the definition and properties of normal extensions in field theory
  • Explore examples of algebraic extensions to solidify understanding
  • Investigate the implications of characteristic zero on field extensions
USEFUL FOR

Students and educators in abstract algebra, particularly those studying field theory and the properties of extensions. This discussion is beneficial for anyone seeking clarity on the application of Theorem 45.5 in the context of normal extensions.

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I am reading Anderson and Feil - A First Course in Abstract Algebra.

I am currently focused on Ch. 45: The Splitting Field ... ...

I need some help with an aspect of the proof of Theorem 45.6 ...

Theorem 45.6 and its proof read as follows:

https://www.physicsforums.com/attachments/6701At the start of the proof of Theorem 45.6 we read the following:

"Suppose that $$K$$ is a normal extension of $$F$$, a field with characteristic zero. Then by Theorem 45.5, $$K = F( \alpha )$$, where $$\alpha$$ is algebraic over $$F$$. ... .. "
Can someone please explain exactly how $$K = F( \alpha )$$ follows in the above statement ... ?The quote mentions Anderson and Feil's Theorem 45.5 and also mentions that $$K$$ is a normal extension so I am providing the statement of Theorem 45.5 and Anderson and Feil's definition of a normal extension as follows ... ...

View attachment 6702

https://www.physicsforums.com/attachments/6703
 
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Peter said:
I am reading Anderson and Feil - A First Course in Abstract Algebra.

I am currently focused on Ch. 45: The Splitting Field ... ...

I need some help with an aspect of the proof of Theorem 45.6 ...

Theorem 45.6 and its proof read as follows:

At the start of the proof of Theorem 45.6 we read the following:

"Suppose that $$K$$ is a normal extension of $$F$$, a field with characteristic zero. Then by Theorem 45.5, $$K = F( \alpha )$$, where $$\alpha$$ is algebraic over $$F$$. ... .. "
Can someone please explain exactly how $$K = F( \alpha )$$ follows in the above statement ... ?The quote mentions Anderson and Feil's Theorem 45.5 and also mentions that $$K$$ is a normal extension so I am providing the statement of Theorem 45.5 and Anderson and Feil's definition of a normal extension as follows ... ...
I am answering my own question aincve I have now noticed that regarding my question on the first semtence of the proof of Theorem 45.6 ... the conditions to apply Theorem 45.5 were actually satisfied ...BUT ... in the way of explanation of the source of my question ... ... I was myopically focused on the first sentence of the proof where we are given(i) $$K$$ is a normal extension of $$F$$
(ii) $$F $$ is a field with characteristic zero and was wondering about how to get from this to the initial conditions of Theorem 45.5 that(i) $$F$$ is a field with characteristic zero
(ii) $$K$$ was a finite algebraic extension of $$ F$$

BUT ... in my tunnel vision I neglected that $$K$$ was given as a finite extension ... sorry for such a simple oversight ...

... HOWEVER ... still concerned that we need to establish that $$K$$ is algebraic over $$F$$ ... that is all elements of $$K$$ are algebraic over $$F$$ ... before applying Theorem 45..5 ...Peter
 

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