Galois Theory - Algebraic extensions

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SUMMARY

The discussion centers on the relationship between algebraic field extensions M/L and L/K, specifically whether M/K is necessarily algebraic. It concludes that if both M/L and M/K are finite extensions, then M/K is also a finite extension and therefore algebraic. The Tower Law is applied to demonstrate this relationship, indicating that if one or both extensions are infinite, further analysis is required. The discussion emphasizes the importance of identifying an element m in M and showing its algebraicity over K using the constructed tower of fields.

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  • Understanding of algebraic field extensions
  • Familiarity with the Tower Law in field theory
  • Knowledge of algebraic elements and their properties
  • Experience with constructing field towers
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  • Study the implications of the Tower Law in more complex field extensions
  • Explore the concept of algebraic closures and their applications
  • Learn about finite and infinite algebraic extensions in detail
  • Investigate examples of algebraic elements over various fields
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Mathematics students, particularly those studying abstract algebra, field theory, and anyone interested in the properties of algebraic extensions in Galois Theory.

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Homework Statement



Let M/L and L/K be algebraic field extensions. Is M/K necessarily algebraic?

Homework Equations



Tower law: [M:K]=[M:L][L:K]

The Attempt at a Solution



If both M/L and M/K are finite extensions then by the tower law M/K is also a finite extension, hence is algebraic. So one or both of them must be infinite. The only infinite algebraic extensions I can think of are similar in construction to the algebraic closure of the rationals.
An element m \in M is algebraic over L so we can write \sum a_{i}m^i=0 for some a_i \in L, where i runs from 0 to some n. Might be able to use the fact that L/K is algebraic now?

Thanks for any help!
 
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You have now singled out an element m and want to show that it's algebraic over K. Consider the tower:
K \subseteq K(a_1,\ldots,a_n) \subseteq K(a_1,\ldots,a_n,m)
You can show that m is algebraic over K(a_1,\ldots,a_n) and you should therefore be able to get back to your finite case.
 
Funnily enough I was just working via this argument on paper to see whether I could delete the thread. At least now I know I was on the right lines, thanks very much!
 

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