Twenty-Three
by djmathman, Dec 28, 2019, 4:05 PM
the passage of time scares me
To "celebrate", here's a problem from one of my final exams last semester. I suppose it's a precursor to the Algebraic Number Theory class I'll be taking during the Spring, which I'm pretty excited for.
Math 500 Fall 2019 Final Exam #10. Prove that
is not a subfield of any cyclotomic extension.
(An alternate wording without the field theory: prove that there do not exist rational numbers
such that
whee
is an
root of unity.)
Solution
To "celebrate", here's a problem from one of my final exams last semester. I suppose it's a precursor to the Algebraic Number Theory class I'll be taking during the Spring, which I'm pretty excited for.
Math 500 Fall 2019 Final Exam #10. Prove that
![$\mathbb Q(\sqrt[3]2)$](http://latex.artofproblemsolving.com/5/1/e/51e1c2d5430f701ec6d8a790a7ac2a675d1ceb55.png)
(An alternate wording without the field theory: prove that there do not exist rational numbers

![\[
\sqrt[3]{2} = a_0 + a_1\zeta_n + a_2\zeta_n^2 + \cdots + a_{n-1}\zeta_n^{n-1},
\]](http://latex.artofproblemsolving.com/1/a/f/1afcd86a25af3310a81fc1d078fb4a23b4b54975.png)


Solution
Suppose, for the sake of contradiction, that there exists a chain of subfields
, where
is an
root of unity. It's well-known that the field extension
is Galois with Galois group equal to
. This group is abelian, and hence every subgroup
of
is normal in
.
Now consider the automorphism group
as per the Fundamental Theorem of Galois Theory. Since
is a subgroup of
,
is normal in
. Thus, again by the Fundamental Theorem, the extension
must be Galois.
But we know this isn't the case: the only automorphism of
fixing
is the identity automorphism, since
is the only root of
lying in
. Hence we've obtained a contradiction and the proof is complete.
![$\mathbb Q\subseteq\mathbb Q(\sqrt[3]{2})\subseteq \mathbb Q(\zeta_n)$](http://latex.artofproblemsolving.com/9/1/8/9187b4e0e83af613ac06673a7d1192558d62b589.png)







Now consider the automorphism group
![$H:=\operatorname{Aut}(\mathbb Q(\zeta_n)/\mathbb Q(\sqrt[3]2))$](http://latex.artofproblemsolving.com/d/f/c/dfcc830521c7bdb4bebd268a123281b10be5278b.png)




![$\mathbb Q(\sqrt[3]2)/\mathbb Q$](http://latex.artofproblemsolving.com/3/b/8/3b81da44d038864f18873cc644253353cff717f2.png)
But we know this isn't the case: the only automorphism of
![$\mathbb Q(\sqrt[3]2)$](http://latex.artofproblemsolving.com/5/1/e/51e1c2d5430f701ec6d8a790a7ac2a675d1ceb55.png)

![$\sqrt[3]2$](http://latex.artofproblemsolving.com/f/d/2/fd2db1af029a3aaaad5a3ee0ea826875cec0022e.png)

![$\mathbb Q(\sqrt[3]2)$](http://latex.artofproblemsolving.com/5/1/e/51e1c2d5430f701ec6d8a790a7ac2a675d1ceb55.png)