Maximal Real Subfields (14.5.7,9)

Dummit and Foote Abstract Algebra, section 14.5, exercises 7 and 9:

7. Show that complex conjugation restricts to the automorphism $σ_{-1}∈\text{Gal}(\mathbb{Q}(\zeta_n)/\mathbb{Q})$. Show that the field $K^+=\mathbb{Q}(\zeta_n+\zeta_n^{-1})$ is the subfield of real elements in $K=\mathbb{Q}(\zeta_n)$, called the maximal real subfield of $K$.

9. Let $K_n=\mathbb{Q}(\zeta_{2^{n+2}})$ for $n ≥ 0$.
(a) Prove that $K_n^+/\mathbb{Q}$ is cyclic of degree $2^n$.
(b) Prove $K_n/K_{n-1}^+$ is biquadratic and that two of the three intermediate subfields are $K_{n-1}$ and $K^+$. Prove the remaining intermediate field is a cyclic extension of $\mathbb{Q}$ of degree $2^n$.

Proof: (7) Note that complex conjugation is an automorphism of $\mathbb{C}$, and also restricts to an automorphism of $K$ since $\overline{\zeta_n}=\zeta_n^{-1}$ ($|\zeta_n^n|=|\zeta_n|^n=1$ implies for $\zeta_n=a+bi$ that $|\zeta_n|=a^2+b^2=\zeta_n\overline{\zeta_n}=1$). Now, when $n > 2$ we note $\zeta_n∉\mathbb{R}$ so $K_n/K^+$ is of degree greater than or equal to $2$. As well, $\dfrac{1}{2}(\zeta_n+\zeta_n^{-1})=a$ and $1-a^2=b^2$, so when $α$ is a root of the equation $x^2+b^2$ we observe $K^+(α)=K_n$ and hence $K_n/K^+$ is of degree exactly $2$. Since the maximal real subfield is necessarily properly contained in $K_n$, we must have $K^+$ is the maximal real subfield.

(9)(a) The degree of the extension follows from above. Here we may observe the isomorphism $\text{Gal}(K/ℚ)≅(ℤ/2^{n+2}ℤ)^×≅ℤ/2ℤ×ℤ/2^nℤ$ shown in Exercises 2.3.22-23. We show $(ℤ/2^{n+2}ℤ)^×$ is generated by $5$ and $-1$. This is clearly the case when $n=0$ so assume $n ≥ 1$. Now $5$ is of order $2^n$ by the mentioned exercises and it suffices to show $5$ doesn't generate $-1$; we observe this by demonstrating $k_n$ such that $5^{k_n}≡2^{n+1}+1~\text{mod }2^{n+2}$, a different element of order $2$. First, we see $k_1=1$. Now, given $k_n$, we have $k_{n+1}=2k_n$ since $(5^{k_1})^2=(2^{n+1}+1+m2^{n+2})^2≡2^{(n+1)+1}+1~\text{mod }2^{(n+1)+2}$ when $n ≥ 1$. Thus $-1$ and $5$ generate $(ℤ/2^{n+2}ℤ)^×$, and since we notice the quotient group of $(ℤ/2^{n+2}ℤ)^×$ over the subgroup generated by $-1$ is the cyclic group generated by the image of $5$, we conclude $\text{Gal}(K^+/ℚ)≅\text{Gal}(K/ℚ)/\text{Gal}(K/K^+)$ is cyclic.

(b) We note that since $K_n/ℚ$ is Galois necessarily $K_n/K_{n-1}^+$ is Galois. This Galois extension by degree must be either cyclic or biquadratic, and since it cannot be cyclic by the presence of two distinct subfields of degree two, it must be biquadratic. We see that $K_{n-1}$, $K_n^+$, and the third subfield manifest as the fixed fields of the (isomorphic pullbacks) of the elements of order $2$ in $(ℤ/2^{n+2}ℤ)^×$, and viewing their Galois groups as the quotient of this group with the subgroups generated by those elements, we see as a result of only one of them intersecting nontrivialy with the order-$2^n$ subgroup generated by $5$ (that is, $2^{n+1}+1$ as shown above, which corresponds to $K_{n-1}/ℚ$ which we see is not cyclic) that the third subfield's Galois group's representation as a quotient is cyclic.$~\square$