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Put $C={\bigcup }_{r\ge 1}\mathbb{Q}({\mu }_{p^r})$, where ${\mu }_{p^r}$ is the group of $p^r$th roots of unity. The compatible Galois groups of the cyclotomic fields giveFor odd $p$, the Teichmüller character givesThe p-adic logarithm identifies $1+p{\mathbb{Z}}_p$ with $p{\mathbb{Z}}_p$, which is topologically isomorphic to ${\mathbb{Z}}_p$. More explicitly, $a\mapsto (1+p{)}^a$ is an isomorphism from the additive p-adic integers onto $1+p{\mathbb{Z}}_p$: the logarithm of $1+p$ has valuation one and generates $p{\mathbb{Z}}_p$. Take the fixed field of ${\mu }_{p-1}$ in $C$.

For $p=2$, use insteadThe p-adic logarithm identifies the second factor with $4{\mathbb{Z}}_2$, and $a\mapsto 5^a$ identifies it with ${\mathbb{Z}}_2$. The fixed field of $\{\pm 1\}$ in $C$ is the union of the maximal real subfields of the cyclotomic fields of $2$-power conductor. Thus in both cases the cyclotomic Zp-extension exists:where ${\Delta }_p={\mu }_{p-1}$ for odd $p$ and ${\Delta }_2=\{\pm 1\}$.

The closed subgroup $p^n{\mathbb{Z}}_p$ is the unique subgroup of index $p^n$, so the Galois correspondence supplies the unique degree-$p^n$ field $\mathbb{Q}[p^n]$. In the odd case it is the ${\Delta }_p$-fixed field in $\mathbb{Q}({\mu }_{p^{n+1}})$; in the even case it is the real subfield of $\mathbb{Q}({\mu }_{2^{n+2}})$. These finite layers are all totally real number fields.

Now set $A=\mathbb{Q}[p^n]$, $B=\mathbb{Q}[p^m]$, and let $H_A$ be the ordinary Hilbert class field of $A$. Thus $H_A/A$ is abelian, unramified at finite primes, and split at every real place, with degree $h(A)$. If $m=n$ there is nothing to prove. Otherwise the unique prime over $p$ is totally ramified in $B/A$, by the permitted total-ramification fact and multiplicativity of the ramification index. Any nontrivial intermediate field of $B/A$ is ramified at that prime. HenceIt follows that the two extensions are linearly disjoint field extensions, andUnramified extensions remain unramified after base change. The real places also remain split: $H_A$ and $B$ are totally real, so their compositum is totally real. Therefore $B{H}_A/B$ lies inside the ordinary Hilbert class field $H_B/B$. Its degree divides $[H_B:B]=h(B)$, provingThis is class number divisibility with split real places; using a class-field statement restricted to imaginary fields would unnecessarily omit the present real layers.

The products $xy$ with $x\in L_1$ and $y\in L_2$ span the field compositum $L_1{L}_2$ over $K$, soThe tower law givesThus the degree is divisible by both $d_1$ and $d_2$. Since they are coprime, it is divisible by $d_1{d}_2$. Combining divisibility with the upper bound givesEquivalently, finite extensions of coprime degrees are linearly disjoint field extensions.

Let $f$ be the monic cubic minimal polynomial of the algebraic integer $\alpha$, let ${\alpha }_1=\alpha ,{\alpha }_2,{\alpha }_3$ be its conjugates, and let ${\beta }_1=\beta ,{\beta }_2$ be the two conjugates of a nonrational $\beta \in {\mathcal{O}}_K$. Because the degrees three and two are coprime, the fields are linearly disjoint, andis a nonzero integer.

AssumeFor large $H(\beta )$ this makes ${\beta }_1$ bounded. Since $\beta$ is a quadratic algebraic integer, the height-Mahler measure formula givesand therefore $|{\beta }_2|\asymp H(\beta {)}^2$. The two remaining factors with $j=1$ are bounded, while the three factors with $j=2$ are $O(|{\beta }_2|)$. Consequently

We now use the standard effective norm form consequence of Part (b). Forwhere $1,\omega$ is an integral basis of $K$, the effective norm-form height estimate supplies effective constants $A,C>0$, depending only on $\alpha$ and $K$, such thatfor every nonzero $z\in V\cap {\mathcal{O}}_L$. Its proof factors $(z)$, balances a generator using the Dirichlet unit theorem, and applies the Baker lower bound for a homogeneous linear form in logarithms to the linear relations defining $V$; the coprime degrees $3$ and $2$ exclude a unit-family degeneracy.

Apply this estimate to $z=\alpha -\beta$. Since $\beta =\alpha -z$, the height inequalities imply $H(\beta )\le 2H(\alpha )H(z)$. HenceChoose the effective value $\epsilon =1/(2A)$. The last inequality bounds $H(\beta )$ effectively. The rational integers $\beta \in \mathbb{Z}$ are already covered by the stronger degree-three Liouville approximation theorem, and the Northcott theorem leaves only finitely many remaining quadratic integers of bounded height. Taking the minimum ofover this effective finite set gives an effective $c>0$ and provesfor every $\beta \in {\mathcal{O}}_K$.