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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 $\beta \in {\mathcal{O}}_K$.

A nonabelian simple group has no nonidentity conjugacy class of prime-power size. Coprime-degree irreducible characters vanish on such an element: the conjugacy-class sum makes the character-to-degree ratio an algebraic integer, and the Kronecker theorem on algebraic integers in the unit disk makes a nonzero ratio a root of unity, forcing a scalar in a faithful group representation. Column character orthogonality then makes $1/p$ an algebraic integer, impossible. This is the character-theoretic ingredient in Burnside's theorem.

Let $E:y^2=(x-\alpha )(x-\beta )(x-\gamma )$ with distinct algebraic integers in a number field $K$. For a rational point $P$, its halving field is unramified outside the finite set $S$ of primes dividing $2(\alpha -\beta )(\alpha -\gamma )(\beta -\gamma )$. At $v\notin S$, the elliptic-curve discriminant is a unit, so there is good reduction. A half of the reduction of $P$ is defined over a finite extension of the residue field. Pass to the corresponding unramified extension of the completion and lift this half using the Hensel lemma. The doubling error lies in the formal kernel of a minimal Weierstrass equation. Since two is a unit, prime-to-residue-characteristic multiplication on a formal group corrects this error uniquely. Every half differs by rational 2-torsion, so all halves are unramified at $v$. The halving cocycle with rational two-torsion has image of order at most four, making the halving field a Galois extension of degree at most four. The bounded-degree extensions with restricted ramification are finite in number; there are only finitely many homomorphisms from their finite Galois groups to $E[2]$. Thus $E(K)/2E(K)$ is finite, without assuming any form of the Mordell-Weil theorem.