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Every identity-preserving biholomorphism of $\mathbb{C}/\Lambda$ lifts to multiplication by a nonzero complex number $\zeta$ satisfying $\zeta \Lambda =\Lambda$. Relative to a $\mathbb{Z}$-basis of $\Lambda$, multiplication by $\zeta$ is represented by a unimodular matrix $A\in {\mathrm{GL}}_2(\mathbb{Z})$. Its real determinant is $|\zeta {|}^2$, while complex multiplication preserves orientation, so $\det A=1$ and $|\zeta |=1$. The Cayley-Hamilton theorem givesConsequently $2\mathrm{Re}\zeta =\mathrm{tr}(A)\in \{-2,-1,0,1,2\}$, and $\zeta$ is a root of unity of order $1$, $2$, $3$, $4$, or $6$.

A cyclic subspace generated by $x$ under a linear operator $T$ is the span of its successive iterates $x,Tx,T^{2}x,\dots$. In finite dimension it is already generated by the first $n$ iterates. The first linear dependence expresses the next iterate in previous ones and proves invariance without using the Cayley-Hamilton theorem.

The Cayley-Hamilton theorem says that every square complex matrix satisfies its own characteristic polynomial: if ${\chi }_A(t)=t^n+c_{n-1}{t}^{n-1}+\cdots +c_0$, thenFor a $2\times 2$ diagonalisable matrix, write $A=SD{S}^{-1}$ with $D=\mathrm{diag}({\lambda }_1,{\lambda }_2)$. The characteristic polynomial is $(t-{\lambda }_1)(t-{\lambda }_2)$, including repeated roots. Evaluating it at $D$ gives zero in each diagonal position, and polynomial evaluation respects matrix similarity: ${\chi }_A(A)=S{\chi }_A(D)S^{-1}=0$. This proves the requested case.

For the final assertion, let $\lambda$ be any complex eigenvalue of $B$, with nonzero eigenvector $\mathbf{v}$. From $B^k=0$ we obtain $0=B^k\mathbf{v}={\lambda }^k\mathbf{v}$, so $\lambda =0$. The characteristic polynomial of this nilpotent linear map is therefore $t^n$. Applying the full Cayley-Hamilton theorem yields $\boxed{B^n=0}$. This argument does not assume $B$ is diagonalisable.

The $n$ spanning vectors form a basis. WriteIn this cyclic basis the matrix is the companion matrixExpanding $\det (tI-C)$ along its last column givesFor example, the term containing $a_j$ has minor $t^j$ and the signs from the subdiagonal cancel the cofactor sign; the diagonal term supplies $t^n$. The defining relation says ${\chi }_T(T)x=0$. Every polynomial in $T$ commutes with $T$, soIt therefore vanishes on a basis, provingThis direct argument for a cyclic vector does not assume the Cayley-Hamilton theorem.

Let $c_1,\dots ,c_d$ be the coefficients of $f$. Each is an algebraic integer. The ring $R=\mathbb{Z}[c_1,\dots ,c_d]$ is a finitely generated $\mathbb{Z}$-module: reduce powers of each generator by its monic integral equation. As $f$ is monic, $R[\alpha ]$ is generated over $R$ by $1,\alpha ,\dots ,{\alpha }^{d-1}$, so is a finite $\mathbb{Z}$-module. It is torsion-free as a subgroup of a characteristic-zero field, hence free of finite rank. Multiplication by $\alpha$ on it is represented by an integer matrix. Its monic characteristic polynomial, by Cayley-Hamilton theorem, annihilates $\alpha$. Thus $\alpha$ is an algebraic integer. This proves the required transitivity of integrality in this setting.

Every root of the monic $h$ is also a root of the monic polynomial $h^n\in {\mathcal{O}}_K[x]$, and is therefore an algebraic integer. The coefficients of $h$ are elementary symmetric polynomials in these roots, hence algebraic integers because algebraic integers form a ring. They also belong to $K$, so belong to ${\mathcal{O}}_K$. Consequently $h\in {\mathcal{O}}_K[x]$.

Choose an integral basis $u_1,\dots ,u_d$ of the nonzero ideal $\mathfrak{a}$. Its rational span is $L$, since it contains a nonzero integer multiple of ${\mathcal{O}}_L$. The condition $x\mathfrak{a}\subseteq \mathfrak{a}$ gives integers $m_{ij}$ with $x{u}_j={\sum }_i{m}_{ij}{u}_i$. Thus multiplication by $x$ on $L$ has an integer matrix $M$ in this basis.

Its characteristic polynomial $P(t)=\det (tI-M)$ is monic with integer coefficients. By the Cayley-Hamilton theorem, $P(M)=0$. On the field $L$, this is multiplication by $P(x)$, so $P(x)=0$. Hence $x$ is an algebraic integer, andThis is the determinant trick for an ideal preserved by multiplication.