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Every complex number is within squared distance at most $3/4$ of an Eisenstein integer: write it as $x+y\omega$ and round $x$ and $y$ to integers. Therefore, for nonzero $\beta$ and arbitrary $\alpha$, a nearest lattice point $q$ to $\alpha /\beta$ givesThe norm makes the Eisenstein integers a Euclidean domain.

The ring $\mathbb{Z}[\sqrt{2}]$ is a Euclidean domain for the absolute field norm. Approximate both rational coefficients of a quotient by nearest integers. If the errors are $u,w\in [-1/2,1/2]$, then $|u^2-2w^2|\le 1/2<1$, giving a remainder of strictly smaller absolute norm. In particular this ring is a principal ideal domain and has trivial ideal class group.

The structure theorem for finitely generated modules over a principal ideal domain applies because a Euclidean domain is a principal ideal domain. For a finitely generated $R$-module,where the $d_i$ are nonzero nonunits. The free rank $r$ and the invariant factors of a finitely generated module $d_i$ are unique, with each $d_i$ determined up to multiplication by a unit. Equivalently, the torsion submodule decomposes into cyclic modules $R/(P^a)$ for irreducible $P$, with uniquely determined prime-power elementary factors.

Let $R$ be a Euclidean domain and $I$ a nonzero ideal. Choose $d\in I\setminus \{0\}$ minimizing the Euclidean size. For any $a\in I$, Euclidean division gives $a=qd+r$, where either $r=0$ or $r$ has smaller size than $d$. But $r=a-qd\in I$, so minimality forces $r=0$. Consequently $I=(d)$. The zero ideal is $(0)$, proving that every Euclidean domain is a principal ideal domain.

A commutative ring is Noetherian if every ideal is finitely generated; equivalently, every ascending chain of ideals stabilizes. To see the equivalence, the union of an ascending chain is an ideal. Finitely many generators for that union already belong to one member of the chain, after which the chain is constant. Conversely, an ideal which is not finitely generated permits successive choices outside the ideals generated by previous choices, producing a strictly ascending chain.

Every principal ideal domain is therefore Noetherian. The given Euclidean domain property of the Gaussian integers yields

Take $\boxed{R=\mathbb{Q}[X]}$ and, for the next clause, its subring $S=\mathbb{Z}[X]$. Division with remainder makes $R$ a Euclidean domain, with the degree as Euclidean size for nonzero polynomials. For completeness, a nonzero ideal of $R$ contains a polynomial $g$ of least degree; division of any member by $g$ leaves a smaller-degree remainder in the ideal, so that remainder must vanish. The ideal is therefore $(g)$; the zero ideal is already principal. Thus $R$ is a principal ideal domain.

A Euclidean domain is an integral domain $R$ with a function $\nu :R\setminus \{0\}\to \mathbb{N}$ such that for each $a\in R$ and $b\ne 0$ there are $q,r\in R$ with $a=qb+r$ and either $r=0$ or $\nu (r)<\nu (b)$. Some conventions add a multiplicative monotonicity condition; it is unnecessary for this implication.

Let $I$ be a nonzero ideal. Choose $d\in I\setminus \{0\}$ minimizing $\nu (d)$, possible because the values lie in the nonnegative integers. For $a\in I$, division by $d$ gives $a=qd+r$, with $r=a-qd\in I$. A nonzero remainder would contradict minimality, so $r=0$ and $a\in (d)$. Since $d\in I$, the reverse inclusion also holds. The zero ideal is $(0)$. Thus

The structure theorem for finitely generated modules over a principal ideal domain applies in particular to a Euclidean domain $R$: every finitely generated module has an isomorphismwith each $d_i$ nonzero and a nonunit. The rank of a free module $r$ and the invariant factors of a finitely generated module $d_i$ are unique up to multiplication by units; empty sums are allowed. This is the classification theorem being used, without a proof of that theorem.

The rational canonical form theorem states that for a linear operator $T$ on a finite-dimensional vector space over any field $F$, there are unique monic nonconstant polynomials $p_1\mid \cdots \mid p_k$ such that some basis givesHere, for $p(t)=t^d+a_{d-1}{t}^{d-1}+\cdots +a_0$, the companion matrix has ones on its subdiagonal and last column $(-a_0,\dots ,-a_{d-1}{)}^T$. The empty matrix covers a zero-dimensional space. Two linear operators have matrix similarity exactly when they have the same invariant factors of a linear operator.

For the proof, make $V$ an $F[t]$-module by $t\cdot v=T(v)$. It is a finitely generated module, since a vector-space basis also generates it as a module. It is a torsion module: the powers of $T$ have linear dependence in the finite-dimensional space ${\mathrm{End}}_F(V)$, so some nonzero polynomial annihilates every vector. Because $F[t]$ is a Euclidean domain, the classification theorem givesThere is no free summand, since a nonzero free module over $F[t]$ is not a torsion module. Choose monic generators of the ideals. On $F[t]/(p_i)$, the classes of $1,t,\dots ,t^{d_i-1}$ form an $F$-basis by polynomial division. Multiplication by $t$ is exactly $C(p_i)$ in that basis, proving existence. Uniqueness in the classification theorem proves uniqueness of the rational canonical form. Finally, an $F[t]$-module isomorphism is precisely an invertible $F$-linear map intertwining the operators, establishing the assertion about matrix similarity. In particular the characteristic polynomial is ${\prod }_i{p}_i$ and the minimal polynomial is $p_k$ for nonzero $V$.

Here the coefficient fields are $\mathbb{R}$ and $\mathbb{C}$. A polynomial ring in one variable over a field is a Euclidean domain, hence a principal ideal domain; its nonzero prime ideals are generated by irreducible polynomials. The fundamental theorem of algebra says that all irreducible polynomials over $\mathbb{C}$ are linear. Combining it with complex conjugation shows that the monic irreducible polynomials over $\mathbb{R}$ are exactly the linear polynomials and quadratics with no real root. ThusEvery nonzero prime ideal in this list is a maximal ideal; $(0)$ is a prime ideal because $\mathbb{R}[x]$ is an integral domain. Likewise,The map is contraction of an ideal along the inclusion. It sends $(0)$ to $(0)$. For $z=a+ib$, the contraction of $(x-z)$ consists of the real polynomials vanishing at $z$. If $b=0$, the factor theorem gives the ideal $(x-a)$. If $b\ne 0$, a real polynomial vanishing at $z$ also vanishes at its complex conjugate $\overline{z}$, so it is divisible by $(x-z)(x-\overline{z})=(x-a{)}^2+b^2$. Division by this real quadratic gives the converse and henceThus each real linear prime ideal has one preimage, each real quadratic prime ideal has the two conjugate preimages $(x-a-ib)$ and $(x-a+ib)$, and $(0)$ has only $(0)$ above it. In particular, $f$ is surjective.

Every Euclidean domain is a principal ideal domain: in a nonzero ideal choose an element of least positive Euclidean value and divide every other element by it; minimality forces every remainder to vanish. Every principal ideal domain is a unique factorization domain: the ascending-chain condition supplies factorizations into irreducibles, and Euclid lemma follows because every irreducible generates a prime ideal. Therefore

The structure theorem for finitely generated modules over a principal ideal domain applies because every Euclidean domain is a principal ideal domain. It says that a finitely generated $R$-module is isomorphic towhere the nonzero nonunits $d_i$ are unique up to multiplication by units. They are the invariant factors.

For $V_{\alpha }$, multiplication by $X$ is the linear map $\alpha$. Since $V$ is finite-dimensional over $F$, $V_{\alpha }$ is a finitely generated torsion module over the polynomial ring $F[X]$, so it has no free summand. Choosing each invariant factor $a_i$ to be monic givesThe basis $1,X,\dots ,X^{\mathrm{deg}{a}_i-1}$ of each cyclic summand makes multiplication by $X$ a companion matrix. Concatenating these bases therefore puts $\alpha$ in rational canonical form.

On $F[X]/(a_i)$, a polynomial annihilates multiplication by $X$ exactly when it is divisible by $a_i$. It follows that the minimal polynomial and characteristic polynomial areThe second identity follows blockwise from the characteristic polynomial of a companion matrix. Since every $a_i$ divides $a_k$, the product ${\chi }_{\alpha }$ annihilates every cyclic summand. Thus ${\chi }_{\alpha }(\alpha )=0$, which is the Cayley-Hamilton theorem.

For the displayed matrix, the given generators of $\ker \theta$ are the columns ofElementary row and column operations over $\mathbb{R}[X]$ give the Smith normal formConsequently the nonunit invariant factors areand, as a check,

A Euclidean domain is an integral domain $R$ equipped with a function $\delta :R\setminus \{0\}\to \mathbb{N}$ such that for $a,b\in R$, $b\ne 0$, there are $q,r\in R$ withFor the Gaussian integers $\mathbb{Z}[i]$, take $\delta (z)=N(z)=|z{|}^2$. Choosing a Gaussian integer nearest to $a/b$ makes the remainder norm smaller than $N(b)$.

The units are precisely the elements of norm one:Unique factorization in this Euclidean domain givesThe displayed factors have prime norms $2$ or $5$, so they are irreducible; factors appearing together are nonassociate.

Now suppose $x^2+4=y^3$. Necessarily $y>0$. First let $y$ be odd. Then $x$ is odd, and $x+2i$ and $x-2i$ are coprime in $\mathbb{Z}[i]$: a common Gaussian prime would divide $4i$, while their product has odd norm. Henceup to a unit, which can be absorbed into the cube. Comparing imaginary parts givesChecking $b\mid 2$ yields only $b=-2$, $a=\pm 1$, and therefore

If $y$ is even, congruence modulo $8$ gives $x=2X$ with $X$ odd and $y=2Y$, whereEach of $X\pm i$ contains exactly one factor $1+i$, so the coprime quotients are cubes up to units. ThusComparing the imaginary part after the four possible units reduces toEach integer factor must have absolute value one. Substitution then gives $a^2+b^2=1$, so $Y=1$ and $X=\pm 1$. HenceAll four pairs satisfy the equation, so the complete answer is