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For a nonzero integral ideal $\mathfrak{a}$ of the full ring of integers of a number field of degree two,Here conjugation is the nonidentity field automorphism and $N(\mathfrak{a})$ is the additive index of the ideal. A direct proof writes $\mathfrak{a}=s⟨d,e+\omega {⟩}_{\mathbb{Z}}$ in an integral basis $1,\omega$. Put $\theta =e+\omega$, $T=\theta +\bar{\theta }$, $U=\theta \bar{\theta }$. The ideal property gives $d\mid U$. Moreover $\gcd (d,T,U/d)=1$: a prime dividing all three would make $\theta /\ell$ an algebraic integer, contradicting its nonintegral coefficient of $\omega$. The product of $⟨d,\theta ⟩$ with its conjugate contains $d^2,dT,U$, so Bezout identity puts $d$ in it; all its generators lie in $(d)$. Scaling back gives $(s^{2}d)$, and the lattice determinant identifies $s^{2}d=N(\mathfrak{a})$. The full ring of integers is essential to the integrality argument.

Two polynomials over a field are coprime if their only common factors are nonzero constants. Equivalently, their greatest common divisor is a unit in the polynomial ring. In one variable this is equivalent to a Bezout identity $Af+Bg=1$ with polynomial coefficients. Independent members of a pencil spanned by coprime polynomials are themselves coprime.

The ascending chain condition supplies existence of factorizations into irreducible elements. The Bezout identity makes every irreducible element prime, which supplies uniqueness up to order and multiplication by units.

A nonzero finitely generated subgroup of the additive rational numbers is an infinite cyclic group. Put the generators over a common positive denominator. Their integer numerators generate the ideal given by their greatest common divisor, and Bezout identity expresses that divisor as an integer combination of them. Dividing back by the denominator gives a single generator.

Write $x=3+18k$. The second modular congruence becomes $18k\equiv -1(\mathrm{mod}23)$. The Bezout identity above says that $9$ is the inverse of $18$ modulo $23$, so $k\equiv -9\equiv 14(\mathrm{mod}23)$. ThereforeThe checks are $255=18\cdot 14+3=23\cdot 11+2$. The Chinese remainder theorem explains uniqueness modulo $18\cdot 23=414$, since the moduli are coprime integers.

Use the Euclidean algorithm: $23=18+5$, $18=3\cdot 5+3$, $5=3+2$, and $3=2+1$. Back-substitution gives the Bezout identity $1=9\cdot 18-7\cdot 23$. Multiplying by $101$ gives one solution of the linear Diophantine equation, and shifting the coefficients by multiples of $(23,-18)$ reduces it toIndeed, $18\cdot 12+23(-5)=216-115=101$. More generally all solutions are $x=12+23k$, $y=-5-18k$, $k\in \mathbb{Z}$: subtracting this particular solution gives $18(x-12)=-23(y+5)$, and the coprime integers $18,23$ force $23\mid (x-12)$.

The space of test functions is $\mathcal{D}(\mathbb{R})=C_c^{\infty }(\mathbb{R})$: its elements are smooth functions with compact support. For a fixed compact set $K$, put ${\mathcal{D}}_K=\{\phi \in C^{\infty }(\mathbb{R}):\mathrm{supp}\phi \subset K\}$ and use the seminorms $p_m(\phi )={\max }_{0\le j\le m}\Vert {\phi }^{(j)}{\Vert }_{\infty }$. The space of test functions carries the usual test-function inductive limit topology of these spaces. In particular, a sequence converges precisely when its supports eventually lie in one compact set and every derivative converges uniformly. Thus ${\phi }_j\to \phi$ means that a common $K$ contains their supports and $p_m({\phi }_j-\phi )\to 0$ for every $m$.

A distribution is a continuous linear functional on this space of test functions, and ${\mathcal{D}}^{\prime }(\mathbb{R})$ denotes their space. We use complex-linear, bilinear pairings $⟨u,\phi ⟩$. Equivalently, for every compact set $K$ there are $C_K$ and a nonnegative integer $m_K$ such thatThe usual weak convergence of distributions is $u_j\to u$ if $⟨u_j,\phi ⟩\to ⟨u,\phi ⟩$ for every test function. This specifies the convergence used below; it does not require convergence in any norm.

The distributional derivative is defined byThe derivative map sends ${\mathcal{D}}_K$ continuously to itself, with $p_m({\phi }^{\prime })\le p_{m+1}(\phi )$. The preceding continuity estimate therefore proves that $u^{\prime }$ is again a distribution. This definition extends the ordinary derivative of a smooth function, by integration by parts.

Choose the translation of a distribution convention ${\tau }_{h}f(x)=f(x-h)$. Its action on a test function isFor fixed $h$, the translated test functions have translated compact support and unchanged derivative sup norms, so this defines a distribution. For the differentiability of distribution translations, apply the Taylor theorem in integral form:All supports lie in one slightly enlarged compact set for $|h|\le 1$, and the same identity for every derivative proves uniform convergence. The continuity of $u$ now givesThe minus sign in the translation parameter is necessary for this convention.

For (a), choose a test function $\eta$ with $\int \eta =1$. If $\int \psi =0$, then $F(x)={\int }_{-\infty }^x\psi (s)ds$ is a test function, since the zero integral makes it vanish beyond both ends of the compact support. If $u_1^{\prime }=0$, then $⟨u_1,\psi ⟩=⟨u_1,F^{\prime }⟩=0$. Decompose $\phi =(\int \phi )\eta +\psi$ to obtainConversely, these constant regular distributions have zero distributional derivative. This also proves the general fact that a distribution with zero derivative is constant.

For (b), choose a cutoff function $\rho$ equal to one near zero. Every test function decomposes asThe quotient extends as a smooth function at zero, and it has compact support. If $x{u}_2=0$, the definition of multiplication of a distribution by a smooth function gives $⟨u_2,\phi ⟩=⟨u_2,\rho ⟩\phi (0)$. Conversely, $x{\delta }_0=0$. Thus the kernel of multiplication by a coordinate is exactlyIn particular, derivatives of the Dirac delta distribution are not additional solutions: $x{\delta }_0^{\prime }=-{\delta }_0$.

Now write $D=d/dx$ and $P(z)=-z^n+{\sum }_{j=0}^{n-1}{a}_j{z}^j$, where $n\ge 1$. The characteristic roots of a constant-coefficient differential equation are the distinct roots of a polynomial ${\lambda }_1,\dots ,{\lambda }_r$ of $P$, with multiplicities $m_1,\dots ,m_r$. The distributional regularity of a constant-coefficient ordinary differential equation can be proved without assuming regularity in advance. Put $q_j(z)=(z-{\lambda }_j{)}^{m_j}$ and $Q={\prod }_j{q}_j=-P$. Since the $q_j$ are coprime polynomials, polynomial division and Bezout identity give polynomials $b_j$ such thatFor example, invert $Q/q_j$ modulo $q_j$, sum the resulting expressions, and absorb a remaining multiple of $Q$ into one coefficient. For $P(D)v=0$, the kernel decomposition for coprime polynomials therefore givesThe Leibniz rule for multiplication of a distribution by a smooth function implies $D^{m_j}(e^{-{\lambda }_{j}x}{v}_j)=0$. Repeatedly using a distribution with zero derivative is constant shows that a distribution with $m$th derivative zero is a polynomial of degree at most $m-1$: subtract the polynomial primitive of its constant $(m-1)$st derivative, and induct on $m$. Consequently the most general exponential polynomial solution of a constant-coefficient differential equation isEvery displayed term is annihilated by $P(D)$, so all coefficients are allowed. For the linear independence of these $n$ functions, on a relation, apply ${\prod }_{i\ne j}(D-{\lambda }_i{)}^{m_i}$; on $e^{{\lambda }_{j}x}$ times a polynomial of degree less than $m_j$, each remaining factor acts invertibly on that polynomial space. Hence the $j$th polynomial must vanish. Every distributional solution is an analytic function, and in particular a classical solution. For real coefficients and real-valued distributions, take complex conjugate coefficients at conjugate characteristic roots of a constant-coefficient differential equation, or equivalently use the corresponding real sine and cosine forms.

For the last equation, the operator is $xP(D)$: the coordinate multiplies the result of differentiation. The kernel of multiplication by a coordinate says precisely thatChoose the retarded fundamental solution of a constant-coefficient ordinary differential operator $E=Hw$, where $H$ is the Heaviside function and $w$ is the analytic solution of $P(D)w=0$ withSuch $w$ exists uniquely by the elementary initial value problem for a constant-coefficient ordinary differential equation; alternatively the residue construction in the next solution gives it explicitly. The distributional jump formula for a Heaviside product isIt follows by induction from $D(Hw)=H{w}^{\prime }+w(0){\delta }_0$, using integration by parts. With the chosen initial derivatives, every lower-order jump term vanishes and the leading coefficient $-1$ gives $P(D)E={\delta }_0$. Subtracting $CE$ reduces the last equation to the homogeneous one. Thus the coordinate-degenerate constant-coefficient differential equation has exactly the solutionsThere are $n+1$ independent constants. For $n\ge 2$, derivatives through order $n-2$ match at zero, while the derivative of order $n-1$ may jump; for $n=1$, the function itself may jump. An arbitrary distribution concentrated at zero cannot be added, because its image under the nonzero leading derivative would contain a nonvanishing highest derivative of the Dirac delta distribution.

Choose a positive common denominator $d$ so that $x=a/d$ and $y=b/d$ with nonzero integers $a,b$. Put $h=\gcd (a,b)>0$. Every element of the generated subgroup has the form $(ma+nb)/d$, so it lies in $(h/d)\mathbb{Z}$. Conversely, Bezout identity gives integers $u,v$ with $ua+vb=h$, so $h/d$ belongs to the generated subgroup. ConsequentlyThis map is a bijective additive group homomorphism. It is the two-generator instance of a finitely generated subgroup of the rational additive group being an infinite cyclic group.

For the real example, take $x=1$ and $y=\sqrt{2}$. The group homomorphism $(m,n)\mapsto m+n\sqrt{2}$ from ${\mathbb{Z}}^2$ onto the generated subgroup is injective because $\sqrt{2}$ is an irrational number. Thus this subgroup is isomorphic to ${\mathbb{Z}}^2$. More directly, if a single real number $t$ generated it, then $1=at$ and $\sqrt{2}=bt$ for integers $a\ne 0,b$, giving $\sqrt{2}=b/a$, a contradiction. The choice $1,\sqrt{2}$ generates a noncyclic additive subgroup.