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The factorization form of Kummer-Dedekind theorem is as follows. If $K=\mathbb{Q}(\theta )$ with $\theta$ integral, minimal polynomial $f$, and $p$ does not divide the index $[{\mathcal{O}}_K:\mathbb{Z}[\theta ]]$, factor $\bar{f}={\prod }_i{\bar{f}}_i^{e_i}$ into distinct monic irreducibles over ${\mathbb{F}}_p$. ThenThe index condition is essential. All integral bases used below make the index one, so it applies to every prime.

For a squarefree quadratic $d\equiv 2,3(\mathrm{mod}4)$, take $\theta =\sqrt{d}$, $f=X^2-d$ and field discriminant $D=4d$. At an odd prime, a repeated factor occurs exactly when $p\mid d$, in which case $\bar{f}=X^2$ and $p{\mathcal{O}}_K=(p,\sqrt{d}{)}^2$. At $p=2$, reduction is $X^2$ if $d$ is even and $(X+1{)}^2$ if $d$ is odd, so $2$ is totally ramified in both cases. Thus the totally ramified primes are exactly those dividing $4d$.

For $d\equiv 1(\mathrm{mod}4)$, take $\theta =(1+\sqrt{d})/2$, with polynomial $X^2-X+(1-d)/4$ and discriminant $D=d$. At an odd prime it has a double root exactly when $p\mid d$, giving a single linear factor squared and total ramification. At $p=2$, its derivative is $1$, so no repeated factor occurs. Hence again

For the prime cyclotomic field, use ${\Phi }_q(X)=1+X+\cdots +X^{q-1}$. Modulo $q$ this is $(X-1{)}^{q-1}$, so $q{\mathcal{O}}_K=(q,\zeta -1{)}^{q-1}$, proving total ramification. For $p\ne q$, the discriminant has no prime factor $p$, so the reduction is squarefree and $p$ is unramified. There is an actual exponent error in the printed discriminant: the correct value is $(-1{)}^{(q-1)/2}{q}^{q-2}$, not $q^{q-1}$. To check it, at a nontrivial root ${\zeta }^j$,and multiplying these derivatives and using ${\prod }_{j=1}^{q-1}(1-{\zeta }^j)=q$ gives the exponent $q-2$. The prime support is still just $\{q\}$, so the requested ramification conclusion is unaffected.

For $\theta =\sqrt[3]{2}$, the integral basis gives $f=X^3-2$ and $D=-108$. Modulo $2$, $\bar{f}=X^3$; modulo $3$, $\bar{f}=(X+1{)}^3$. Dedekind factorization gives $(2)=(2,\theta {)}^3$ and $(3)=(3,\theta +1{)}^3$. For other primes the polynomial is squarefree because its discriminant is not divisible by the prime. Therefore the equivalence also holds in this cubic field. This is a conclusion for these specific fields, not the false general assertion that every ramified prime in every higher-degree number field is totally ramified.

Let $\zeta =u+iv$. Direct differentiation givesThese are holomorphic polynomials and their squared sum isMoreover direct calculation of the metric gives $E=G=(1+u^2+v^2{)}^2>0$ and $F=0$, so the map is genuinely an immersion everywhere. The preceding characterizations prove that it is an isothermal parametrization of a minimal surface, the Enneper surface.

An elementary geometric-series construction suffices, without assuming a general polynomial-approximation theorem. Rotate the square by a multiple of $\pi /2$ so that $\Re w>1$. Put $c=-R$ with $R>0$. Every point of the square then satisfies $|z-c{|}^2\le (R+1{)}^2+1$, whereasfor sufficiently large $R$. Thus the whole square lies inside a disk centred at $c$ whose radius is less than $|w-c|$. The other three ways $w$ can lie outside the square are handled by the indicated rotations, after which rotating back preserves polynomial form.

Now expandLet $\rho ={\max }_{z\in K}|z-c|/|w-c|<1$. The tail after degree $m$ is uniformly bounded by ${\rho }^{m+1}/[|w-c|(1-\rho )]$, tending to zero. Hence the displayed partial sums are the required uniform polynomial approximations. This is the Taylor series about a carefully chosen distant centre, rather than a potentially inadequate series about $0$.

The fixed points are $(0,0)$ and $(\pm 1,1)$. The Jacobian matrix isAt the origin its eigenvalues are $1,-1$, so it is a saddle with stable direction the y-axis. At either other fixed point, the characteristic polynomial is ${\lambda }^2+\lambda +2$, giving eigenvalues $(-1\pm i\sqrt{7})/2$. Thus both are stable foci. The nullclines are $x=0$, $y=1$ and $y=x^2$. In the right half-plane $x$ grows below $y=1$ and decreases above it; in the left half-plane these horizontal arrows reverse. Vertically, $y$ increases below $y=x^2$ and decreases above it.

A global argument determines the requested omega-limit sets, rather than guessing from the local phase portrait. For $x\ne 0$, use the quadratic-logarithmic Lyapunov function for a feedback systemIts sublevel sets are compact and bounded away from $x=0$, so trajectories starting in either half-plane exist for all positive time and stay in such a set. Since $L$ decreases, it tends to a limit, and is constant on the omega-limit set. Every point of that set therefore has $y=1$. Invariance then requires $\dot{y}=x^2-1=0$, so only the corresponding focus can occur. Also $x(t)=x(0)\exp ({\int }_0^t(1-y(s))ds)$ preserves its sign.

ConsequentlyOn $x=0$, the exact solution is $y(t)=y(0)e^{-t}$. The phase portrait below shows this stable axis, the nullclines and the spiral attraction in each half-plane.

The minimal polynomial $m_A$ is the monic polynomial of smallest degree satisfying $m_A(A)=0$. An annihilating polynomial exists, for example because $I,A,\dots ,A^{n^2}$ are linearly dependent in the $n^2$-dimensional space of matrices. A nonzero constant cannot annihilate $A$, so the minimal degree is positive.

If $p$ and $q$ are two such monic polynomials of minimal degree $d$, then $(p-q)(A)=0$ and $\mathrm{deg}(p-q)<d$ unless $p=q$. Minimality therefore forces $p=q$. Polynomial division also shows that $m_A$ divides every annihilating polynomial: the remainder after division would otherwise be an annihilator of smaller degree. This proves uniqueness rather than merely choosing one of several minimal-degree polynomials.

If $A$ is real, coefficientwise complex conjugation of $m_A(A)=0$ gives $\overline{m_A}(A)=0$. The conjugated polynomial is monic of the same degree, hence uniqueness gives $\overline{m_A}=m_A$. The minimal polynomial of a real matrix has real coefficients.

For $n\ge 3$, use the block diagonal matrixThe two-dimensional Jordan block is annihilated by $(t-1{)}^2$ but not by $t-1$, while the scalar block requires the factor $t+1$. A polynomial annihilates the direct sum exactly when it annihilates every block. Thus its minimal polynomial is the least common multiple $(t-1{)}^2(t+1)$. For $n=3$ the final identity block is omitted.

Let $\mathcal{S}$ be the set of all subsets of $G$ with $p^n$ elements. First show that $|\mathcal{S}|=\left(\genfrac{}{}{0ex}{}{p^{n}r}{p^n}\right)$ is not divisible by $p$. In the polynomial ring over the field of $p$ elements, $(1+X{)}^p=1+X^p$, since the intermediate binomial coefficients are divisible by $p$. Iterating givesComparison of coefficients of $X^{p^n}$ yields $\left(\genfrac{}{}{0ex}{}{p^{n}r}{p^n}\right)\equiv r\not\equiv 0(\mathrm{mod}p)$.

Let $G$ act on $\mathcal{S}$ by left translation. Because its orbit sizes sum to a number not divisible by $p$, some orbit has size prime to $p$. Choose a subset $S$ in that orbit, with stabilizer $H$. The orbit-stabilizer theorem gives the orbit size $[G:H]$, so $p\nmid [G:H]$. On the other hand $H$ acts freely on the elements of $S$: if $hs=s$, cancellation in the group gives $h=1$. Every $H$-orbit in $S$ therefore has $|H|$ elements, and $|H|$ divides $|S|=p^n$. Sinceand the second factor has no factor of $p$, the first must contain all $p^n$. Hence $\boxed{|H|=p^n}$, proving the first of the Sylow theorems. This is Sylow existence by subset action and includes the trivial case $n=0$.

For a bound state take $b>0$ and set $\psi (r)=e^{-br}v(r)$. Substitution into the radial equation and cancellation of the common exponential giveSeek a power series $v(r)={\sum }_{j\ge 0}{c}_j{r}^j$ with $c_0\ne 0$. Equating coefficients of $r^j$ yields the polynomial construction of hydrogen S states recurrenceFor any integer $N\ge 1$, choose $b=a/(2N)$. The numerator vanishes at $j=N-1$, while it is nonzero for $j<N-1$, so $v$ is a nonzero polynomial of degree $N-1$. Therefore $\psi$ is continuous at $r=0$ and decays exponentially as $r\to \infty$. The three-dimensional normalization integral isIt is finite near zero because $\psi$ is bounded, and finite at infinity because an exponential dominates every polynomial. Multiplying by its positive reciprocal square root gives a normalizable, nonzero wavefunction. The recurrence ensures the differential equation for $r>0$; at the singular endpoint the regular continuous extension has $2{\psi }^{\prime }(0)+a\psi (0)=0$, so the potentially singular coefficient cancels. No nonexistent value of $1/r$ at zero is substituted.

Since $E=-{\hslash }^2{b}^2/(2m)$, the choice of $b$ givesFor $N=1$, the normalized example is $\psi (r)=\sqrt{b^3/\pi }{e}^{-br}$. For $N=2$, $v(r)$ is proportional to $1-br$. These illustrate the polynomial-times-exponential Coulomb bound states produced for every required energy. The radial derivative at the origin is generally nonzero, a Coulomb cusp consistent with the requested continuity.

For node $x_i$, let ${\ell }_i$ be the Lagrange interpolation polynomial with ${\ell }_i(x_j)={\delta }_{ij}$. It has degree $n$, so ${\ell }_i^2$ has degree $2n$. The Gaussian quadrature with $n+1$ nodes is exact through degree $2n+1$, henceThe integrand is nonnegative. Since ${\ell }_i$ is a nonzero polynomial, it is nonzero on some open subinterval; the positive weight has positive integral there. Consequently $\boxed{a_i>0}$ for every node. This proves the positive weights of Gaussian quadrature property without assuming any coefficient sign beforehand.

Write $W={\int }_{-1}^{1}w(x)dx$, finite and positive. Exactness on the constant polynomial gives ${\sum }_i{a}_i^{(n)}=W$. Positivity therefore implies the uniform operator bound $|I_n(h)|\le W\Vert h{\Vert }_{\infty }$, and the integral has the same bound.

For a continuous $f$ and $ϵ>0$, the Weierstrass approximation theorem gives a fixed polynomial $Q$ of degree $m$ with $\Vert f-Q{\Vert }_{\infty }<ϵ$. For every $n\ge m$, polynomial exactness gives $I_n(Q)=I(Q)$. ThusSince $ϵ$ is arbitrary, $\boxed{I_n(f)\to I(f)}$ for every continuous $f$. This is convergence of positive quadrature on continuous functions. The argument requires nodes in $[-1,1]$, implicit in a rule defined on all continuous functions there; otherwise their function values would not even be specified by the stated domain. Exactness alone would not provide the norm bound if arbitrarily signed weights were allowed.

Suppose $v=(v_1,\dots ,v_r{)}^t$ belongs to the kernel of the displayed matrix $\Lambda$. Define the polynomial $q(z)={\sum }_{k=1}^r{v}_k{z}^k$. Then $q({\lambda }_j)=0$ for all $j$, and $q(0)=0$. These are $r+1$ distinct roots of a polynomial of degree at most $r$. Repeated application of the factor theorem therefore makes $q$ identically zero, so every $v_k=0$. Hence $\Lambda$ is injective and, by finite-dimensional rank-nullity theorem, invertible. Its transpose is invertible with inverse $({\Lambda }^{-1}{)}^t$, proving both kernels are trivial.

For the second matrix, membership of the all-ones vector in the kernel meansGroup the distinct nonzero values as ${\mu }_1,\dots ,{\mu }_r$, with multiplicities $m_1,\dots ,m_r>0$. Zero values contribute nothing to these positive powers. The first $r$ equations areor $\Lambda (\mu {)}^{t}m=0$. The proved transpose-kernel statement implies $m=0$, contradicting its positive entries if $r>0$. Therefore $r=0$ andConversely, all zero values plainly give a zero matrix. This proves the zero power sums force a finite complex multiset to vanish criterion, including repetitions and zero values rather than assuming all eigenvalues distinct.