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We use these countability facts: the integers are countable; a finite Cartesian product of countable sets is countable; a countable union of countable sets is countable; and every subset of a countable set is countable. Finite products can be enumerated by listing index tuples in increasing maximum coordinate, and countable unions by diagonal enumeration of pairs of indices.

Fix $d\ge 1$. For each degree bound $D$, there are finitely many monomials $X_1^{{\alpha }_1}\cdots X_d^{{\alpha }_d}$ with ${\alpha }_1+\cdots +{\alpha }_d\le D$. An integer polynomial of total degree at most $D$ is specified by an integer coefficient for each of them, so these polynomials form a finite Cartesian product of copies of $\mathbb{Z}$ and are countable. Every polynomial has some finite degree bound. Taking the union over $D=0,1,2,\dots$ proves the countability of integer polynomial rings:It is infinite because it contains all integer constant polynomials.

For $d=1$, each nonzero polynomial has only finitely many real roots: repeated application of the factor theorem bounds their number by its degree. The real algebraic numbers are the union of these finite root sets over the countably many nonzero integer polynomials. Thus they are countable. If the real transcendental numbers were countable too, their union with the algebraic numbers would make $\mathbb{R}$ countable, contrary to the allowed uncountability assumption. Hence there are uncountably many real transcendental numbers.

To construct an algebraically independent real sequence, choose $x_1$ outside the algebraic numbers. Suppose $x_1,\dots ,x_d$ already satisfy the required nonvanishing condition. For each nonzero $P\in \mathbb{Z}[X_1,\dots ,X_d,T]$, writeAt least one coefficient polynomial $P_j$ is nonzero, and the inductive hypothesis makes its value at $(x_1,\dots ,x_d)$ nonzero. Consequently $P(x_1,\dots ,x_d,T)$ is a nonzero one-variable polynomial and has only finitely many real roots. There are countably many choices of $P$, so the union of all these forbidden roots is countable. Choose $x_{d+1}$ outside it, which is possible because $\mathbb{R}$ is uncountable. This extends the nonvanishing condition to $d+1$. Recursion therefore givesIn particular the chosen numbers are distinct, since a relation $X_i-X_j$ is forbidden. Clearing rational denominators shows that every finite initial segment consists of algebraically independent elements over $\mathbb{Q}$.

A nonzero polynomial of degree $n$ over a field has at most $n$ distinct roots in that field. For a proof, the factor theorem gives $(t-a)\mid f(t)$ whenever $f(a)=0$. Distinct linear factors are coprime in $K[t]$, a unique factorization domain, so for distinct roots $a_1,\dots ,a_k$ their product divides $f$. Its degree is $k$, giving $k\le n$. One may equivalently remove one linear factor at a time and argue by induction.

If the field has at least $n$ elements, this bound is attained by a product of $n$ distinct linear factors. For a fixed finite field with fewer than $n$ elements, the largest possible number of distinct roots is $\min (n,|K|)$ instead; multiplicities do not create more distinct roots. The zero polynomial is excluded since it has no finite degree.

Put $h_0=1$ and $h_r=0$. The Markov property and first-step conditioning give $h_i={\sum }_j{P}_{ij}{h}_j$. Consider a finite history ending at state $i$ and having positive conditional probability given $A$. The chance of eventual absorption at zero after that history is $h_i$: the future depends only on the present state, and if zero has already been reached, absorption keeps the chain there. If the next state is $j$, this chance is $h_j$. Bayes' rule therefore givesThis depends on the history only through $i$, proving the conditional Markov property. The conditional transition matrix isIts row sums are one by harmonicity of $h$, and $Q_{ir}=0$. State $r$ is never visited under the conditioning and is naturally removed; any row assigned there on an enlarged formal state space is immaterial, since conditioning starting at $r$ is undefined. State zero stays absorbing. For a random initial law $\nu$, its conditional law is ${\nu }_i{h}_i/{\sum }_j{\nu }_j{h}_j$. This is the Doob h-transform with the absorption probability as harmonic function.

For the simple symmetric walk, $h_i=(r-i)/r$ solves the linear harmonic recurrence and its boundary values. ThusIn particular, the conditioned chain moves downward with probability one from $r-1$.

Let $\tau$ be the first hit of $\{0,r\}$ in the original walk and let $g_i={\mathbb{E}}_i[\tau 1_A]$. Its absorption time has finite expectation: from any transient state a run of at most $r$ downward steps hits zero with probability at least $2^{-r}$, giving a geometric tail bound by blocks. First-step conditioning then givesEquivalently, $g_{i+1}-2g_i+g_{i-1}=-2(r-i)/r$. The cubic polynomialhas exactly these second differences and boundary values. It is the unique solution, since the difference of any two solutions is a linear sequence vanishing at both endpoints. Conditional on $A$, $\tau$ is the first time in zero, so division by $h_i$ provesThis is the duration of gambler's ruin conditioned on a chosen boundary.

For nonnegative integers $m,n$, Fourier orthogonality givesEvery polynomial is a finite linear combination of these monomials, and distinct monomials are linearly independent. ThereforeThis is an algebraic basis: no infinite linear combinations are needed in the polynomial vector space.

The integral is linear in its first argument and conjugate-linear in its second, and complex conjugation of the integrand shows $⟨g,f⟩=\overline{⟨f,g⟩}$. Thus it is a Hermitian form.

If $f(z)={\sum }_{j=0}^N{a}_j{z}^j$, Fourier orthogonality givesThis is nonnegative and vanishes only when every coefficient vanishes, that is, only for the zero polynomial. Hence the Hermitian form is positive definite.

Hilbert's basis theorem: if $R$ is a commutative Noetherian ring with identity, then the polynomial ring $R[x]$ is Noetherian. Iteration gives the same assertion for every finite number of variables.

Here is a bounded-degree coefficient proof of Hilbert basis theorem. Let $I$ be any ideal of $R[x]$. For $n\ge 0$, define $J_n\subseteq R$ to be the coefficients of $x^n$ in the polynomials of $I$ of degree at most $n$, including the zero coefficient. Closure under addition and scalar multiplication makes $J_n$ an ideal of $R$. Multiplication of a polynomial by $x$ shows $J_n\subseteq J_{n+1}$. By the ascending chain condition, there is $N$ such that $J_N=J_{N+1}=\cdots$.

Each $J_n$ for $0\le n\le N$ is finitely generated. Choose generators $a_{n,1},\dots ,a_{n,r_n}$ and corresponding polynomials $f_{n,j}\in I$ of degree at most $n$ whose coefficient of $x^n$ is $a_{n,j}$. There are only finitely many chosen polynomials. We prove they generate $I$ by induction on degree.

Take a nonzero $f\in I$ of degree $d$ and leading coefficient $a$. If $d\le N$, express $a={\sum }_j{c}_j{a}_{d,j}$ and subtract ${\sum }_j{c}_j{f}_{d,j}$. If $d>N$, use $a\in J_d=J_N$, express $a={\sum }_j{c}_j{a}_{N,j}$, and subtract ${\sum }_j{c}_j{x}^{d-N}{f}_{N,j}$. In both cases the leading coefficient cancels, leaving a polynomial in $I$ of smaller degree. For degree zero the remainder is zero, and the induction completes the claim.

Thus every ideal of $R[x]$ is finitely generated, proving the Hilbert basis theorem. The proof uses no cancellation of nonzero elements of $R$, so it remains valid when $R$ has zero divisors.

A field, for example $\mathbb{Q}$, is a Noetherian ring: its only ideals are $(0)$ and the whole field, both finitely generated.

For a non-Noetherian example, take $R=\mathbb{Q}[x_1,x_2,\dots ]$, the polynomial ring in countably many variables, each individual polynomial using only finitely many variables. The ascending chain of idealsis strictly increasing. Indeed, if $x_{n+1}$ belonged to $(x_1,\dots ,x_n)$, setting the first $n$ variables to zero would give the impossible equality $x_{n+1}=0$. The chain never stabilizes, contradicting the ascending chain condition. Thus $\mathbb{Q}$ is Noetherian, while $\mathbb{Q}[x_1,x_2,\dots ]$ is not.

Each endomorphism has three distinct eigenvalues in a three-dimensional complex vector space. Choose a nonzero eigenvector for each eigenvalue; these vectors are linearly independent and hence form an eigenbasis. Both operators are therefore diagonalizable with diagonal matrix $D=\mathrm{diag}(1,2,3)$.

Their characteristic polynomials areIf a polynomial $p$ annihilates either operator, applying it to each eigenvector gives $p(1)=p(2)=p(3)=0$. Thus its minimal polynomial must contain all three factors. Conversely, their product annihilates the diagonal matrix, soFinally, send an ordered eigenbasis of $S$ to the correspondingly ordered eigenbasis of $T$. The resulting invertible map $C$ satisfies $CS=TC$, and hence $T=CS{C}^{-1}$: the two endomorphisms are conjugate.

Seek a power series $y(x)={\sum }_{k\ge 0}{a}_k{x}^k$ in the Legendre differential equation. Equating coefficients of $x^k$ givessoThe even and odd coefficients form separate chains. Choose a nonzero seed of the same parity as $n$ and set the other seed to zero. In the chosen chain, none of the factors vanishes before $k=n$, and the factor $k-n$ vanishes at $k=n$. Therefore this chain terminates with a nonzero $a_n$: it gives a polynomial of degree exactly $n$.

To impose the normalization, its value at $x=1$ must be nonzero. Suppose a nonzero polynomial solution were $y=(x-1{)}^{m}q(x)$ with $m\ge 1$ and $q(1)\ne 0$. The lowest-order coefficient in $(1-x^2)y^{\prime \prime }-2x{y}^{\prime }$ would be $-2m^{2}q(1)(x-1{)}^{m-1}$, whereas $n(n+1)y$ has order $m$. This cannot satisfy the equation. Thus divide by $y(1)$ to obtain the Legendre polynomial $P_n$ with $P_n(1)=1$. Substitution or the coefficient recurrence givesFor an axisymmetric solution of the Laplace equation regular at the two polar axes, the separated modes and their superposition in spherical polar coordinates areThis is the usual separation of variables expansion; the singular angular solutions are excluded by regularity at the axes.

Here ${\sin }^2\theta =\frac{2}{3}(P_0(\cos \theta )-P_2(\cos \theta ))$, so only degrees $0$ and $2$ are required. The degree-zero radial function $A_0+B_0/r$ takes the same value $2/3$ at $r=a,b$; since $a<b$, this forces $B_0=0$, $A_0=2/3$. Write the degree-two coefficient as $-(2/3)T(r)$, where $T=A{r}^2+B{r}^{-3}$ and $T(a)=T(b)=1$. Multiplying these boundary equations by $a^3,b^3$ and subtracting givesConsequentlyEach term is harmonic in the shell, and the radial factor is $1$ at both boundaries, so this satisfies both Dirichlet boundary conditions. The maximum principle for harmonic functions, applied to the difference of two solutions, proves uniqueness among solutions continuous on the closed shell.

The Cayley-Hamilton theorem states that every square complex matrix satisfies its own characteristic polynomial: if ${\chi }_A(t)=\det (tI-A)=t^n+c_{n-1}{t}^{n-1}+\cdots +c_0$, thenFor a proof, expand the adjugate matrix as $\mathrm{adj}(tI-A)={\sum }_{k=0}^{n-1}{B}_k{t}^k$. The adjugate identity givesComparing coefficients yields $-A{B}_0=c_{0}I$, $B_{k-1}-A{B}_k=c_{k}I$ for $1\le k\le n-1$, and $B_{n-1}=I$. Multiply the middle identities on the left by $A^k$, include the first identity, and add $A^n{B}_{n-1}=A^n$. The terms involving the $B_k$ cancel consecutively:Their right sides sum to $c_{0}I+{\sum }_{k=1}^{n-1}{c}_k{A}^k+A^n$. This proves the theorem without any substitution of noncommuting matrices into a scalar polynomial identity.

Use the unitary normalization of the Fourier transform,The Schwartz space $\mathcal{S}({\mathbb{R}}^n)$ consists of smooth functions for which every Schwartz seminorm ${\sup }_x|x^{\alpha }{\partial }^{\beta }f(x)|$ is finite. Such functions and all their derivatives are integrable. Differentiating under the integral and integrating by parts giveConsequentlyExpand the derivative by the product rule and insert the integrable weight $(1+|x|{)}^{-n-1}$; the right side is bounded by finitely many Schwartz seminorms of $f$. This proves continuity $\mathcal{F}:\mathcal{S}\to \mathcal{S}$. The identical argument with the opposite sign proves continuity of $\mathcal{G}$.

For surjectivity and injectivity an inversion proof is needed. If $f\in \mathcal{S}$, multiply $\mathcal{F}f(\xi )$ by $e^{-\epsilon |\xi {|}^2/2}$. Fubini's theorem and the elementary Gaussian integral giveThe Gaussian has total integral one and its mass outside any fixed ball tends to zero. The right side therefore tends to $f(x)$, by uniform continuity of $f$. On the left, $\mathcal{F}f\in \mathcal{S}\subset L^1$, so dominated convergence gives $\mathcal{GF}f(x)$. This proves $\mathcal{GF}=I$. Reversing the signs in the same calculation proves $\mathcal{FG}=I$. ThusIts inverse is continuous by the seminorm estimate already proved. This establishes the requested bijection without assuming Fourier inversion as an unexplained theorem.

Inversion gives ${\mathcal{F}}^{2}f(x)=f(-x)$ and the Parseval identity. Indeed, inserting the integral for $\mathcal{F}f$ into $\int \mathcal{F}f\overline{\mathcal{F}g}$ and using $f,\mathcal{F}g\in L^1$ gives $\int f\overline{g}$. Density of Schwartz functions in $L^2$ extends $\mathcal{F}$ to a unitary operator on $L^2$, the Plancherel theorem. Translation of $f$ becomes multiplication by a phase, modulation becomes frequency translation, andThese formulas explain why the Fourier transform diagonalizes constant-coefficient differential operators. By transpose it also defines a bijection of the tempered distributions: with the complex-linear distributional pairing, $⟨\mathcal{F}T,\varphi ⟩=⟨T,\mathcal{F}\varphi ⟩$, since the Fourier kernel is symmetric in its variables.

The connection with creation and annihilation operators is especially transparent for the harmonic oscillator. On the common invariant dense domain $\mathcal{S}$, setIntegration by parts shows that these are formal Hilbert-space adjoints. They are closable unbounded operators; commutators below are computed on $\mathcal{S}$, avoiding undefined products on arbitrary $L^2$ vectors. The identity $[{\partial }_j,x_k]={\delta }_{jk}I$ gives the canonical commutation relationsIt also givesThe normalized vacuum $h_0(x)={\pi }^{-n/4}{e}^{-|x{|}^2/2}$ satisfies $a_j{h}_0=0$. Define the Hermite functionsMoving annihilation operators past creation operators with the commutation relations provesThese same relations show that the Hermite functions are orthonormal: unequal occupations are orthogonal eigenvectors of one of the commuting symmetric number operators $a_j^{†}{a}_j$, and the ladder relations recursively give norm one.

For completeness, let $g\in L^2$ be orthogonal to every $h_{\alpha }$. The Hermite polynomials have nonzero leading monomials and triangular lower-degree terms, so these orthogonality relations say that all polynomial moments of $h(x)=g(x)e^{-|x{|}^2/2}\in L^1$ vanish. Its Fourier integral extends to an entire function of $\zeta \in {\mathbb{C}}^n$:The Cauchy-Schwarz inequality, with the Gaussian absorbing every exponential linear weight and every polynomial, justifies differentiation uniformly on compact sets of $\zeta$. Every Taylor coefficient at zero is a vanishing moment, so the entire function is zero. Fourier uniqueness on $L^1$ follows here without circularity: if an $L^1$ function has zero Fourier transform, convolving it with ${\rho }_{\epsilon }$ gives zero by the Gaussian integral and Fubini, and the approximate-identity limit in $L^1$ gives the original function zero. Thus $h=0$ and $g=0$. The Hermite functions form an orthonormal basis of $L^2$.

Finally, the elementary Gaussian integral gives $\mathcal{F}{h}_0=h_0$. The derivative and multiplication identities above implywhere the operators on the right act on the frequency variable. HenceThus creation and annihilation organize both the oscillator's spectral decomposition and the four possible Fourier eigenvalues. On Schwartz functions the oscillator powers control weighted derivatives, and Hermite coefficients decay faster than every power of $1+|\alpha |$; conversely this rapid coefficient decay implies all Schwartz seminorms are finite. The Fourier phases preserve this decay, providing another explanation of the Fourier automorphism of Schwartz space.