Set $I=(X^3{)}_T$ and $J=(Y^3{)}_{\mathbb{R}[Y]}$. Part (b)(i) gives a unique representative $a+b{X}^2+c{X}^4$ for each element of the quotient ring $T/I$. By polynomial division, each element of the quotient ring $\mathbb{R}[Y]/J$ likewise has a unique representative $a+bY+c{Y}^2$. Thus the proposed map $\Phi$ is well-defined and a bijection.

It preserves addition and $1$. To verify multiplication, all terms of polynomial degree at least $5$ lie in $I$. HenceIn the other quotient ring, $Y^3=0$, giving exactly the same coefficient formula:Thereforeis a ring isomorphism. Both sides are examples of a truncated polynomial ring; the generator on the left is the class of $X^2$, whose square is nonzero and whose cube vanishes.

All principal ideals in this part are taken inside the subring $T$, rather than inside $\mathbb{R}[X]$. First, $T$ is a subring: sums and additive inverses retain zero coefficient of $X$, and the coefficient of $X$ in a product $uv$ is $u(0)[X]v+v(0)[X]u=0$. It contains $1$.

The principal ideal $I=(X^3{)}_T=X^{3}T$ is preciselyIndeed, multiplying an element of $T$ by $X^3$ gives a possible term of polynomial degree $3$ and terms of polynomial degrees at least $5$, with no $X^4$ term. Conversely, dividing any displayed polynomial by $X^3$ gives a polynomial with no $X$ term, hence an element of $T$.

Given $p\in T$, let $a,b,c$ be its coefficients of $1,X^2,X^4$, respectively. Subtracting these terms leaves an element of $I$, soFor uniqueness, if two such representatives differ by an element of $I$, their difference has possible monomials only $1,X^2,X^4$, whereas every element of $I$ has zero coefficients at all three. Comparing coefficients gives $a=a^{\prime }$, $b=b^{\prime }$, $c=c^{\prime }$.

In particular, $X^4\notin I$. Treating $I$ as the principal ideal of $\mathbb{R}[X]$ instead would incorrectly discard this surviving term.

Regard $\mathbb{R}[X,Y]$ as the polynomial ring $(\mathbb{R}[X])[Y]$. The coefficient ring $\mathbb{R}[X]$ is a Euclidean domain, hence a principal ideal domain. The element $X$ is a prime element: the quotient ring $\mathbb{R}[X]/(X)\cong \mathbb{R}$ is an integral domain.

The monic polynomial $Y^2+X$ has its other coefficients divisible by $X$, while $X^2\nmid X$. Part (a)(i), with $X$ playing the role of $p$ and $Y$ the indeterminate, proves that $Y^2+X$ is an irreducible polynomial.

Now use the automorphism of a polynomial ringIts inverse fixes $X$ and sends $Y$ to $Y-1$. A ring isomorphism preserves units and factorizations, hence preserves irreducible elements. Sincewe obtainApplying the Eisenstein criterion directly to the second expression with the prime element $X$ would fail, since $X\nmid 2$; the change of indeterminate supplies the required argument.

We prove the stated Eisenstein criterion directly. Assume $n\ge 1$. Suppose $f=gh$ in the polynomial ring $R[X]$. Since the principal ideal domain $R$ is an integral domain, polynomial degree is additive under polynomial multiplication, and the product of the leading coefficients of $g,h$ is the leading coefficient $1$ of $f$. These two coefficients are therefore units. Multiplying $g$ and $h$ by inverse units, without changing their product, makes both factors monic polynomials.

If either factor is a constant polynomial, that factor is now $1$, so the original factor was a unit. It remains to rule out $r=\mathrm{deg}g>0$ and $s=\mathrm{deg}h>0$.

The prime element $p$ generates a maximal ideal in the principal ideal domain $R$. Indeed, if $(p)\subseteq (d)$, then $p=de$, and the irreducible element property of $p$ forces either $(d)=R$ or $(d)=(p)$. Consequently $k=R/(p)$ is a field. Apply reduction modulo an ideal to every coefficient. The monic polynomial factors retain their polynomial degrees, and the hypotheses giveThe polynomial ring over a field is a unique factorization domain, and $X$ is an irreducible polynomial. Thus its only monic polynomial divisors in $X^n$ are powers of $X$. Hence $\overline{g}=X^r$ and $\overline{h}=X^s$. Both have zero constant term, so $p\mid g(0)$ and $p\mid h(0)$. This gives $p^2\mid g(0)h(0)=a_0$, a contradiction.

The monic polynomial $f$ has positive polynomial degree, so it is a nonzero nonunit. Every factorization therefore has a unit factor, proving

Interpret the source's missing variable in the limit arrow as $t\to \infty$. We prove convergence in distribution by showing that the angle of the Brownian motion becomes uniform on the circle group.

Let $Z_t=e^{iB(t)}$. For every integer $m$, its angular Fourier coefficients satisfyThese are precisely the Fourier coefficients of the uniform distribution on the unit circle. Thus expectations of every trigonometric polynomial converge to its uniform angular average. By the Stone-Weierstrass theorem, trigonometric polynomials approximate continuous functions on the unit circle uniformly. Given an approximation error $\epsilon$, the discrepancy of expectations is at most the polynomial discrepancy plus $2\epsilon$. Taking $t\to \infty$ and then $\epsilon \to 0$ proves weak convergence of probability measures to uniform angular measure.

The function $z\mapsto (\mathrm{Im}z{)}^2$ is continuous on the unit circle, so the continuous mapping theorem and part (a)(ii) implyEquivalently, the limiting arcsine distribution has cumulative distribution function $2\arcsin \sqrt{x}/\pi$ for $0\le x\le 1$, and probability density function $1/(\pi \sqrt{x(1-x)})$ for $0<x<1$. This proves the limiting law rather than inferring it merely from the limiting mean.

The characteristic function of $B(t)\sim N(0,t)$ gives $\mathbb{E}{e}^{2iB(t)}=e^{-2t}$. Taking real parts and using ${\sin }^{2}b=(1-\cos 2b)/2$,There is also a direct check from the stochastic differential equation. Its Itô integral has zero expectation because its integrand is bounded. Thus $m(t)=\mathbb{E}\Delta (t)$ satisfies $m^{\prime }(t)=1-2m(t)$ and $m(0)=0$, whose solution is the same expression.

Here $\Delta (t)={\sin }^{2}B(t)$. For $f(b)={\sin }^{2}b$, $f^{\prime }(b)=\sin 2b$ and $f^{\prime \prime }(b)=2\cos 2b$. Since standard Brownian motion has quadratic variation $dt$, Itô's formula givesThereforeThe squared diffusion coefficient is $V(t{)}^2=4\Delta (t)(1-\Delta (t))$. Its sign depends on $B(t)$: one cannot replace $\sin (2B(t))$ by the nonnegative square root while retaining the same driving Brownian motion.

Apply the same characteristic polynomial calculation with $U$ replaced by $B(t)$. Pathwise,The real-valued processes $\pm B(t)$ give continuous lifts of these angles. The eigenvalues coincide whenever $B(t)\in \pi \mathbb{Z}$; the exponential formulas remain valid at those times.

The difference of the two eigenvalues is $2i\sin U$, hence $\Delta ={\sin }^{2}U$. Its expectation isFor $0\le x\le 1$, put $\alpha =\arcsin \sqrt{x}\in [0,\pi /2]$. The set of angles satisfying $|\sin u|\le \sqrt{x}$ consists of $[0,\alpha ]$, $[\pi -\alpha ,\pi +\alpha ]$ and $[2\pi -\alpha ,2\pi ]$, of total length $4\alpha$. Thus the full cumulative distribution function isDifferentiating on $(0,1)$ gives the probability density functionThis is the arcsine distribution, equivalently the beta distribution with both parameters $1/2$. It has no atoms at either endpoint.

The characteristic polynomial of the rotation matrix isIts roots are $\cos U\pm i\sin U$, soFor example $(1,i{)}^T$ is an eigenvector for $e^{iU}$ and $(1,-i{)}^T$ for $e^{-iU}$. At $U=0$ or $\pi$ the two eigenvalues coincide; these events have probability zero under the uniform distribution.

The correlation function of a point process is even and tends to zero as $|\lambda |\to \infty$. Its derivative factors asConsequently its local maxima occur at $0,\pm \sqrt{3}$, and its local minima at $\pm 1$. The heights are

For an interval $I$, ${\int }_I{R}_3^{(1)}(\lambda )d\lambda$ is the expected value of the number of eigenvalues in $I$. Thus equally short intervals have the highest expected counts nearwith the central peak highest. Each exchangeably labelled eigenvalue has the same marginal probability density function $R_3^{(1)}/3$. The three peaks should not be interpreted as three separate marginal densities for ordered eigenvalues.

In this case the most likely joint configuration also lies at these three positions. On the ordered chamber, the logarithm of the joint probability density function, up to a constant, isIts stationary equations are ${\lambda }_i=2{\sum }_{j\ne i}({\lambda }_i-{\lambda }_j{)}^{-1}$, satisfied by $(-\sqrt{3},0,\sqrt{3})$. For every nonzero real vector $v$, the second directional derivative isThus $L$ is strictly concave on this convex chamber, proving that the stationary configuration is its unique global maximum. Permuting the labels gives the other maxima on ${\mathbb{R}}^3$.

Evaluate the orthogonal polynomial projection kernel on its diagonal:The ${\lambda }^2$ terms cancel. Integrating against Lebesgue measure gives $(\mathbb{E}{Z}^4+3)/2=3$ for a standard normal random variable $Z$, checking the correlation function of a point process normalization.

The correlation functions of a point process use factorial normalization:Start with ${\rho }_3=D_3/6$. By Gaudin's lemma, integrating $D_3$ once gives $D_2$, and integrating $D_2$ once gives $2D_1$. The factorial factors cancel, so the eigenvalues form a determinantal point process withandThe last equality uses the real symmetry of this kernel. In particular, $\int R_3^{(1)}=3$ and $\iint R_3^{(2)}=6$, as required for three points and six ordered pairs of distinct points.

Use the symmetric joint probability density function of exchangeably labelled eigenvalues, normalized over all of ${\mathbb{R}}^3$. The finite-rank projection kernel hasSuccessive applications of Gaudin's lemma therefore giveIt follows thatAs a normalization check, the weighted Vandermonde determinant yieldsIf instead one uses ordered eigenvalues on ${\lambda }_1<{\lambda }_2<{\lambda }_3$, the normalized density on that chamber is six times the labelled density. In that convention the determinant prefactor on the chamber is one.

The squared norm of $p_{\ell }(x)e^{-x^2/4}$ in $L^2(\mathbb{R},dx)$ is $\sqrt{2\pi }{h}_{\ell }$, where $(h_0,h_1,h_2)=(1,1,2)$. Choose positive normalizing constantsThen ${\varphi }_0,{\varphi }_1,{\varphi }_2$ are an orthonormal set for Lebesgue measure. Expanding the finite sums givesThus this is an orthogonal polynomial projection kernel, explicitlyThe positive constants above are forced by the reproducing identity: before normalization, its convolution multiplies each independent term ${\varphi }_{\ell }(x){\varphi }_{\ell }(z)$ by $\sqrt{2\pi }{h}_{\ell }/{\kappa }_{\ell }^2$. Each multiplier must equal one. Changing the sign of a real ${\kappa }_{\ell }$ leaves the kernel unchanged.

Gaudin's lemma is the projection kernel determinant integration identity. Suppose an integral kernel $K$ satisfieswith the relevant integrals absolutely convergent. Define $D_k(x_1,\dots ,x_k)=\det [K(x_i,x_j){]}_{i,j=1}^k$ and $D_0=1$. ThenIn particular, a finite-rank projection kernel of rank $N$ has $\int D_{N}d{\mu }^N=N!$.

One can see the factor directly from the permutation expansion of the determinant. Terms whose permutation fixes $k$ integrate to $N{D}_{k-1}$. Every permutation on the other $k-1$ indices admits $k-1$ insertions of $k$ into a cycle. Each insertion reverses its sign, and the reproducing identity contracts its two adjacent kernel factors into one. These terms integrate to $-(k-1)D_{k-1}$, proving the formula.

Use the inner product $⟨f,g⟩=\mathbb{E}[f(Z)g(Z)]$ for a standard normal random variable $Z$. The degree-zero monic polynomial is $p_0=1$. Write $p_1=x+a$ and $p_2=x^2+bx+c$. The conditionsuse $\mathbb{E}Z=\mathbb{E}{Z}^3=0$ and $\mathbb{E}{Z}^2=1$. Hence these first three Probabilists' Hermite polynomials areThe Gaussian fourth moment is $\mathbb{E}{Z}^4=3$, so their squared norms are

Let $P$ be the $3$-by-$3$ evaluation matrix $P_{i,\ell +1}=p_{\ell }({\lambda }_i)$. The kernel evaluation matrix is the Gram matrix $P{P}^T$, so multiplicativity of the determinant givesBecause the polynomials are monic of consecutive degrees, their coefficient matrix relative to $1,x,x^2$ is a triangular matrix with unit diagonal. It has determinant one, so the Vandermonde determinant invariance under monic basis change givesSquaring removes the ordering sign. Thus the Gaussian unitary ensemble joint probability density function has precisely the kernel formNo orthogonality of the polynomials is needed for this step.

Let $g_i=(X_{i1},\dots ,X_{in})$ be row $i$ of $X$, viewed as a real vector. These are independent standard Gaussian random vectors, and $(X{X}^T{)}_{ij}=g_i\cdot g_j$. The trace expression for the empirical spectral measure givesWe need the norm momentsFor completeness, $S=\Vert g_i{\Vert }^2$ has the chi-squared distribution with $n$ degrees of freedom. By independence and the Gaussian integral, its moment-generating function is $\mathbb{E}{e}^{sS}=(1-2s{)}^{-n/2}$ for $s<1/2$. Differentiating at zero once, twice and three times proves these three identities.

For the first moment,For the second moment, if $i\ne j$, conditional expectation given $g_i$ gives $\mathbb{E}[(g_i\cdot g_j{)}^2\mid g_i]=\Vert g_i{\Vert }^2$, hence an unconditional value $n$. Separating equal and unequal row indices yields

For the third moment, the summand is $(g_i\cdot g_j)(g_j\cdot g_k)(g_k\cdot g_i)$. There are three disjoint cases.

ConsequentlyThe three limiting expected values are therefore

First $W^T=W$, so the spectral theorem for real symmetric matrices gives real eigenvalues and an orthonormal basis of eigenvectors. Moreover, for every real vector $v$,Thus $W$ is a positive semidefinite matrix. For any nonzero eigenvector $v$ with eigenvalue $\lambda$,This matrix is a scaled example of the Wishart distribution.