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Write a polynomial $f\in W$ by its coefficients. Evaluation of $f$ and of each first partial derivative at $p$ is polynomial in these coefficients and in the coordinates of $p$. Consequently the bad incidence locusis Zariski closed, and $U=(W\times k^3)\setminus B$ is Zariski open. Here a hypersurface is regarded as smooth at an ambient point outside it automatically.

In fact $B$ is empty. For homogeneous $f$ of degree $d>0$, the Euler homogeneous function theorem saysIf $(f,p)\in B$, its left side is zero while its right side is $d$, which is nonzero because $\mathrm{char}k=0$. Thus every nonzero level is covered by the smoothness of a nonzero homogeneous level set, andwhich is certainly nonempty and Zariski open.

For an affine variety $X\subseteq {\mathbb{A}}_k^n$ with vanishing ideal $I(X)$, its Zariski tangent space at $p$ isThe dimension from minimum tangent dimension is

Now let $X=Z(f)$ with $f$ irreducible. The Strong Hilbert Nullstellensatz gives $I(X)=(f)$, soIts dimension is at least $n-1$, and it is exactly $n-1$ wherever $\nabla f(p)\ne 0$.

It remains to find such a point in positive characteristic $q$. If every partial derivative of $f$ were the zero polynomial, every exponent occurring in $f$ would be divisible by $q$. Since the algebraically closed field $k$ is perfect, all coefficients have $q$th roots and $f$ would be a $q$th power, contradicting irreducibility. Thus some partial derivative $f_{x_i}$ is nonzero. If the gradient nevertheless vanished at every point of $Z(f)$, the Nullstellensatz would imply $f_{x_i}\in (f)$. This is impossible because $0\le \mathrm{deg}{f}_{x_i}<\mathrm{deg}f$. The smooth point on an irreducible hypersurface in positive characteristic therefore exists, and

Normalize the mutually orthogonal eigenvectors so that $(x_j)$ is an orthonormal sequence. If ${\lambda }_j$ did not tend to zero, some subsequence would satisfy $|{\lambda }_j|\ge \epsilon >0$. For distinct indices in this subsequence, orthogonality givesThus $(T{x}_j)$ would have no Cauchy subsequence, contradicting compactness of $T$. Therefore the orthogonal eigenvectors of a compact operator satisfy

Let $K=\overline{T(B_X)}$, which is compact. Every $y^{\ast }$ in the unit ball of the continuous dual space $Y^{\ast }$ restricts to a function on $K$. These functions are uniformly bounded and equicontinuous, sinceBy the Arzelà-Ascoli theorem, every sequence $(y_n^{\ast })$ in the unit ball has a subsequence that converges uniformly on $K$. Along this subsequence,The continuous dual space $X^{\ast }$ is complete, so $(T^{\ast }{y}_n^{\ast })$ converges in norm. Every sequence in the image of the unit ball therefore has a convergent subsequence, proving the forward direction of the Schauder theorem for compact operators:

A bounded linear operator $T:X\to Y$ is a compact operator when the image of the closed unit ball has compact closure, equivalently when every bounded sequence $(x_n)$ has a subsequence for which $(T{x}_n)$ converges.

Each $T_i$ is compact because its image lies in a finite-dimensional subspace. Let $B_X$ be the unit ball and fix $\epsilon >0$. Choose $i$ with $\Vert T-T_i\Vert <\epsilon /3$. Since $T_i(B_X)$ is relatively compact, it has a finite $\epsilon /3$-net $y_1,\dots ,y_N$. For each $x\in B_X$, choose $y_j$ with $\Vert T_{i}x-y_j\Vert <\epsilon /3$; thenThus $T(B_X)$ is totally bounded. Its closure is complete because $Y$ is a Banach space, so it is compact. This proves the norm-closedness of compact operators and hence

Triangulate $S^n$ and $S^m$. By the Simplicial approximation theorem, after subdividing the domain, a continuous map $f:S^n\to S^m$ is homotopic to the realization of a simplicial map $g$. Since $n<m$, the image of $|g|$ lies in the $n$-skeleton of the target triangulation and therefore misses an interior point of every $m$-simplex. A sphere with one point removed is homeomorphic to the Euclidean space ${\mathbb{R}}^m$, which is contractible. Hence $|g|$, and therefore $f$, is homotopic to a constant map. Thus

The Simplicial approximation theorem states that if $K$ is a finite simplicial complex, $L$ is a simplicial complex, and $f:|K|\to |L|$ is continuous, then for some $r\ge 0$ there is a simplicial approximation $g:K^{(r)}\to L$ to $f$.

The target open stars form an open cover of $|L|$, so their inverse images under $f$ form an open cover of the compact metric space $|K|$. The Lebesgue number lemma supplies $\delta >0$ such that every subset of diameter below $\delta$ lies in one inverse image. Choose $r$ so large that every vertex star in $K^{(r)}$ has diameter below $\delta$; this is possible because its diameter is at most twice $\mu (K^{(r)})$ and the mesh tends to zero.

For each vertex $v$ choose a vertex $w(v)$ of $L$ such thatIf $v_0,\dots ,v_q$ span a simplex, a point in its relative interior belongs to all their open stars. Its image belongs to all the open stars of $w(v_0),\dots ,w(v_q)$, so these target vertices span a simplex. Thus $v\mapsto w(v)$ extends to a simplicial map $g$, and its defining inclusion says exactly that $g$ approximates $f$. This proves the theorem.

A simplicial map $g:K\to L$ is a simplicial approximation to $f:|K|\to |L|$ whenfor every vertex $v$, where $\mathrm{st}$ denotes the open star in a simplicial complex.

Fix $x$ in a simplex with vertices $v_0,\dots ,v_q$, retaining only vertices with positive barycentric coordinate at $x$. Then $x\in {\mathrm{st}}_K(v_i)$ for every $i$, so $f(x)$ lies in every ${\mathrm{st}}_L(g(v_i))$. The vertices $g(v_i)$ and the vertices carrying the barycentric coordinates of $f(x)$ consequently span a common simplex of $L$. Both $f(x)$ and $|g|(x)$ lie in its convex realization. The straight-line homotopy from a simplicial approximationis therefore well-defined and continuous, with endpoints $f$ and $|g|$.

The mesh of a simplicial complex is the largest diameter of one of its realized simplices, or the supremum when $K$ is not finite:If $\dim K=n$, thenIterating gives

For every nonempty simplex $\sigma$ of $K$, let $b_{\sigma }$ be its barycentre. The first barycentric subdivision $K^{\prime }$ has the $b_{\sigma }$ as vertices, andis a simplex exactly when, after reordering, ${\sigma }_0⊊\cdots ⊊{\sigma }_q$. Its realization is the same polyhedron as that of $K$. The iterated barycentric subdivision is defined by

Put $L=\mathbb{Q}(\omega )$ and $F=\mathbb{Q}(\omega +{\omega }^{-1})$. The maximal real subfield of the fifth cyclotomic field is $F=\mathbb{Q}(\sqrt{5})$. The roots of unity in a rational cyclotomic field show that the roots of unity in $L$ are exactly $\{\pm {\omega }^a\}$.

Let $u\in {\mathcal{O}}_L^{\times }$. For every embedding $\sigma :L\to \mathbb{C}$,Thus the kernel of the logarithmic unit embedding implies that $u/\bar{u}$ is a root of unity. Since conjugation acts trivially modulo the prime element $1-\omega$, we have $u/\bar{u}\equiv 1(\mathrm{mod}1-\omega )$. Among the tenth roots of unity this excludes the elements $-{\omega }^a$, so $u/\bar{u}\in ⟨\omega ⟩$. Squaring is a bijection on this group of order five, so choose a root of unity $\eta$ with $\eta /\bar{\eta }=u/\bar{u}$. Then $v=u/\eta$ is fixed by conjugation and is therefore a unit of $F$.

The units of the quadratic field Q square root of five are $\pm {\varphi }^b$, where $\varphi =(1+\sqrt{5})/2$. Absorbing the sign and $\eta$ into $\pm {\omega }^a$ proves

The conjugates of $\omega$ are $\omega ,ldots,{\omega }^{p-1}$. Fromwe obtain, at every nontrivial $p$th root $\alpha$,The product of the roots is one and ${\prod }_{a=1}^{p-1}(1-{\omega }^a)={\Phi }_p(1)=p$. The derivative formula for a polynomial discriminant therefore yields the discriminant of a prime cyclotomic power basis:

Since $p\equiv 1(\mathrm{mod}4)$, the sign is positive. The Vandermonde determinantbelongs to $L$ and satisfies ${\Delta }^2=p^{p-2}$. Hence

The cyclotomic polynomial of the primitive $p$th root $\omega$ isBy the irreducibility of cyclotomic polynomials, this is the minimal polynomial of $\omega$ over $\mathbb{Q}$. The prime cyclotomic field degree therefore gives

The normis a positive integer, because it is the sum of the squares of the multiplicities of the irreducible constituents. The character restriction norm bound and $[G:K]=3$ make it at most three. It is therefore $1$, $2$, or $3$.

All three values occur:

By character orthogonality, the squared multiplicities are the character inner product norm of the restriction of a character:Since $\chi$ is an irreducible character, its norm on $G$ is one, and henceThe omitted summands are nonnegative, so the character restriction norm bound is an equality precisely when

Quantum no-signalling says that, without communicating a measurement outcome, any local trace-preserving operation performed on subsystem $A$ leaves all outcome probabilities for measurements on subsystem $B$ unchanged; equivalently, it leaves the reduced density matrix ${\rho }_B$ unchanged.

The two Bell states are orthogonal. Applying the inverse Bell-state preparation circuit, namely a Controlled-NOT gate from $A$ to $B$ followed by a Hadamard gate on $A$, givesA computational-basis measurement therefore distinguishes them perfectly, soIf only $B$ is accessible, however,Every measurement on $B$ consequently has the same outcome distribution under both hypotheses. With equal prior probabilities no information is available, andthe error probability of a random guess.

Let $|\alpha ⟩=a|0⟩+b|1⟩$ be Alice's unknown qubit, and label her half of the shared Bell state by $A$ and Bob's by $B$. Alice applies a Controlled-NOT gate from the unknown qubit to $A$, then a Hadamard gate to the unknown qubit. The three-qubit state becomesAlice measures her two qubits in the computational basis and sends the two classical outcomes $(j,k)$ to Bob. Bob applies $Z^j{X}^k$, recovering $|\alpha ⟩$ with certainty. This is quantum teleportation; Alice never needs to know the state.

For the shared GHZ state, Charlie first applies a Hadamard gate to his qubit. ThenCharlie measures in the computational basis. Outcome $0$ leaves Alice and Bob with $|{\varphi }^{+}⟩$; outcome $1$ leaves $|{\varphi }^{-}⟩$, which either party converts to $|{\varphi }^{+}⟩$ by applying $Z$ after Charlie reports the outcome. Thus