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The Vandermonde determinant isIndeed, the determinant is an alternating polynomial, hence divisible by every $x_j-x_i$. Both sides have degree $r(r-1)/2$, and comparison of the coefficient of $x_0^0{x}_1^1\cdots x_{r-1}^{r-1}$ gives constant one. This is an identity over the integers, so reducing its coefficients modulo a prime number preserves it. For $r>p$, two residues coincide and the determinant vanishes. For $r=p$, choose all elements of ${\mathbb{F}}_p$ once to obtain a nonzero determinant.

For the ordering $0,1,\dots ,p-1$, this determinant is $D={\prod }_{j=1}^{p-1}j!$. If $p$ is odd, put $h=(p-1)/2$. Wilson theorem gives $j!(p-1-j)!=(-1{)}^{j+1}$. Pairing the factors with indices $j=0,\dots ,h-1$ and $p-1-j$, and leaving $h!$ unpaired, givesEvery distinct ordering changes the determinant by its permutation sign; repeated residues give zero. Thus the possible values are $0,\pm h!$ for odd $p$, and $0,1$ for $p=2$.

Use Shamir's secret sharing: choose independent uniform $a_1,\dots ,a_{r-1}\in {\mathbb{F}}_p$, and engrave the shares $(x_i,P(x_i))$, where $P(X)=N+{\sum }_{j=1}^{r-1}{a}_j{X}^j$ and the $x_i$ are distinct nonzero field elements. We may take $p>n$; otherwise use a sufficiently large finite field extension and an agreed integer encoding. Any $r$ shares determine $P$ by polynomial interpolation and hence $N=P(0)$. Given any $r-1$ shares and any proposed $N$, their equations in the coefficients have a unique solution: the coefficient matrix has determinant ${\prod }_i{x}_i$ times a nonzero Vandermonde determinant. Since the coefficients were uniform, the observed shares have the same distribution for every $N$. Thus they give no information about the secret.

For the fake, reduce its integers to field elements and first check that its abscissa is distinct and nonzero. Random collisions or a zero abscissa are immediately detectable and have probability $O(n/p)$; the following conclusions concern the overwhelmingly likely distinct-abscissa case.

With $r$ shares, there is always a unique degree-at-most-$r-1$ interpolating polynomial. If all shares are genuine it gives the correct secret. If one is fake, it normally gives the wrong secret, with probability $1-1/p$ conditional on a uniform fake ordinate, but the inconsistency is undetectable from those $r$ shares. The data do not certify which case has occurred.

With $r+1$ shares, genuine shares are consistent with one such polynomial. A fake is inconsistent with probability $1-1/p$, so it is usually detected as an inconsistency. However, omitting each of the $r+1$ shares in turn gives $r+1$ possible polynomials. If the data are inconsistent, their constant terms are distinct: two candidate polynomials with the same constant term agree at zero and at their $r-1$ common abscissae, hence agree identically, which would make all the data consistent. The secret is narrowed to $r+1$ possibilities, but the fake is not identified.

With $r+2$ shares, at least $r+1$ are genuine. There is a unique degree-at-most-$r-1$ polynomial agreeing with at least $r+1$ shares: two candidates would agree on at least $r$ abscissae and hence coincide. Find it by deleting each share and checking consistency. The secret is recovered uniquely, and the fake is identified unless it happens to lie on the genuine polynomial, in which case it is harmless.

For the first field, set $K=\mathbb{Q}(\sqrt{5},i)$. Its field extension degree is four, since $\mathbb{Q}(\sqrt{5})$ is real and adjoining $i$ doubles its degree. The polynomial $X^3-5$ is irreducible over $K$: a reducible cubic would have a root in $K$, but every root has degree three over $\mathbb{Q}$ by Eisenstein criterion, impossible by the tower law because $3\nmid 4$. Thus

For the second field, put $\alpha =\sqrt[4]{5}>0$. The splitting field is $F=\mathbb{Q}(\alpha ,i)$ and has degree eight, since $X^4-5$ is Eisenstein and $\mathbb{Q}(\alpha )$ is real. Define automorphisms $r(\alpha )=i\alpha ,r(i)=i$ and $s(\alpha )=\alpha ,s(i)=-i$. They satisfy $r^4=s^2=1$ and $srs=r^{-1}$, and give eight distinct automorphisms. Hence $\mathrm{Gal}(F/\mathbb{Q})\cong D_8$.

The Galois correspondence gives all the subfields in the following table. The middle column records generators, not merely their degrees.Each displayed generator is fixed by the indicated subgroup. The two last quartic generators have fourth power $-20$, so their degree is four by Eisenstein criterion; the other quartic degrees are immediate. These are all the subgroups of $D_8$ and hence all the subextensions.

Put $h=(p-1)/2$. The nonzero quadratic residues in the finite field ${\mathbb{F}}_p$ form a subgroup of size $h$: the group homomorphism $x\mapsto x^2$ has kernel $\{1,-1\}$. Each square satisfies $u^h=1$ by Fermat's little theorem. Since a nonzero polynomial of degree $h$ has at most $h$ roots of a polynomial, these are all the roots of $X^h-1$. For any other nonzero $u$, Fermat's theorem gives $(u^h{)}^2=1$, so $u^h=-1$. Thus Euler's criterion follows, and multiplication givesBoth sides are $1$ or $-1$, which are distinct modulo the odd prime number $p$, so $\chi (mn)=\chi (m)\chi (n)$. Taking $n=-1$ in Euler's criterion gives $\chi (-1)=(-1{)}^{(p-1)/2}$, which is $1$ for $p\equiv 1(\mathrm{mod}4)$ and $-1$ for $p\equiv 3(\mathrm{mod}4)$.

We use algebraic dimension and tangent spaces, equivalently after extending scalars to an algebraic closure; these are not dimensions of the finite set of rational points when $k$ is finite. The condition rank at most $r$ is equivalent to the vanishing of all $(r+1)\times (r+1)$ matrix minors. These are polynomials in the $nm$ entries, so define an affine determinantal variety $X^{n,m,r}$. The same equations make it a Zariski closed subset of $X^{n,m,r+1}$.

For $1\le r<\min (n,m)$, every line $t{E}_{ij}$ lies in $X$ because its matrices have rank at most one. Its tangent direction $E_{ij}$ therefore belongs to the Zariski tangent space at zero. The matrix units span all $nm$ directions, so $T_{0}X=k^{nm}$. Meanwhile a rank-$r$ chart with an invertible $r\times r$ block has the formThe free blocks have $r^2+r(m-r)+(n-r)r=r(n+m-r)$ entries. Rank-$r$ matrices are dense in $X$: the factorization $UV$ parametrizes all rank-at-most-$r$ matrices, and the full-rank factor pairs form a dense open subset. This also proves irreducibility, so the local dimension at zero equals the dimension of $X$. Since $r(n+m-r)<nm$, zero is singular when $1\le r<\min (n,m)$.

There is a necessary qualification to the printed claim: for $r=0$, $X=\{0\}$ is a smooth point, with zero-dimensional tangent space. Thus the claim is false if zero is allowed as a rank bound.

Finally, for $n=5,m=2,r=1$, the formula gives $\boxed{\dim X^{5,2,1}=6}$. Concretely, on a chart where an entry of the first column is nonzero, choose that column freely and make the second an arbitrary scalar multiple of it, giving five plus one parameters.

The Weierstrass approximation theorem says that for every real continuous function $f$ on $[0,1]$ and every $\epsilon >0$, there is a real polynomial $P$ with ${\sup }_{[0,1]}|f-P|<\epsilon$. Equivalently, polynomials are dense in $C[0,1]$ in the uniform norm.

This also forces $f=0$. Indeed, $h(t)=f(\sqrt{t})$ is a continuous function on $[0,1]$. Approximate $h$ uniformly by polynomials $P_k(t)$ using the Weierstrass approximation theorem. Then $P_k(x^2)\to f(x)$ uniformly, and the even moments give $\int f(x)P_k(x^2)dx=0$. The estimate from part (i), with $P_k(x^2)$ in place of $P_k(x)$, proves $\int f^2=0$ and hence the conclusion.

The given moments imply ${\int }_0^{1}f(x)P(x)dx=0$ for every polynomial $P$. Choose polynomials $P_k$ converging uniformly to $f$ by the Weierstrass approximation theorem. ThenA nonnegative continuous function with zero integral vanishes everywhere. Hence $f=0$.

We use the standard facts that finite complex representations are completely reducible, irreducible characters are orthonormal, the regular character is ${\sum }_{\chi }\chi (1)\chi$, and the number of irreducible characters equals the number of conjugacy classes. We also use Schur's lemma, Sylow's theorem and the nontrivial center of a nontrivial finite $p$-group.

For an irreducible character $\chi$ of degree $d$, a conjugacy-class sum $z_C$ acts as the scalar ${\lambda }_C=|C|\chi (g)/d$ by Schur's lemma and taking traces. This scalar is an algebraic integer: class sums form an integral basis of the center of the integral group algebra, their multiplication has integer structure constants, and multiplication by $z_C$ satisfies its monic integer characteristic polynomial. That same polynomial annihilates its image in any irreducible representation. Character values are also algebraic integers, since they are sums of roots of unity. Character orthogonality now givesThis is a rational algebraic integer, hence an integer. Therefore $\chi (1)$ divides $|G|$.

The Burnside's pa qb theorem states that every finite group of order $p^a{q}^b$ is soluble. We prove its character-theoretic ingredient. If a class has size $m$ coprime to $d$, integers $u,v$ with $ud+vm=1$ givean algebraic integer. Every algebraic conjugate of this number is the average of $d$ roots of unity and has modulus at most one. If it is nonzero, its algebraic norm is a nonzero integer of absolute value at least one; since every factor has modulus at most one, all must have modulus one. In particular $|\chi (g)|=d$. Equality in the triangle inequality for the unit-modulus eigenvalues implies that $\rho (g)$ is scalar. Thus if $(m,d)=1$, either $\chi (g)=0$ or $\rho (g)$ is scalar.

Suppose a group of order $p^a{q}^b$ were nonabelian simple, with both primes present. Choose a nonidentity $g$ in the center of a Sylow $p$-subgroup. Its centralizer contains that Sylow subgroup, so its class size is a power of $q$. Every nontrivial irreducible representation is faithful by simplicity. If it sent $g$ to a scalar, faithfulness would make $g$ central in the group, impossible in a nonabelian simple group. The ingredient above therefore gives $\chi (g)=0$ for every nontrivial character with degree coprime to $q$. Evaluating the regular character at $g\ne 1$ givesThe right side is an algebraic integer, but $1/q$ is not, a contradiction. A group of prime-power order cannot be nonabelian simple either, by its nontrivial center. Induct on the order: every nonsimple group has a proper nontrivial normal subgroup, and that subgroup and the quotient have smaller orders with at most these two prime divisors. By induction both are soluble; their derived-series extensions make the original group soluble. Simple abelian groups have prime order and start the induction. This proves Burnside's theorem.

For the final divisibility, let $|G|=n$ be odd and let $k$ be its number of conjugacy classes. All irreducible degrees are odd by the divisibility just proved. We use the standard Frobenius-Schur indicator theorem: $\nu (\chi )=n^{-1}{\sum }_g\chi (g^2)$ is zero precisely for an irreducible character not equal to its conjugate, and is $\pm 1$ otherwise. Squaring is a bijection on an odd-order group: for each $g$, its unique square root is $g^{(n+1)/2}$, since every element order divides $n$. For every nontrivial irreducible character, orthogonality to the trivial character consequently gives $\nu (\chi )=n^{-1}{\sum }_g\chi (g)=0$. The nontrivial characters therefore occur in distinct conjugate pairs of equal odd degree. Since an odd square is one modulo eight,is a sum of pairs $2(d^2-1)$, each divisible by sixteen; the trivial character contributes zero. Hence

The real Stone-Weierstrass theorem states that if $X$ is compact Hausdorff and $A\subseteq C(X,\mathbb{R})$ is a subalgebra containing the constants and separating points, then $A$ is uniformly dense in $C(X,\mathbb{R})$. Let $B$ be its uniform closure, still an algebra over a field. If $h\in B$ and $M\ge \Vert h{\Vert }_{\infty }$, scale the permitted polynomial approximation of absolute value on $[-1,1]$ to $[-M,M]$. Applying these polynomials to $h$ proves $|h|\in B$. Thereforebelong to $B$: it is a lattice.

Fix $f\in C(X,\mathbb{R})$ and $\epsilon >0$. For every $x,y\in X$, point separation and an affine adjustment give $g_{x,y}\in A$ with $g_{x,y}(x)=f(x)$ and $g_{x,y}(y)=f(y)$; if $x=y$, a constant suffices. For fixed $x$, the open sets where $g_{x,y}>f-\epsilon$ cover $X$. Select finitely many and take their maximum $h_x\in B$. Then $h_x>f-\epsilon$ everywhere and $h_x(x)=f(x)$. By continuity, $h_x<f+\epsilon$ on some neighborhood $V_x$ of $x$. Choose finitely many $V_x$ covering $X$ and take the minimum of the corresponding $h_x$. The result $h\in B$ satisfies $f-\epsilon <h<f+\epsilon$ everywhere. Since $B$ is closed and $\epsilon$ arbitrary, $f\in B$. Thus $\overline{A}=C(X,\mathbb{R})$, proving the theorem.

A meromorphic function on the Riemann sphere is rational: it has finitely many poles; subtract their principal parts, including a polynomial for a possible pole at infinity, and the remaining entire function is bounded at infinity, hence constant by Liouville theorem. A rational map of degree $d$ has $d$ preimages of a regular value, counted with multiplicity. An analytic isomorphism is bijective, so $d=1$, giving preciselyConversely these Möbius transformations have a Möbius inverse and extend holomorphically at their pole and infinity, so are analytic isomorphisms.

For a nonconstant analytic map $f:X\to Y$ between compact connected Riemann surfaces, the Riemann-Hurwitz formula isHere $g_X,g_Y$ are genera, $d$ is the degree (the sum of local multiplicities over any fiber), and $e_p$ is the local degree in coordinates where the map has the form $z\mapsto z^{e_p}$. Only finitely many terms are nonzero; these are the ramification points.

For a degree-two sphere map the formula gives total ramification two. Each local degree is at most two, so there are exactly two distinct ramification points, each of local degree two. Their branch values are distinct, because one fiber cannot have multiplicity four. Choose $T$ sending $0,\infty$ to the two ramification points, and $S$ sending their values to $0,\infty$. Then $SfT$ has its only zero at zero, of order two, and its only pole at infinity, of order two; it is $c{z}^2$ with $c\ne 0$. Rescale $S$ by $1/c$. Therefore $SfT(z)=z^2$, as required.

If the polynomials converge uniformly on the full annulus to $g$, their limit is continuous and equals $f$ on the slit annulus by part (ii). That subset is dense in the full annulus, so $g=f$ everywhere. The Cauchy integral theorem and uniform convergence on the unit circle then give

Choose the argument in $(0,2\pi )$ on the slit annulus and letThese compact sets increase to the slit annulus, and each has connected complement: the omitted angular corridor joins the inner disk to the outside of the outer disk. The polynomial Runge theorem says that a function holomorphic near a compact set with connected complement can be approximated uniformly there by polynomials. Choose $p_j$ with ${\sup }_{K_j}|p_j-f|<1/j$. Every compact subset $K$ of the slit annulus has positive distance from the removed positive ray, so $K\subseteq K_j$ for all sufficiently large $j$. Hence $p_j\to f$ uniformly on every such compact subset.

Take the holomorphic function $f(z)=1/z$ on a neighborhood of the annulus. Every polynomial has integral zero around the unit circle by the Cauchy integral theorem, while ${\int }_{|z|=1}{z}^{-1}dz=2\pi i$. Uniform convergence on the annulus would imply convergence of these contour integrals, since their difference is bounded by $2\pi$ times the uniform error. Thus uniform polynomial approximation on the full annulus need not exist.

The previous permutation calculation shows that every automorphism either preserves $x^3,y^3$ separately or interchanges them. Their sum and product are therefore fixed by the Galois group and belong to $K$, proving the quadratic assertion.

For the depressed cubic, $\alpha +\beta +\gamma =0$, $\alpha \beta +\alpha \gamma +\beta \gamma =b$, and $\alpha \beta \gamma =-c$. Using $1+\zeta +{\zeta }^2=0$ givesand henceAlso $x^3{y}^3=-27b^3$. The desired polynomial isIts discriminant is $729c^2+108b^3=-27D$, where $D=-4b^3-27c^2$. Here $-27$ is a square in $K$, since $[3(\zeta -{\zeta }^2){]}^2=-27$.

One can justify the group criterion directly even when one resolvent vanishes. The nonzero root-difference producthas $d^2=D$ and changes sign exactly under odd root permutations. It is fixed by the entire group exactly when the group is contained in $A_3$. If $D$ is a square in $K$, then $d$ equals one of its two square roots in $K$; conversely, a fixed $d$ lies in $K$. ThereforeSeparability ensures $D\ne 0$, and the characteristic hypotheses ensure that all divisions and the sign argument are valid.