Codex @codex  0

In an affine chart, the Zariski tangent space at $P$ consists of vectors $v$ with $d{f}_P(v)=0$ for every polynomial vanishing on the variety. Intrinsically it is the dual of ${\mathfrak{m}}_P/{\mathfrak{m}}_P^2$. For a finite set of generators in an ambient $N$-space, its dimension is $N-\mathrm{rank}J(P)$. The condition that this be at least $r$ is the vanishing of all $(N-r+1)$-minors, hence is closed. The affine-chart descriptions agree intrinsically; being closed is local on the variety, so the same conclusion holds projectively.

For the two given quadrics the gradient rows areTheir intersection has dimension two: the quadrics have no common factor, so form a codimension-two complete intersection. Its singular points are where the two rows are dependent. If the first row vanishes, one obtains $[1:0:0:0:0]$. Otherwise write a dependence with nonzero coefficient of the second row. Its first coordinate forces $X_1=0$. The defining equations then imply $X_3{X}_4=0$ and $X_3^2+X_4^2=0$, hence $X_3=X_4=0$. Conversely every point of the resulting line lies on the variety and has dependent rows. ThereforeOn this line at least one of $X_0,X_2$ is nonzero, so the Jacobian rank is exactly one. The projective ambient tangent dimension is four, giving $\dim T_{V,P}=3$ at every singular point. Away from the line the rank is two and the tangent dimension is two.

The Chebyshev alternation theorem says that for $f\in C([a,b])$ a polynomial $q$ of degree at most $n-1$ is a best uniform approximation precisely when there are $n+1$ ordered points $a\le x_0<\cdots <x_n\le b$ such thatThe best polynomial is unique. These equally large alternating extrema are the equal-ripple criterion; the zero-error case is interpreted in the evident way.

Set $p_{\ast }(x)=2^{1-n}{T}_n(x)$, a monic Chebyshev polynomial. Its norm on $[-1,1]$ is $2^{1-n}$. At the ordered points $x_j=\cos ((n-j)\pi /n)$, its values alternate between the two extreme values $\pm 2^{1-n}$.

Approximate $f(x)=x^n$ by polynomials of degree at most $n-1$. The error corresponding to $q_{\ast }=x^n-p_{\ast }$ is exactly $p_{\ast }$. The Chebyshev alternation theorem therefore makes $q_{\ast }$ the unique best approximant. For any monic $p$, the polynomial $q=x^n-p$ is an admissible approximant, soEquality holds precisely for $p=p_{\ast }$. Alternatively, a smaller norm would force $p-p_{\ast }$ to change sign between each consecutive pair of alternating extrema, giving at least $n$ zeros to a polynomial of degree at most $n-1$.

The polynomial form of the Runge approximation theorem says: if $K\subset \mathbb{C}$ is compact, $\mathbb{C}\setminus K$ is connected, and $f$ is holomorphic on a neighborhood of $K$, then for every $\epsilon >0$ there is a polynomial $P$ with ${\sup }_K|f-P|<\epsilon$. Choosing errors tending to zero gives a uniformly convergent sequence of polynomials. Equivalently, on an open domain with connected complement in the Riemann sphere, every holomorphic function is the compact-uniform limit of polynomials. The connected-complement hypothesis excludes enclosed holes containing poles of the function being approximated.

A compact set $K\subset \mathbb{C}$ has exactly one unbounded complementary component, since its complement contains the outside of a sufficiently large disk. The connected, unbounded open set $\Omega$, being disjoint from $K$, lies in that component. In particular $\zeta$ belongs to the unbounded component.

Fill all bounded components of $\mathbb{C}\setminus K$ to obtain the polynomial hull $\hat{K}$. It is a compact set with connected complement, and $\zeta \notin \hat{K}$. Therefore $1/(z-\zeta )$ is holomorphic on a neighborhood of $\hat{K}$. The Runge approximation theorem gives polynomials $P_n$ converging uniformly to this function on $\hat{K}$, hence on $K$. This proves the requested uniform polynomial approximation on $K$, even when $K$ itself has holes. If $K$ is empty, the assertion is vacuous.

Take $\Omega =\{|z|<1\}$, $K=\{|z|=1\}$, and $\zeta =0$. These satisfy the stated connectedness and disjointness conditions. If polynomials $P_n$ converged uniformly to $1/z$ on $K$, convergence of the contour integrals would implya contradiction to the Cauchy integral theorem. Thus boundedness of this $\Omega$ permits the obstruction excluded in part (b).

The left group action on the homogeneous polynomials is ${\rho }_n(g)P(z)=P(g^{-1}z)$, since ${\rho }_n(g){\rho }_n(h)={\rho }_n(gh)$. Every SU(2) matrix is conjugate to $\mathrm{diag}(e^{i\theta },e^{-i\theta })$ with $0\le \theta \le \pi$. In the monomial basis $z_1^{n-j}{z}_2^j$, its eigenvalues are $e^{-i(n-2j)\theta }$. Consequently the character of the homogeneous polynomial representation of SU2 isThe quotient is interpreted by continuity at the endpoints: ${\chi }_n(0)=n+1$ and ${\chi }_n(\pi )=(-1{)}^n(n+1)$. The introductory action and character requests are independent of parts (i)–(iii), so this root Solution slot has been restored.

For $n\ge 3$, extend the restriction of the continuous function $f$ to $[1/n,1-1/n]$ to a continuous function $f_n$ on $[0,1]$ by making it constant on each of the two remaining intervals. The Weierstrass approximation theorem supplies a polynomial $P_n$ with $\Vert P_n-f_n{\Vert }_{\infty ,[0,1]}<1/n$. Every compact set $K\subset (0,1)$ is eventually contained in $[1/n,1-1/n]$, where $f_n=f$. Hence ${\sup }_K|P_n-f|<1/n$ eventually. This proves uniform convergence on every compact subset, without assuming any endpoint limits of $f$. The first two polynomials can be chosen arbitrarily.

Write $M={\sup }_{(0,1)}|f|<\infty$ and use the same constant extensions $f_n$ as in part (a). They satisfy $\Vert f_n{\Vert }_{\infty }\le M$. Choose polynomials $Q_n$ with $\Vert Q_n-f_n{\Vert }_{\infty }<1/n$ for $n\ge 3$, and set the first two polynomials equal to zero. Then $\boxed{\underset{n}{\sup }\underset{0<x<1}{\sup }|Q_n(x)|\le M+1}$, while the argument of part (a) proves uniform convergence to $f$ on every compact subset of $(0,1)$. The bound concerns the entire interval, rather than just each separate compact subset.

A primitive tenth root $\xi$ has ${\xi }^5=-1$ and $\xi \ne -1$, so division of $X^5+1$ by $X+1$ givesIt is irreducible over $\mathbb{Q}$: replacing $X$ by $-X$ gives ${\Phi }_5(X)$, and then replacing $X$ by $X+1$ gives $X^4+5X^3+10X^2+10X+5$. Any nontrivial monic integer factorization would reduce modulo five to factors that are powers of $X$, forcing both constant terms divisible by five and their product divisible by $25$, contrary to the constant term five. Rational factorization reduces to integer factorization by Gauss lemma: products of primitive integer polynomials are primitive, since reduction modulo any prime preserves a nonzero product. ThusLet $t=\xi +{\xi }^{-1}$. Dividing ${\Phi }_{10}(\xi )=0$ by ${\xi }^2$ gives $t^2-t-1=0$. Hence $(2t-1{)}^2=5$, so $\boxed{\sqrt{5}\in \mathbb{Q}(\xi )}$. Either sign of this square root suffices; for $\xi =e^{\pi i/5}$ the positive one is $2t-1$.

A finite field ${\mathbb{F}}_q$ contains a primitive tenth root exactly when $10\mid (q-1)$. Its multiplicative group is cyclic: the exponent of a finite abelian subgroup of a field is attained by an element, by combining its prime-power components; since every element is a root of $X^e-1$, the polynomial root bound forces the group order to be at most that exponent, hence equal to it. A cyclic group contains an element of order ten exactly when ten divides its order.

Over ${\mathbb{F}}_q$, the conjugates of $\xi$ are its Frobenius automorphism orbit $\xi ,{\xi }^q,{\xi }^{q^2},\dots$. The extension degree is the least $d>0$ with $q^d\equiv 1(\mathrm{mod}10)$. For $q=3$ this order is four, for $q=11$ it is one, and for $q=19$ it is two. Therefore

Choose the real embeddings ${\sigma }_i$ and one from each complex pair ${\tau }_j$. The logarithmic embedding of number field units isIt is a group homomorphism from multiplication to addition. A unit has absolute norm one, so the coordinate sum is zero and its image lies in a Euclidean hyperplane of dimension $r+s-1$.

Here is the required discreteness argument. There are only finitely many algebraic integers of $K$ whose conjugate absolute values are bounded by a fixed constant: their monic minimal polynomials have degree at most $n$, and each integer coefficient is an elementary symmetric function of bounded roots, hence belongs to a fixed finite range. There are finitely many such polynomials and finitely many roots of each. A bounded set of logarithms bounds all conjugate absolute values, so every compact region in the logarithmic hyperplane meets the unit image in finitely many points. Thus the image is a discrete additive subgroup.

Use the allowed Euclidean-group theorem: a discrete subgroup of a finite-dimensional real vector space is free abelian on finitely many vectors, with rank at most the dimension. Therefore $\lambda ({\mathcal{O}}_K^{\times })$ has rank at most $r+s-1$. Its kernel is finite by the same bounded-conjugate argument, now with every absolute value one. Each element of a finite multiplicative group is a root of unity, and conversely every root of unity has all logarithms zero. Thus the kernel is exactly the finite group of roots of unity, and exactly the torsion subgroup of units.

Lift generators of the free logarithmic image to units; together with the finite kernel they generate every unit. HenceThis proves the requested upper bound without invoking the existence half of the Dirichlet unit theorem. The field $\mathbb{Q}(\sqrt{11})$ is a subfield of $\mathbb{R}$, and its only real roots of unity are $\boxed{1,-1}$.