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The circle group is $U(1)=\{z\in \mathbb{C}:|z|=1\}$ under multiplication. For continuous finite-dimensional complex representations, its irreducibles are the one-dimensional charactersA group representation is a continuous homomorphism to the invertible linear maps; it is irreducible when it has no proper nonzero invariant subspace. Compactness permits an invariant Hermitian inner product by averaging. Since the resulting unitary operators commute for $U(1)$, simultaneous diagonalization reduces an irreducible to a character, and the continuous characters of the circle have the stated integer winding number.

The group $SU(2)$ consists of unitary two-by-two complex matrices of determinant one. Each has the unique formidentifying it homeomorphically with the unit 3-sphere in ${\mathbb{R}}^4$. This is the Spin group $\mathrm{Spin}(3)$. A unitary matrix is conjugate to $\mathrm{diag}(e^{i\theta },e^{-i\theta })$, and its conjugacy class is determined by $0\le \theta \le \pi$, or equivalently its trace $2\cos \theta$.

The irreducible representations of SU2 are $V_n={\mathrm{Sym}}^n({\mathbb{C}}^2)$ for integers $n\ge 0$, with dimension $n+1$. The symmetric power can be realized as homogeneous degree-$n$ polynomials in two variables, with the induced linear action. On the diagonal torus its eigenvalues are $e^{i(n-2j)\theta }$ for $j=0,\dots ,n$, so its character, the trace of the representing operator, isAt the endpoints use the continuous values ${\chi }_n(0)=n+1$ and ${\chi }_n(\pi )=(-1{)}^n(n+1)$. The highest-weight classification says this list is complete.

For the final action, multiplication of block matrices gives $(AB{)}_1=A_1{B}_1$, so conjugation is a linear group representation. Decompose the underlying three-dimensional space as $V_1\oplus V_0$. Its endomorphism space isThe two off-diagonal blocks are each $V_1$, since the invariant alternating form identifies $V_1^{\ast }$ with $V_1$. The scalar part of $\mathrm{End}(V_1)$ is $V_0$ and its traceless part is $V_2$: the same alternating form identifies traceless endomorphisms with symmetric tensors of degree two. The lower scalar block is a second $V_0$. ThusThe dimensions add to $2+4+3=9$, and its character is $2{\chi }_0+2{\chi }_1+{\chi }_2$.

A rational root of the monic cubic would be one of $\pm 1,\pm 3$, and substitution excludes all four. A reducible cubic over $\mathbb{Q}$ would have a rational root, so the polynomial is irreducible andIts polynomial discriminant is $-4(-1{)}^3-27(3{)}^2=-239$. The integer 239 is prime: it has no prime divisor among $2,3,5,7,11,13$, the primes not exceeding its square root.

The discriminant-index formula for an integral lattice for an integral primitive element isBoth discriminants are integers, and the left side is $-239$. Its squarefreeness forces the positive index to be one. Since $\alpha$ is an algebraic integer by its monic equation,

A rational function with no poles away from $P_{\infty }$ is regular on the smooth affine curve. Its ring of regular functions is the coordinate ringso it has a unique expression $A(x)+yB(x)$ with polynomials $A,B$. This regular-function assertion uses smoothness, hence normality, of the affine curve, not Riemann-Roch theorem.

The nonconstant leading term of $A$ has even pole order $2\mathrm{deg}A$ at infinity, while $yB$ has odd pole order $3+2\mathrm{deg}B$. These different parities prevent cancellation. Thus no nonconstant such function has a pole of order at most one. Therefore

The fourth Chebyshev polynomial is $T_4(x)=8x^4-8x^2+1$. Therefore takeIndeed $x^4-p(x)=T_4(x)/8$, and $|T_4(x)|\le 1$ on $[-1,1]$, as follows from $T_4(\cos \theta )=\cos 4\theta$.

Here is a direct Chebyshev alternation theorem argument proving optimality, without assuming the theorem. At the five ordered pointsthe error $x^4-p(x)$ takes the alternating values $1/8,-1/8,1/8,-1/8,1/8$. If a polynomial $q$ of degree at most three had strictly smaller uniform error, then $q-p$ would have alternating strictly positive and strictly negative values at these points. The intermediate value theorem would give a distinct root in each of the four intervening intervals. A nonzero polynomial of degree at most three cannot have four distinct roots. This contradiction proves the required minimax inequality.

For an explicit Runge-Kutta method, write its strictly lower-triangular stage matrix as $A$, weight vector as $b$, and $1=(1,\dots ,1{)}^T$. Applied to the scalar test equation, its stage vector satisfiesSince $A^s=0$, inversion is a finite Neumann series. Putting $z=\lambda h$ givesThis proves degree at most $s$. The printed statement of degree exactly $s$ needs a nondegeneracy assumption: a two-stage explicit scheme with $b=(1,0)$ is simply Euler's method with an unused second stage, and has polynomial $1+z$ of degree one.

Order $p$ applied to this particular differential equation forces the local stability polynomial to agree with the exact exponential through degree $p$:As its degree is at most $s$, the nonzero coefficient $1/p!$ cannot be present for $p>s$. ThereforeThe implication is a necessary stability-polynomial condition; it does not assert that every such polynomial automatically gives the full nonlinear order conditions.

An A-stable method must satisfy $|P_s(z)|\le 1$ for every $z$ with nonpositive real part. A consistent explicit Runge-Kutta method has $P_s^{\prime }(0)=1$, so its stability polynomial is nonconstant. Along the negative real axis the leading term of any nonconstant polynomial has magnitude tending to infinity. HenceThe positive order assumption in the question supplies the required consistency.

Substitution of $y_n={\lambda }^n$ into the linear recurrence relation gives ${\lambda }^{n}p(\lambda )=0$. Since $\lambda \ne 0$, division by ${\lambda }^n$ proves the sequence is a solution exactly when $p(\lambda )=0$.

For a nonzero triple root of a polynomial $\mu$, the repeated characteristic root of a linear recurrence rule gives the three independent solutions ${\mu }^n,n{\mu }^n,n^2{\mu }^n$. One way to see the polynomial factors is to write $y_n={\mu }^n{z}_n$: the recurrence becomes the third forward difference operator applied to $z_n$ equal to zero, so $z_n$ is a quadratic polynomial. ThusThese three constants give arbitrary initial values $y_0,y_1,y_2$, hence the full third-order solution. If $\mu =0$, the recurrence instead reduces to $y_{n+3}=0$: the first three values are arbitrary and every subsequent value is zero. The nonzero-root formula must not be used to discard those initial values.

For each integer $n\ge 0$, put $Z_n=\{z:f^{(n)}(z)=0\}$. Every derivative $f^{(n)}$ is an entire function and cannot be identically zero: otherwise repeated integration would make $f$ a polynomial. By the identity theorem for holomorphic functions, its zero set has empty interior, and continuity makes $Z_n$ a closed set. Thus $Z_n$ is nowhere dense.

The complex plane with its usual metric is a complete metric space, so the Baire category theorem gives $z_0\notin {\bigcup }_{n\ge 0}{Z}_n$. The Taylor series there has coefficients

Set $p=t-\lambda r$ using part (a). At every node,If $\lambda =0$, the error vanishes. Otherwise any degree-at-most-$n$ polynomial $q$ with strictly smaller error would make $q-p$ have alternating nonzero signs at the increasingly ordered nodes (reversed globally if $\lambda <0$). It would then have at least $n+1$ distinct roots by the intermediate value theorem, impossible. Hence

Let $E=\Vert f-p{\Vert }_{\infty }$. If $E=0$, minimality is immediate. Otherwise suppose a polynomial $q$ of degree at most $n$ has strictly smaller error. At each alternation point,Thus $q-p$ changes sign between every pair of consecutive $x_j$. By the intermediate value theorem it has at least $n+1$ distinct roots of a polynomial, contradicting its degree bound. ThereforeThis is the sufficiency direction of the Chebyshev alternation theorem.

Lagrange theorem for polynomial congruences says that a nonzero polynomial of degree $d$ over a field has at most $d$ distinct roots of a polynomial. It follows by repeatedly factoring out $x-a$ at a root. Over the finite field ${\mathbb{F}}_p$, the only solutions of $x^2=1$ are $1$ and $-1$: these are distinct because $p$ is odd, and the root bound excludes any others.

Every nonzero element has a unique multiplicative inverse. Pair each element of ${\mathbb{F}}_p^{\times }$ with its inverse. All pairs of distinct elements contribute $1$ to the product; the only unpaired elements are $1$ and $-1$. Thus Wilson theorem follows:If Lagrange's theorem is formulated for finite groups, it states that the order of a subgroup divides the order of the group; in particular every element of ${\mathbb{F}}_p^{\times }$ has order dividing $p-1$. The inverse-pairing proof above establishes the required congruence directly.

Yes. Take exterior poles ${\zeta }_n=1+1/n$. By part (i), choose a polynomial $p_n$ withFor any fixed $z\in \Gamma$, $z\ne 1$, and therefore $(z-{\zeta }_n{)}^{-1}\to (z-1{)}^{-1}$. The triangle inequality provesThis is pointwise convergence. It is not uniform convergence on $\Gamma$: every polynomial is bounded on the closed diamond, whereas $(z-1{)}^{-1}$ is unbounded as $z\to 1$ along $\Gamma$.

Let $K=\{z:|\mathrm{Re}z|+|\mathrm{Im}z|\le 1\}$. Choose a unit complex number $u$ normal to a supporting side of this convex set, withSuch a choice exists explicitly: take $u=(s+it)/\sqrt{2}$, where $s,t\in \{-1,1\}$ are the signs of the real and imaginary parts of $\zeta$; then $h=1/\sqrt{2}$ and the strict inequality is exactly the hypothesis. Set $c=-Ru$. Since $|z|\le 1$ on $K$,For sufficiently large $R$, the former is strictly smaller than the latter uniformly on $K$. Consequently $q={\max }_K|z-c|/|\zeta -c|<1$.

The geometric series now supplies actual approximating polynomials:Choose $N$ so the right-hand side is below $ϵ$. This proves the required uniform approximation without assuming a general approximation theorem.

For $P(x)=x^3+4x-7$, the derivative $P^{\prime }(x)=3x^2+4$ is positive everywhere, so the mean value theorem makes $P$ strictly increasing. Since a polynomial is a continuous function, the intermediate value theorem and the values $P(1)=-2$, $P(2)=9$ locate its unique real root in $(1,2)$. By the rational root theorem, a rational root of a monic integer polynomial must be an integer. There is no integer in $(1,2)$, so the unique real root is an irrational number. The denominator part of that theorem also follows directly: if $a/b$ in lowest terms were a root, $a^3+4a{b}^2-7b^3=0$ would force $b\mid a^3$, hence $b=1$.

Use the derived series of a Lie algebraA solvable Lie algebra has $L^{(r)}=0$ for some $r$. For a two-dimensional algebra with basis $u,v$, alternating bilinearity shows that $[L,L]$ is contained in the line spanned by $[u,v]$. A one-dimensional Lie algebra is abelian, so $[[L,L],[L,L]]=0$. Every two-dimensional complex Lie algebra is solvable, with derived length at most two. This dimension argument actually works over any field with the alternating definition of a Lie bracket.

There is a genuine qualification in the last request: the assertion about all Irreducible Lie algebra representations is true for finite-dimensional representations. Without that restriction, the infinite-dimensional simple module for the two-dimensional affine Lie algebra is a counterexample. Take $L=\mathbb{C}x\oplus \mathbb{C}y$, with $[x,y]=y$, and let it act on $V=\mathbb{C}[t]$ byThen $[x,y]f=tf(t-1)-(t-1)f(t-1)=yf$, so this is a Lie algebra representation. Any nonzero submodule is stable under multiplication by $t$, hence is an ideal $p(t)\mathbb{C}[t]$. Stability under $y$ implies $p(t)\mid p(t-1)$. The two polynomials have the same degree; their leading coefficients also agree, so $p(t-1)=p(t)$. In characteristic zero this forces $p$ to be constant. Thus $V$ is infinite-dimensional and irreducible, while $L$ is solvable. The literal unrestricted assertion is false.

Here is a proof of the intended finite-dimensional assertion, including the essential Lie theorem argument. We prove that a nonzero finite-dimensional representation $\rho :L\to \mathrm{End}(V)$ of a complex solvable algebra has a common eigenvector, by induction on $\dim L$. If $L=0$, any nonzero vector works. Otherwise $[L,L]\ne L$; choose a hyperplane $K$ containing $[L,L]$. It is a solvable ideal, and $L=K\oplus \mathbb{C}x$. Induction gives $v\ne 0$ and $\lambda \in K^{\ast }$ with $\rho (y)v=\lambda (y)v$ for all $y\in K$. Write $X=\rho (x)$ and suppress $\rho$ on elements of $K$.

Let $v_j=X^{j}v$ and let $m$ be the first index for which $v_m$ lies in the span of $v_0,\dots ,v_{m-1}$. The space $W=\mathrm{span}(v_0,\dots ,v_{m-1})$ is $X$-invariant. Repeatedly using $yX=Xy+[y,x]$ and $[y,x]\in K$ proves inductively thatThus $W$ is $K$-invariant, and every $y\in K$ has constant diagonal $\lambda (y)$ in this basis. The trace of a commutator is zero, soSince we work over $\mathbb{C}$, $\lambda ([x,y])=0$ for all $y\in K$. Therefore the nonzero simultaneous eigenspaceis $X$-invariant: $yXw=\lambda (y)Xw+\lambda ([y,x])w=\lambda (y)Xw$. A complex linear operator on a nonzero finite-dimensional space has an eigenvector. An eigenvector of $X{|}_{V_{\lambda }}$ is consequently a common eigenvector for $L$.

Its span is a one-dimensional submodule. If $V$ is irreducible, that span must be all of $V$. Hence every finite-dimensional irreducible representation of a finite-dimensional complex solvable Lie algebra has dimension one. Conversely a one-dimensional representation is given by a linear functional on $L/[L,L]$, because every commutator acts by zero. Passing repeatedly to quotients also gives the simultaneous triangularization of a Lie algebra representation. Neither the eigenvector step nor the trace argument extends to the infinite-dimensional counterexample above.

Take the three-dimensional affine quadric coneThe polynomial $xy-zw$ is an irreducible polynomial: as a polynomial in $x$ over $k[y,z,w]$, it is primitive because $y$ and $zw$ are coprime elements of a unique factorization domain, and it is a linear irreducible polynomial over $k(y,z,w)$. Gauss lemma for polynomials then applies. Thus $X$ is an irreducible variety that is an affine variety, of algebraic dimension $3$ by the principal hypersurface dimension lemma in affine space ${\mathbb{A}}^4$.

The coordinate rings of $Y$ and $Z$ are respectively $k[y,w]$ and $k[x,z]$. Both are affine planes, hence irreducible closed subsets of algebraic dimension $2$. Their intersection is precisely the origin. ThereforeThe origin is the singular point of an algebraic variety of this three-dimensional affine quadric cone. The example shows why a smoothness of an algebraic variety hypothesis matters in intersection dimension estimates.