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The $k$th elementary symmetric polynomial adds every product obtained by choosing $k$ distinct positions from $x_1,\dots ,x_n$:The conventions are $e_0=1$ and $e_k=0$ for $k>n$. The positions must be distinct, but their values may coincide. Each permutation of the variables merely rearranges the summands, so this is a symmetric polynomial.

For $x_1=1$, $x_2=2$, $x_3=3$, we obtain $e_1=1+2+3=6$, $e_2=1\cdot 2+1\cdot 3+2\cdot 3=11$, and $e_3=1\cdot 2\cdot 3=6$. Expanding the product givesIn general, ${\prod }_{i=1}^n(z-x_i)=z^n-e_1{z}^{n-1}+e_2{z}^{n-2}-\cdots +(-1{)}^n{e}_n$. Thus the coefficients of a monic polynomial collect elementary symmetric polynomials of its roots of a polynomial, with alternating signs. These are the Vieta formulas.^[ref]

In a discrete-time model with independent event indicators whose odds are $a{u}_i$, condition on exactly $d$ events in the risk set $R$. The common factor $a^d$ cancels and the conditional probability of the event set $D$ is the displayed ratio. The denominator is an elementary symmetric polynomial and can be evaluated by recursion rather than enumerating every subset. This conditional model is distinct from summing a continuous-time rank likelihood over all unobserved within-tie orders.

For an explicit linear multistep method, the highest-degree coefficient of $\rho (w)-z\sigma (w)$ is fixed while at least one lower coefficient grows unboundedly with negative real $z$. If all roots stayed in the unit disk, the elementary symmetric polynomials of those roots would keep every monic coefficient bounded. This contradiction proves that no nontrivial consistent explicit method is A-stable.

For order $k$ B-splines, comparison of coefficients in the Marsden identity giveson the basic knot interval. Here $e_m$ is the elementary symmetric polynomial, with $e_0=1$. The $m=0$ case is partition of unity; the $m=1$ coefficients are the Greville abscissae.

The polynomial identities remain true in positive characteristic of a field, but recovering elementary symmetric polynomials by division can fail. In the prime field ${\mathbb{F}}_2$ of characteristic two, the root lists $(0,0)$ and $(1,1)$ have the same power sum zero for every positive exponent: $0^k+0^k=0$ and $1^k+1^k=0$. Their monic polynomials differ, however: $z^2$ and $(z-1{)}^2=z^2+1$. Thus even all positive power sums need not determine the polynomial when its degree meets the characteristic. The relation $2e_2=e_1{p}_1-p_2$ becomes $0=0$ and supplies no value for $e_2$.

Newton's identities let two different summaries of the same polynomial roots talk to each other. Elementary symmetric polynomials record products of distinct root positions; power sums record sums of individual powers, $p_r={\sum }_{i=1}^n{x}_i^r$. Either can be easier to obtain in a particular problem. The identities turn one into the other, often avoiding any need to find the individual polynomial roots.

Write $e_0=1$ and $e_m=0$ for $m>n$. For every positive integer $m$, the identity isIt is a polynomial identity over the integers, so it remains true on substitution in any commutative ring. To solve it for $e_m$ by division, however, $m$ must have a multiplicative inverse. Over a field of characteristic zero, the first $n$ power sums recover every elementary symmetric polynomial. The same recovery works when the positive characteristic of a field is greater than $n$; small characteristic needs care.^[ref]

The first three relations are $e_1=p_1$, $2e_2=e_1{p}_1-p_2$, and $3e_3=e_2{p}_1-e_1{p}_2+p_3$. When division by two and three is permitted, this becomesFor the three roots $1,2,3$, their power sums are $p_1=6$, $p_2=14$, and $p_3=36$. The formulas give $e_2=(36-14)/2=11$ and $e_3=(216-252+72)/6=6$, recovering the polynomial $z^3-6z^2+11z-6$ from aggregate data.

The beauty is that the summaries do not depend on the order of the polynomial roots. In symmetric polynomials, one can work with quantities that survive every relabeling; in linear algebra, traces of matrix powers give the same kind of aggregate information about eigenvalues. The identities convert those summaries into coefficients of a characteristic polynomial.

A short derivation packages the identities into a formal power series. Let $E(t)={\prod }_i(1+x_{i}t)={\sum }_{m\ge 0}{e}_m{t}^m$. Since its constant coefficient is one, it has an inverse as a formal power series, and its formal derivative satisfiesMultiply by $E(t)$ and compare the coefficient of $t^{m-1}$. This gives Newton's identities without a convergence assumption or division by $m$. The division issue arises only when using them to recover $e_m$. The logarithmic-derivative approach is also developed in Newton's Identities (January 2019 notes) by Milan Mossé (§4, pp. 6–8).

The alternate name Newton-Girard identities recalls Albert Girard's earlier work. A modern historical account credits formulas through sixth powers to Albert Girard in 1629 and discusses their later appearance in Isaac Newton's Arithmetica universalis.^[ref] This citation refers to that account; the original historical editions have not been inspected here.

Chapter authors: Ranjan Roy, Frank W. J. Olver, Richard A. Askey, Roderick S. C. Wong, William P. Reinhardt. Their names are given on the official chapter landing page.

Online text: dlmf.nist.gov/1

Cited section: dlmf.nist.gov/1.10

Cited passages: §1.10(iv), equation 1.10.9, states the argument principle; equation 1.10.10 gives the multiplicity-weighted sum of zero locations minus pole locations used in the weighted argument principle.

Cited passage: §1.11(ii), equation 1.11.10, gives the coefficient relations used in elementary symmetric polynomial.

Additional cited passages: §1.13(i), equations 1.13.1–1.13.6, common equation, Wronskian, Abel identity and fundamental-pair equivalence; §1.13(iii), equations 1.13.8–1.13.10, normalized inhomogeneous equation and variation of parameters. Inspected online release: version 1.2.8, 15 September 2026; this is a reference release date, not the original chapter publication date. The source uses analytic coefficients on a simply connected complex domain.

Additional cited section: §1.2(ii) Finite Series, equation 1.2.10 and last-term definition. The five contributors credited above are Chapter 1 authors, not whole-reference authors. Inspected release: version 1.2.8, 15 September 2026; this is an online reference release date.

Additional cited section: §1.7(i), equation 1.7.2, and §1.7(ii), equation 1.7.5, including equality conditions. The existing five chapter contributors remain the Chapter 1 authors; no new publication or release date is inferred.

Additional cited section: §1.4(v), Functions of Bounded Variation, equations 1.4.33–1.4.34 and the conditions on the derivative integral. The existing five chapter contributors remain the Chapter 1 authors; no new chapter publication date is inferred.

A maximal torus in a compact connected Lie group is a compact connected abelian Lie subgroup maximal among such subgroups. By the connected abelian classification it is isomorphic to a product of circles. Such subgroups exist: choose one of largest dimension among compact connected abelian subgroups; a proper inclusion of connected Lie subgroups would increase dimension. Write $\mathfrak{t}$ for its Lie algebra.

Here are the main steps of the conjugacy assertion, with the critical arguments supplied. Average an inner product on $\mathfrak{g}$ over the adjoint action to obtain an $\mathrm{Ad}(G)$-invariant positive definite inner product. It defines a bi-invariant Riemannian metric. Compactness makes this metric complete, and the Hopf-Rinow theorem joins the identity to any $g$ by a geodesic. For a bi-invariant metric the Levi-Civita connection on left-invariant fields is ${\nabla }_{X}Y=[X,Y]/2$, so the geodesic starting with $X$ is $\exp (tX)$. Consequently every $g$ has the form $\exp X$.

The Lie-algebra centralizer of $\mathfrak{t}$ is exactly $\mathfrak{t}$. If $X$ commutes with $\mathfrak{t}$, its one-parameter subgroup commutes with $T$. The closure of the subgroup generated by $T$ and $\exp (\mathbb{R}X)$ is compact, connected and abelian, so it is a torus; maximality forces it to equal $T$, giving $X\in \mathfrak{t}$. The torus action on ${\mathfrak{g}}_{\mathbb{C}}$ decomposes into finitely many character weight spaces. The zero-weight space is ${\mathfrak{t}}_{\mathbb{C}}$. Choose $H\in \mathfrak{t}$ outside the finitely many kernels of the nonzero weight differentials. Then the centralizer of $H$ is exactly $\mathfrak{t}$.

For an arbitrary $X\in \mathfrak{g}$, the continuous function $g\mapsto ⟨{\mathrm{Ad}}_{g}X,H⟩$ attains a maximum. Put $Y={\mathrm{Ad}}_{g_0}X$ at a maximizer. Differentiating along $\exp (sZ)g_0$ givesHence $[Y,H]=0$, so $Y\in \mathfrak{t}$. Exponentiating gives $g_0(\exp X)g_0^{-1}=\exp Y\in T$. This adjoint-orbit criterion for maximal-torus conjugacy proves

Restriction of representations is a ring homomorphism $\mathrm{Res}:R(G)\to R(T)$. If a virtual representation restricts to zero, its character vanishes on $T$. Since characters are class functions and every element is conjugate into $T$, it vanishes on all of $G$. The injectivity of the character map proved in Question 3 then givesThe Weyl group is $W=N_G(T)/T$. For $n\in N_G(T)$, the operator $\rho (n)$ intertwines the restriction of $\rho$ with its twist by $t\mapsto nt{n}^{-1}$. Therefore every restricted representation, and every virtual difference of them, is fixed by $W$:

For the special orthogonal group $G=SO(2m+1)$, takewhere $R_{\theta }$ is a real planar rotation. It is a maximal torus: its Lie-algebra centralizer consists of the independent infinitesimal rotations in those planes. Its character lattice is ${\mathbb{Z}}^m$, so, writing $z_i=e^{i{\theta }_i}$,Indeed commuting unitary torus matrices diagonalize simultaneously into one-dimensional characters, and the characters of the circle are exactly the integral powers.

The normalizer permutes the real rotation planes and can reverse each angle independently. Permutations of whole two-dimensional blocks have determinant one. A reflection in one rotation plane reverses its angle and has determinant minus one; combining it with reflection of the remaining fixed line gives an element of $SO(2m+1)$. Thus all independent inversions occur. Conversely a normalizer must permute the distinct nontrivial torus character pairs $\{\pm e_i\}$ and preserve the fixed line, so there are no additional actions. The kernel of this action is exactly $T$, giving

It follows that the restriction identifies $R(SO(2m+1))$ with a subring of the signed-permutation-invariant Laurent polynomials. In this case we can prove equality. Invariance under individual inversions first givesFor one variable, every invariant is an integral combination of $1$ and $z^k+z^{-k}$, and these satisfy the recurrence $s_{k+1}=x{s}_k-s_{k-1}$. Applying this one coordinate at a time proves the assertion over the other Laurent coefficient rings. Permutation invariance and the Fundamental theorem of symmetric polynomials then givewhere $e_k$ are the elementary symmetric polynomials.

Let $V$ be the complexification of the defining $(2m+1)$-dimensional real representation. Its torus weights are $1,z_i,z_i^{-1}$. The characters of its exterior powers obeyFor $k\le m$, choosing $j$ linear terms and $r$ quadratic terms shows that the coefficient $c_k={\chi }_{{\Lambda }^{k}V}$ isIts leading term is $e_k$ with coefficient one; all other terms use $e_j$ with $j<k$. By induction $e_k$ lies in the restriction image, because $c_k$ is the character of a genuine representation and all smaller $e_j$ already lie in the image. Hence every invariant polynomial lies there, proving the representation ring of an odd-dimensional special orthogonal group descriptionThe triangular change of generators is integral, so $[{\Lambda }^{1}V],\dots ,[{\Lambda }^{m}V]$ are algebraically independent polynomial generators. For $m=1$, $c_1=1+z+z^{-1}$ and this reduces to $R(SO(3))=\mathbb{Z}[V_3]$, with the three-dimensional defining representation as generator. These are genuine representations of the orthogonal group, not the half-integral spin weights of its simply connected double cover.

Let $d=k-1$ and let $e_m$ denote the elementary symmetric polynomial of degree $m$ in the $d$ interior knots $t_{i+1},\dots ,t_{i+d}$, with $e_0=1$. Expand the two sides of the Marsden identity as polynomials in $x$:andComparing the coefficient of $x^{d-m}$ and cancelling the common sign gives the monomial B-spline coefficientsEquivalently,In particular $a_i^{(0)}=1$, giving partition of unity on the interior interval. For $k\ge 2$, $a_i^{(1)}=(t_{i+1}+\cdots +t_{i+k-1})/(k-1)$, and the highest-degree coefficient is the product of those interior knots. The formulas also cover $k=1,m=0$ through empty products.

Put $d=k-1$ and let $e_m$ be the elementary symmetric polynomial of degree $m$ in the $d$ interior spline knots $t_{i+1},\dots ,t_{i+d}$, with $e_0=1$. Expanding the knot polynomial givesCompare the coefficient of $x^{d-m}$ in the Marsden identity. The left coefficient is $(-1{)}^m(\genfrac{}{}{0ex}{}{d}{m})t^m$; the right coefficient is $(-1{)}^m{\sum }_i{e}_m(t_{i+1},\dots ,t_{i+d})N_i(t)$. Cancelling the common sign therefore gives the monomial B-spline coefficients:Equivalently, for $m\ge 1$ the numerator is ${\sum }_{1\le r_1<\cdots <r_m\le k-1}{t}_{i+r_1}\cdots t_{i+r_m}$. In particular $a_i^{(0)}=1$, so the order-zero case reproduces constants, and for $k\ge 2$ the linear coefficients are $a_i^{(1)}=(t_{i+1}+\cdots +t_{i+k-1})/(k-1)$, the Greville abscissae. At $m=k-1$, the coefficient is the product of all interior knots. Each formula holds on the basic knot interval used in the Marsden identity.

A fixed-step linear multistep method for $y^{\prime }=f(t,y)$ uses several previous solution and derivative values:The coefficients are fixed independently of the step size. It is explicit when ${\beta }_s=0$ and implicit otherwise. Its characteristic polynomials of a linear multistep method are $\rho (w)={\sum }_j{\alpha }_j{w}^j$ and $\sigma (w)={\sum }_j{\beta }_j{w}^j$. The contraction mapping theorem gives a uniquely solvable implicit step under a Lipschitz continuity bound $L$ whenever $k|{\beta }_s|L<|{\alpha }_s|$, by contraction of the new-value equation on the appropriate solution region. A suitable starting procedure must supply the first $s$ values; the method alone does not determine them.

Order is defined by inserting a smooth exact solution into the recurrence. If the residual is $O(k^{p+1})$ and its first nonzero coefficient occurs at that degree, the method has order $p$; dividing the residual by $k$ gives an $O(k^p)$ local truncation error. Taylor expansion gives the exact order conditions for a linear multistep methodwith the right side interpreted as zero at $q=0$, and failure at $q=p+1$ for exact order. Equivalently,The basic consistency of a numerical method conditions are $\rho (1)=0$, ${\rho }^{\prime }(1)=\sigma (1)$; for the ordinary nondegenerate consistent formulation this common value is nonzero. Exactness on constants and linear functions does not control propagation of numerical perturbations.

That propagation is the role of zero-stability. On $y^{\prime }=0$ the homogeneous recurrence is $\rho (E)y_n=0$. Its modes are $n^r{\xi }^n$, with powers of $n$ determined by root multiplicities. The theorem on the root condition for a multistep method states that zero-stability is equivalent to every root of $\rho$ having modulus at most one, with every unit-modulus root simple. Interior repeated roots are permitted, since geometric decay dominates their polynomial factors. An exterior root amplifies perturbations geometrically; a repeated unit root amplifies them polynomially and violates a step-independent perturbation bound.

The Dahlquist equivalence theorem states that a fixed-step consistent linear multistep method is convergent exactly when it is zero-stable. This concerns ordinary differential equations with a uniformly Lipschitz continuous right-hand side on the solution region over a fixed finite interval, convergent starting values, and the well-defined nearby branch of each small-step implicit solve. For order $p$ convergence of a numerical method, the solution must have the required smoothness and the starting errors must be $O(k^p)$.

The mechanism can be sketched explicitly. The root condition for a multistep method bounds the recurrence's discrete Green function. For an order-$p$ exact-solution defect $d_n=O(k^{p+1})$, the Lipschitz continuity error equation then gives an estimate of the formThe current-step term can be absorbed for small $k$. The discrete Gronwall inequality says that a bound $E_n\le A+Bk{\sum }_{j<n}{E}_j$ implies $E_n\le A(1+Bk{)}^n\le A{e}^{B{t}_n}$. There are $O(k^{-1})$ defects on a fixed interval, so their sum is $O(k^p)$ and this yields the claimed global order. Conversely, testing $y^{\prime }=0$ with vanishing starting perturbations in exterior-root or repeated-unit-root modes proves that convergence of a numerical method fails without the root condition for a multistep method.

For example, the two-step Adams-Bashforth method hasIt is explicit, order two and zero-stable, hence convergent with order-two starting values. By contrast, a recurrence with $\rho =(w-1{)}^2$ has solutions $y_n=y_0+n(y_1-y_0)$ on the zero differential equation. Starting differences $\sqrt{k}$ tend to zero yet become unbounded at a fixed positive time, since $n$ is of order $1/k$. A small local defect or formal high order cannot rescue this mode. Canceling a common factor may remove it only by changing the allowed starting relations, as the parameter endpoint in Question 2 illustrates.

Absolute stability addresses a different limit: apply the method to the Dahlquist test equation $y^{\prime }=\lambda y$ and put $z=k\lambda$. The amplification roots satisfyAll roots must satisfy the disk condition, with unit roots simple; one cannot inspect only the root that tends to one as $z\to 0$. The degree and solvability of the recurrence must also be preserved. The resulting set of $z$ is the domain of absolute stability. A-stability means that it includes the whole closed left half-plane, so the numerical solutions of every Dahlquist test equation with $\mathrm{Re}\lambda \le 0$ are stable for every positive step size. It is a linear test-equation property, not a proof of arbitrary nonlinear contractivity or high accuracy at large steps.

Forward Euler method has order one and factor $1+z$, so its stability region is the disk $|1+z|\le 1$. Backward Euler method has factor $(1-z{)}^{-1}$ and is A-stable. The trapezoidal rule is order two with factor $(1+z/2)/(1-z/2)$; its modulus is at most one throughout the left half-plane. These one-step methods are included as the simplest multistep examples. For the two-step Adams-Bashforth method, the negative-real stability interval is $-1\le z\le 0$: its polynomial is $w^2-(1+3z/2)w+z/2$, and the root reaches $-1$ at $z=-1$. It therefore imposes a step restriction when applied to strongly decaying modes.

The genuinely two-step BDF2 method isIt has order two, roots $1,1/3$ at zero step and is A-stable. For its normalized polynomials $\rho =3w^2/2-2w+1/2$, $\sigma =w^2$, the unit-circle boundary locus hasThe leading coefficient has no zero in the left half-plane, and the roots at a small negative $z$ are both inside the disk. Root continuity then proves open-half-plane stability; on the imaginary boundary the only possible unit root is the simple root one at $z=0$. This gives a direct proof of the stated A-stability rather than relying on a boundary plot.

Two sharp order barriers explain the tradeoff. The first Dahlquist barrier states that a consistent zero-stable real linear $s$-step method using only first derivatives has order at most $s+1$ for odd $s$ and $s+2$ for even $s$; an explicit method has order at most $s$. The Second Dahlquist barrier states that an irreducible A-stable real linear multistep method has order at most two. These statements concern fixed constant coefficients and the ordinary first-derivative multistep class, not multiderivative methods or arbitrary variable-step formulations. Thus higher order alone is not a route to unconditional stiff stability. Question 2 provides a concrete illustration: its third-order member is zero-stable but not A-stable.

No nontrivial consistent explicit multistep method is A-stable. Its amplification polynomial of a multistep method has a fixed highest coefficient, while at least one lower coefficient grows without bound as real $z\to -\infty$. Were all roots uniformly in the unit disk, their elementary symmetric polynomials would keep all the monic coefficients bounded, a contradiction. This is the coefficient argument behind explicit multistep methods cannot be A-stable.

In practice, selecting a method therefore involves accuracy, startup, the presence of a stiff differential equation, implicit-solve cost and step-size control. Zero-stability handles propagation as the step tends to zero; consistency of a numerical method supplies a vanishing defect; A-stability handles the numerical solutions of all decaying Dahlquist test equations at finite steps. None substitutes for the others, and variable steps or PDE discretizations require additional uniform stability arguments appropriate to their setting.

The coordinate ring is $\mathbb{C}[W]=\mathrm{Sym}(W^{\ast })$. The induced action is contragredient on functions:It is a group action by degree-preserving algebra automorphisms, extending the dual action on linear polynomial functions. Its polynomial invariant ring isFor a monic polynomial $f(t)={\prod }_{i=1}^d(t-r_i)$, its polynomial discriminant is $\mathrm{disc}(f)={\prod }_{i<j}(r_i-r_j{)}^2$. This is symmetric in the roots, hence polynomial in the coefficients, and is zero exactly when a root is repeated.

For the alternating group action, let $e_1,\dots ,e_n$ be the elementary symmetric polynomials and put $\Delta ={\prod }_{i<j}(X_i-X_j)$. If $p$ is $A_n$-invariant and $\tau$ is any transposition, decomposeNormality and index two of $A_n$ show that $p_{+}$ is symmetric and $p_{-}$ transforms by the sign character of $S_n$. Any alternating polynomial vanishes when $X_i=X_j$, so every $X_i-X_j$ divides it. These pairwise nonassociate prime factors have product $\Delta$, so $p_{-}=\Delta q$ with $q$ symmetric. The Fundamental theorem of symmetric polynomials yieldsThe sum is direct, since a polynomial that is both symmetric and alternating is zero in characteristic zero. Also ${\Delta }^2=D(e_1,\dots ,e_n)$, where $D$ is the discriminant polynomial ofThus, more precisely than the requested quotient assertion,Surjectivity follows from the direct-sum expression. Divide any putative kernel element by the monic quadratic in $Z$; its remainder is $a(T)+Zb(T)$. The direct sum forces $a(e)=b(e)=0$, and algebraic independence of the $e_i$ gives $a=b=0$. Hence the displayed relation is the entire kernel. The argument also covers $n=2$, where $A_2$ is trivial.

Now let $G$ be finite with no nontrivial linear characters, and write $R=\mathbb{C}[W]$, $S=R^G$. The polynomial ring $R$ is a unique factorization domain. Factor a nonzero invariant $f$ in $R$; invariance permutes the associate classes of its irreducible factors and makes their exponents constant on each orbit. For an orbit $\mathcal{O}$, choose representatives and form its orbit product of polynomial factorsFor every $g$, $g\cdot q_{\mathcal{O}}=\theta (g)q_{\mathcal{O}}$ for a nonzero scalar $\theta (g)$. Applying two group elements proves that $\theta :G\to {\mathbb{C}}^{\times }$ is a homomorphism. The hypothesis forces $\theta =1$, so $q_{\mathcal{O}}\in S$.

This orbit product is prime in $S$. If it divides $ab$ with $a,b\in S$, one factor $p\in \mathcal{O}$ divides $a$ or $b$ in $R$. Invariance of that chosen polynomial makes every factor in the orbit divide it. Their product therefore divides it in $R$, and the quotient is invariant because numerator and denominator are invariant and cancellation is valid in the integral domain $R$. Thus $q_{\mathcal{O}}$ divides $a$ or $b$ in $S$. Every nonzero $f\in S$ is a scalar times a product of these prime orbit products, and the only units of $S$ are nonzero constants, as they are units in $R$. Therefore $\mathbb{C}[W{]}^G$ is a unique factorization domain. This gives a direct proof, without assuming a divisor-class-group theorem.

For a failure in characteristic zero, let the cyclic group of order two act on ${\mathbb{C}}^2$ by $-I$. Its coordinate invariants areEvery invariant monomial has even total degree: its exponents are either both even, or both odd, giving the indicated generators. Reducing powers of $b$ to at most one shows that $ac-b^2$ is the only relation, since $a^i{c}^j$ and $b{a}^i{c}^j$ map to distinct monomials. The three quadratic invariants are irreducible in $S$: each nonconstant invariant has degree at least two, so a product of two nonunits has degree at least four. They are pairwise nonassociate, butgives two different irreducible factorizations. Hence this invariant ring is not a unique factorization domain. The nontrivial sign character is precisely the kind of character excluded in the preceding theorem.

Set $d=k-1$ and let $e_m(t_{i+1},\dots ,t_{i+d})$ be the elementary symmetric polynomial of degree $m$ in the interior knots, with $e_0=1$. Expand the Marsden identity in powers of $x$. The coefficient of $x^{d-m}$ on the left is $(-1{)}^m(\genfrac{}{}{0ex}{}{d}{m})t^m$; that coefficient in ${\omega }_i(x)$ is $(-1{)}^m{e}_m(t_{i+1},\dots ,t_{i+d})$. Comparing and cancelling the sign proves the monomial B-spline coefficients:For $m=0$ this gives partition of unity on the basic interval. For $m=1$ it gives the Greville abscissae, the arithmetic means of the $k-1$ interior knots. For the top degree it gives their product. When $k=1$, only $m=0$ occurs and the empty symmetric polynomial is one.

Take the Vandermonde determinant $\Delta ={\prod }_{i<j}(x_i-x_j)$. Its square is a symmetric polynomial, hence a polynomial in the elementary symmetric polynomials and belongs to $K$. It is nonzero because the variables are algebraically independent. A transposition of two variables fixes every $e_i$ and therefore fixes $K$, but sends $\Delta$ to $-\Delta \ne \Delta$ in characteristic zero. Thus $\boxed{\alpha =\Delta \notin K,{\alpha }^2\in K}$.

Work first where all three coordinates are finite and the joining line is not vertical. By the chord-and-tangent group law, the third intersection of $y=\lambda x+\mu$ with the elliptic curve is $-(P_1+P_2)$, whose first coordinate is also $x_3$. Substitution gives the monic polynomialIts roots are $x_1,x_2,x_3$, counted with intersection multiplicity. Comparing its coefficients with the elementary symmetric polynomials provesThis symmetric-coordinate identity for elliptic-curve addition is an identity of rational functions. It includes tangent cases by specialization. Literally assigning finite $s_i$ when a point is $O$ would be meaningless; such cases must be interpreted on the projective curve. A nonsingular equation in the printed short form has characteristic different from two.

Here is how the identity yields the quadratic form for the degree of an isogeny. With $u=x(P)$, $v=x(Q)$, write $s=u+v$ and $q=uv$. As an equation for $t=x(P+Q)$, it becomesBoth $x(P+Q)$ and $x(P-Q)$ satisfy it, because replacing $Q$ by $-Q$ does not change $v$. Thus its two branches are the sum and difference maps. Their common denominator is the square of $u-v$. The following divisor on an algebraic curve calculation keeps track of cancellations and exceptional points rigorously, instead of assuming degrees of displayed numerators always add.

On $E_2\times E_2$, equality of the two first coordinates means $Q=P$ or $Q=-P$. Each coordinate has a double pole at $O$. Consequently the principal divisor identity iswhere $D_{+}$ is the diagonal and $D_{-}$ is the graph of negation. Equivalently, the two zero divisors are the inverse images of $O$ under difference and sum, exactly the two branches identified above. Their generic multiplicities are one; the divisor identity also accounts for their intersections at 2-torsion points.

For nonzero isogenies of elliptic curves $\varphi ,\psi :E_1\to E_2$ with $\varphi \ne \pm \psi$, pull this identity back by $(\varphi ,\psi )$. Taking degrees of the resulting principal divisor on $E_1$ gives the divisor proof of the degree parallelogram law:If one map is zero, this is immediate. If $\varphi =\pm \psi$, it follows from $\mathrm{deg}([2]\circ \varphi )=4\mathrm{deg}\varphi$. This also proves the parallelogram law for a general Weierstrass equation of an elliptic curve in characteristic two: the same divisor identity uses the quotient by negation, even though the particular symmetric-coordinate formula above is unavailable.

Put $q(\varphi )=\mathrm{deg}\varphi$ and $q(0)=0$. Applying the parallelogram law to $n\varphi ,\varphi$ gives the recurrence $q((n+1)\varphi )+q((n-1)\varphi )=2q(n\varphi )+2q(\varphi )$, and hence $q(n\varphi )=n^{2}q(\varphi )$ for every integer $n$. Its polarization $B(\varphi ,\psi )=q(\varphi +\psi )-q(\varphi )-q(\psi )$ is symmetric and integer-valued. For fixed $\psi$, the parallelogram law makes $B(\cdot ,\psi )$ odd and gives $B(x+y,\psi )+B(x-y,\psi )=2B(x,\psi )$. Interchanging $x,y$ and adding proves additivity, since integers have no two-torsion. Thus $B$ has bilinearity, and $q$ is a positive quadratic form on $\mathrm{Hom}(E_1,E_2)$, with $q(\varphi )>0$ whenever $\varphi \ne 0$.

A suitable version of the Hensel lemma is this: if $R$ is a complete discrete valuation ring, $h\in R[X]$, and $\overline{a}$ is a root of $\overline{h}$ with ${\overline{h}}^{\prime }(\overline{a})\ne 0$, there is a unique $a\in R$ reducing to $\overline{a}$ with $h(a)=0$. Starting with any lift, Newton iteration over a valued field $a_{j+1}=a_j-h(a_j)/h^{\prime }(a_j)$ keeps the derivative a unit and doubles the error valuation, giving convergence. Uniqueness follows by factoring $h(u)-h(v)$ when two roots have the same residue.

The field in the rest of this question is not assumed complete, so that version cannot simply be applied to it. Instead use the stated uniqueness of the extension of an absolute value. Let ${\alpha }_i,{\alpha }_j$ be any two roots of a polynomial $f$, which is irreducible. The field embedding sending ${\alpha }_i$ to ${\alpha }_j$ identifies $K({\alpha }_i)$ with $K({\alpha }_j)$. Pulling the latter field's absolute value back gives an extension on $K({\alpha }_i)$, which must coincide with the given one. More generally, for every $h\in K[X]$,This proves equal absolute values of algebraic conjugates without assuming separability or a transitive action of a Galois group; repeated roots cause no difficulty.

Every root of the monic $f\in {\mathcal{O}}_K[X]$ has absolute value at most one. Indeed, if $|\alpha |>1$, then each lower term satisfies $|c_j{\alpha }^j|\le |\alpha {|}^j<|\alpha {|}^n$. The ultrametric inequality makes the leading term dominate their sum, contradicting $f(\alpha )=0$. Thus

All roots lie in the valuation ring ${\mathcal{O}}_M$, so their residues are defined in the residue field $k_M$. Let $\phi \in k_K[X]$ be the monic minimal polynomial of an algebraic element of ${\overline{\alpha }}_1$; this residue is algebraic because it satisfies the monic $\overline{f}$. Choose a coefficientwise lift $\Phi \in {\mathcal{O}}_K[X]$. Then $|\Phi ({\alpha }_1)|<1$, and the conjugacy argument gives $|\Phi ({\alpha }_i)|<1$ for every $i$. Hence every ${\overline{\alpha }}_i$ is a root of $\phi$.

Reduction of the splitting-field factorization, with multiplicities, givesEvery monic irreducible factor of $\overline{f}$ over $k_K$ has a root among these residues. That root also satisfies $\phi$, so its minimal polynomial is $\phi$. Unique factorization therefore proves the pure-power reduction of a monic irreducible polynomial:This does not assert that $\phi$ is separable or that the reduction is square-free.

Finally, factor the monic $g$ over $K$ as $g={\prod }_j{h}_j^{e_j}$ with distinct monic irreducible polynomials $h_j$. The same leading-term argument bounds every root of $g$ by one. Coefficients of any $h_j$ are elementary symmetric polynomials in some of these roots, so they lie in ${\mathcal{O}}_K$. By the preceding result each ${\overline{h}}_j$ is a power of a single monic irreducible residue polynomial ${\phi }_j$.

Since the prescribed ${\overline{g}}_1,{\overline{g}}_2$ are coprime, every ${\phi }_j$ occurs on exactly one side. Assign the entire factor $h_j^{e_j}$ to that side and form the two products. Comparing irreducible-factor multiplicities in the reduction proves the coprime factor lifting from valuation-extension uniqueness:Empty products are one, covering constant prescribed factors. The construction proves the required factor-lifting form of the Hensel lemma directly from extension uniqueness, without adding completeness as an unstated hypothesis.

For the First Chern class identity, use finite-dimensional classifying maps. A complex line bundle on a compact Hausdorff space embeds in a finite-dimensional trivial vector bundle, so it is a pullback vector bundle of the tautological bundle on some Complex projective space. For the two complex line bundles, the universal tensor product of vector bundles is classified by the mapIts restriction to either factor pulls the tautological bundle back to the tautological bundle on that factor. The degree-two cohomology of the product is the direct sum of the degree-two cohomology of its factors, so these two restrictions determine its First Chern class. Pulling back to $X$ provesThis proves the integral identity, including possible torsion elements in the cohomology of $X$; an argument using only curvature forms would not detect those torsion elements.

The splitting principle for complex vector bundles says that for a vector bundle $E\to X$ there is an iterated projective bundle $p:F(E)\to X$ for which $p^{\ast }$ is injective on integral cohomology, and $p^{\ast }E$ is a direct sum of complex line bundles. The same assertion holds with rational coefficients, and a common such space can split any finite collection of vector bundles. The injectivity follows at each stage from the projective bundle formula for complex vector bundles; a Hermitian metric splits the resulting filtration into a direct sum.

Write $x_i=c_1(L_i)$ for the Chern roots of the pulled-back vector bundle. Define the degree-$2j$ component of its Chern character byThe Fundamental theorem of symmetric polynomials expresses each power-sum symmetric polynomial ${\sum }_i{x}_i^j$ uniquely as an integral polynomial in the elementary symmetric polynomials, which are the Chern classes of $E$. Thus the definition makes sense on $X$ itself, without choosing actual Chern roots there. On a compact Hausdorff space, each vector bundle is classified by a finite-dimensional Grassmannian; its high-degree Chern character components therefore vanish. Hence this defines a class in the direct sum $H^{2\ast }(X;\mathbb{Q})$ even if $X$ has no finite dimension. On more general spaces an infinite Chern character is instead interpreted in the product of the even-degree cohomology groups.

On a common splitting space, the Chern roots of a direct sum are the combined root lists, whereas the First Chern class identity makes the Chern roots of a tensor product of vector bundles the pairwise sums $x_i+y_j$. ConsequentlyThe injectivity in the splitting principle for complex vector bundles descends both identities to $X$. Additivity extends the Chern character to the Grothendieck group by $\mathrm{ch}([E]-[F])=\mathrm{ch}(E)-\mathrm{ch}(F)$. Multiplicativity and $\mathrm{ch}(1)=1$ then show that the Chern character is a unital ring homomorphism.

For Complex K-theory of complex projective space, let $\gamma$ be the tautological bundle, put $t=1-[\gamma ]$, and set $h=-c_1(\gamma )$. The cohomology ring of complex projective space is $\mathbb{Z}[h]/(h^{n+1})$. The CW filtration gives a cofibrationThe K-theory six-term exact sequence and Complex K-theory of a sphere give, inductively,Here $b_n$ is the image of a Bott element from $S^{2n}$, normalized so that $\mathrm{ch}(b_n)=h^n$. This normalization follows from the multiplicative form of Bott periodicity: the Chern character of the degree-two Bott element is an integral generator, and multiplication yields the top-degree generator on $S^{2n}$. In particular, the abelian group $K^0({\mathbb{CP}}^n)$ is a free abelian group of rank $n+1$.