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We prove the finite-dimensional spectral theorem for normal operators by induction on dimension. A complex linear operator has an eigenvector $v$ with eigenvalue $\lambda$, by the fundamental theorem of algebra. For a normal operator $T$, the operator $T-\lambda I$ is also normal, andfor every $w$. Applying this to $v$ shows $T^{\ast }v=\overline{\lambda }v$.

Normalize $v$. Its orthogonal complement $W=v^{\perp }$ is invariant under both $T$ and $T^{\ast }$: the adjoint operator identities pair $Tw$ with $v$ through $T^{\ast }v$, and $T^{\ast }w$ with $v$ through $Tv$. The restricted operator on $W$ remains normal, and its adjoint is the restriction of $T^{\ast }$. Induction supplies an orthonormal eigenbasis of $W$, which together with $v$ gives one for $V$.

Conversely, in an orthonormal basis of eigenvectors, $T$ is diagonal and $T^{\ast }$ is the diagonal matrix of the conjugate eigenvalues. The two diagonal operators commute. Thus normality is equivalent to an orthonormal eigenbasis. The zero-dimensional case is immediate.

The Weak Hilbert Nullstellensatz states that if $k$ is an algebraically closed field, every maximal ideal of $k[Y_1,\dots ,Y_s]$ isEquivalently, every proper ideal has a common zero. We prove the more general field statement, the Zariski lemma: a field finitely generated as a $k$-algebra is a finite field extension of $k$.

Write such a field as $L=k[b_1,\dots ,b_s]$. Choose a maximal set of algebraically independent elements $t_1,\dots ,t_r$ among these generators and denote the remaining generators by $c_1,\dots ,c_h$. Each $c_j$ is an algebraic element over $K=k(t_1,\dots ,t_r)$, so $L/K$ is a finite field extension. Clearing the denominators in their monic equations, choose a nonzero $d\in k[t_1,\dots ,t_r]$ such that each $c_j$ is an integral element overBecause $L$ is a field and $d^{-1}\in L$, we have $L=D[c_1,\dots ,c_h]$. Thus $L$ is an integral extension of $D$.

A subring over which a field is integral must itself be a field. Indeed, for $0\ne a\in D$, an integral equation for $a^{-1}$ multiplied by $a^{m-1}$ expresses $a^{-1}$ as an element of $D$. But if $r>0$, the ring $D$ is not a field: choose an irreducible polynomial $q(t_1)\in k[t_1]$ which does not divide $d$. There are infinitely many monic irreducible polynomials in $k[t_1]$, by the usual product-plus-one argument, whereas only finitely many can divide the nonzero $d$. The prime ideal $(q)$ stays proper after inverting $d$, so $q$ has no inverse in $D$. Therefore $r=0$, proving that $L/k$ is a finite field extension, and in particular an algebraic field extension.

Apply the Zariski lemma to the residue field $k[Y_1,\dots ,Y_s]/\mathfrak{m}$ of a maximal ideal. When $k$ is an algebraically closed field, this field is $k$ itself. The images $a_i$ of $Y_i$ then give $\mathfrak{m}=(Y_1-a_1,\dots ,Y_s-a_s)$, proving the Weak Hilbert Nullstellensatz.

For the real Laurent polynomial ring, the same argument applies because it is the finitely generated algebraEvery residue field at a maximal ideal is consequently a finite field extension of $\mathbb{R}$, and so is $\mathbb{R}$ or $\mathbb{C}$. Here we use the fundamental theorem of algebra, which makes $\mathbb{C}$ the algebraic closure of $\mathbb{R}$ of degree two. The images of $X_1,X_2$ must be nonzero since these elements are units. It follows that the answer iswith $(\alpha ,\beta )$ and $(\overline{\alpha },\overline{\beta })$ defining the same maximal ideal. If both coordinates are real, evaluation has image $\mathbb{R}$; otherwise its image is $\mathbb{C}$. Thus every displayed kernel is maximal, and the Zariski lemma shows that this list is exhaustive.

For explicit generating sets of an ideal, there are three cases. For real nonzero $a,b$,If $\alpha =a\in {\mathbb{R}}^{\times }$ and $\beta =s+it$ with $t\ne 0$,If $\alpha =u+iv$ with $v\ne 0$ and $\beta =s+it$, thenThe last two quotients are $\mathbb{C}$; the nonzero coordinate condition ensures that the Laurent inverses exist in them. There are no further coincidences in the classification: equal kernels induce an $\mathbb{R}$-isomorphism between the evaluation image fields. The only $\mathbb{R}$-automorphisms of $\mathbb{C}$ are the identity and complex conjugation, so the two pairs must agree or be conjugate.

To prove the intersection assertion, fix a prime ideal $P$ and an element $f\notin P$. The integral domain $B=R/P$ has nonzero image $\bar{f}$. Its localization of a ring $B[{\bar{f}}^{-1}]$ is nonzero and still a finitely generated algebra over $\mathbb{R}$, so it has a maximal ideal $\mathfrak{n}$. By the Zariski lemma, the field $L=B[{\bar{f}}^{-1}]/\mathfrak{n}$ is finite over $\mathbb{R}$.

The image of $B$ in $L$ is a subring containing $\mathbb{R}$ and is a finite-dimensional $\mathbb{R}$-vector space. It is an integral domain, hence a field: multiplication by any nonzero element is an injective linear map of a finite-dimensional vector space and therefore surjective. The kernel of $B\to L$ is thus a maximal ideal which avoids $\bar{f}$, since $\bar{f}$ is invertible in $L$. Its inverse image in $R$ is a maximal ideal containing $P$ but avoiding $f$. As this works for every $f\notin P$,This is the Jacobson ring property for this Laurent polynomial ring.

For a nonzero finitely generated module $M$, its set of associated primes isHere ${\mathrm{Ann}}_R(m)=\{a\in R:am=0\}$ is the annihilator of the element. The Laurent polynomial ring $R$ is a Noetherian ring, so the nonempty collection of annihilators of nonzero elements of $M$ has a maximal member, say $P={\mathrm{Ann}}_R(m)$. It is proper because $m\ne 0$.

If $ab\in P$ and $b\notin P$, then $bm\ne 0$ andMaximality forces ${\mathrm{Ann}}_R(bm)=P$, hence $a\in P$. Thus $P$ is a prime ideal, proving $\boxed{{\mathrm{Ass}}_R(M)\ne \varnothing }$. This argument in fact only needs that $R$ is Noetherian, not the finite generation of $M$.

A proper ideal $Q$ is $P$-primary when its radical of an ideal is $P$ andEquivalently, $ab\in Q$ and $b\notin P$ imply $a\in Q$; equivalently, every zero divisor of $R/Q$ is nilpotent. These formulations distinguish the actual primary ideal property from the weaker assertion that there is only one minimal prime ideal.

For the final example, set $u=X_1-1$ and $v=X_2-1$. ThenSince $R/P\cong \mathbb{R}[X_2,X_2^{-1}]$ is an integral domain, $P$ is a prime ideal. Moreover $I\subseteq P$ and $u^2\in I$, so $\sqrt{I}=P$. Consequently $P$ is the unique minimal prime ideal over $I$.

Nevertheless $uv\in I$, while $u\notin I$ and no power of $v$ belongs to $I$. For the first assertion, reduce modulo $v$: the image of $I$ is $(u^2)$ in $\mathbb{R}[X_1,X_1^{-1}]$, and $u$ is not a multiple of $u^2$ because $u$ is a nonunit in this integral domain. For the second, $v\notin P$, so $v^j\notin P\supseteq I$ for every $j\ge 1$. Thus $I$ is not a $P$-primary ideal: a unique minimal prime does not imply primary. In $R/I$, the nonzero class of $u$ is killed by $v$, a zero divisor which is not nilpotent.

Orient both spheres in the standard way and normalize their area forms to total area $4\pi$. The degree of a map between oriented manifolds can be obtained in two ways. For a regular value $y$, use the degree as a sum of local degrees:Each inverse image is isolated by the inverse function theorem, and compactness makes the set finite. The determinant is computed in consistently oriented local coordinates. A second method is spherical degree by area pullback:where the last formula represents $g$ as a unit vector in ${\mathbb{R}}^3$. The pullback of a differential form already contains the signed Jacobian determinant; no extra $\sin \theta$ is to be inserted in that last coordinate expression.

To relate the methods, replace $\omega$ by a smooth top-degree differential form with the same total integral supported in a small neighbourhood of a regular value. Two such top-degree forms with equal integral differ by an exact differential form on $S^2$, by its top-degree de Rham cohomology. Their pullbacks therefore have the same integral by Stokes theorem. Over the chosen neighbourhood, $g$ splits into local inverse branches; the change of variables formula makes the contribution of each branch its orientation sign times $4\pi$. Their sum is precisely the first formula. Thus the area integral is an integer and agrees with the signed inverse-image count.

For a nonconstant rational map, first use common-factor reduction of a rational map so $p$ and $q$ are coprime. Write $k=\max (\mathrm{deg}p,\mathrm{deg}q)$ for these reduced polynomials. A generic finite target value $w$ has inverse images at the roots of $p-wq$: avoiding exceptional values makes its degree $k$ and its roots simple. The fundamental theorem of algebra supplies $k$ roots. A holomorphic map has positive real Jacobian determinant $|R^{\prime }{|}^2$ at a regular point, so every local sign is $+1$. HenceThe source leaves coprimality implicit. In an unreduced representation the answer is $\max (\mathrm{deg}p,\mathrm{deg}q)-\mathrm{deg}\gcd (p,q)$, including degree zero for a constant reduced map. For example $(z^2-1)/(z-1)$ extends to $z+1$ and has degree one, although the unreduced maximum degree is two. Exceptional inverse images at infinity or multiple roots do not change the degree of a rational map of the Riemann sphere.

For the rational map approximation for Skyrmions, use stereographic projection $z=\tan (\theta /2)e^{i\phi }$ and the unit target vectorCombine this rational map with a radial profile to form a special unitary group field:where $\mathbf{\sigma }$ are the Pauli matrices. The endpoint values make $U(0)=-1$ independent of angle and $U(\infty )=1$. Appropriate radial behaviour gives an admissible finite-energy field configuration. With $L_i=U^{†}{\partial }_{i}U$, choose the topological baryon number in the Skyrme model conventionSeparating the radial and angular factors givesThus the degree of a rational map of the Riemann sphere supplies the Skyrmion charge.

In conventional dimensionless massless Skyrme model units, its static energy reduces towith the angular Jacobian of a rational mapThe Cauchy-Schwarz inequality gives $\mathcal{I}\ge k^2$. These formulas follow from the radial strain $|f^{\prime }|$ and the two equal angular strains $\sin f\sqrt{J_R}/r$: the quadratic energy sums their squares and the quartic Skyrme term sums their pairwise products of squares. Minimize the angular integral in the rational map approximation over degree-$k$ maps, then minimize the remaining radial energy with the stated endpoints. This replaces a three-dimensional field minimization by finitely many map coefficients and an ordinary differential equation for $f$.

The method constructs a charge-$k$ variational approximation, with topology built in and with rotational symmetry of a rational map translated into combined spatial and isospin rotations. It is efficient for identifying shapes and providing initial data for unrestricted numerical relaxation. Its restrictions are equally concrete: it uses one radial profile and a holomorphic angular map independent of radius, so it cannot represent arbitrary radial-angular correlations, separated clusters, or all deformations. Apart from the degree-one Skyrmion hedgehog ansatz, it generally does not solve the full field equation exactly. Massive-pion terms can be included in the radial functional but do not remove these restrictions, and multi-shell or unrestricted fields may be needed for larger charges. Approximate energy minima and a final collective-coordinate quantization are distinct steps.

Again use $D_j=-i{\partial }_j$ and distribution dual pairings with complex bilinearity. The Malgrange–Ehrenpreis theorem asserts that every nonzero constant-coefficient differential operator defined by a polynomial has a fundamental solution of a linear differential operator:If $P=a\ne 0$ is constant, take $E={\delta }_0/a$. For positive degree $m$, write $P_m$ for the highest homogeneous part. A nonzero polynomial cannot vanish on all of ${\mathbb{R}}^n$, so choose a real unit vector $e$ with $P_m(e)\ne 0$. A change of coordinates by an orthogonal matrix makes it the last coordinate direction. In these coordinates,Crucially the leading coefficient $a$ is constant and never vanishes as ${\xi }^{\prime }\in {\mathbb{R}}^{n-1}$ varies.

The finite-height polynomial root avoidance construction gives the needed Hörmander staircase. Set $h_{\ell }=3\ell$ for $0\le \ell \le m$. For each fixed ${\xi }^{\prime }$, the fundamental theorem of algebra supplies $m$ roots $z_1,\dots ,z_m$, with multiplicity. A root's imaginary part can be within distance strictly less than one of at most one candidate height, since those heights are separated by three. The pigeonhole principle therefore leaves at least one height with $|h_{\ell }-\mathrm{Im}{z}_j|\ge 1$ for all roots. At that height, for every real $s$,No continuous labelling of the roots is required. To select the heights measurably, defineEach $B_{\ell }$ is closed, since the coefficients are continuous in ${\xi }^{\prime }$. Continuity in $s$ makes the rational test equivalent to the bound for all $s\in \mathbb{R}$. The $A_{\ell }$ are therefore disjoint Borel sets covering ${\mathbb{R}}^{n-1}$. The union of horizontal fibresis a Hörmander staircase, with bounded imaginary height and a uniform nonzero denominator. For $n=1$, the transverse space has one point, and this is simply the choice of one horizontal line.

For example, $P({\xi }_1,z)={\xi }_1^2+z^2$ has roots $z=\pm i{\xi }_1$. One may select $h=0$ for $|{\xi }_1|\ge 1$, and $h=3$ for $|{\xi }_1|<1$. Then the first region has $|P|\ge 1$, while in the second each root is at least two units from the chosen height. The figure shows the height projection of these horizontal fibres, rather than additional connecting contours.

Define a linear functional on test functions byTo establish that this is a distribution, fix a compact set containing the support of $\phi$, inside a radius-$R$ ball. The Fourier decay in a bounded complex strip estimate follows by applying $(1-\Delta {)}^L$ to $e^{-h_{\ell }{x}_n}\phi (x)$:The finite bound on the heights absorbs all exponential factors and powers of $h_{\ell }$ into $C_{R,L}$. Taking $2L>n$ and using $|P|\ge |a|$ proves absolute convergence and the required finite-order continuity on each fixed support set. Hence $E\in {\mathcal{D}}^{\prime }$.

Now the formal transpose identity and the Fourier derivative identity giveThe polynomial denominator cancels in the pairing for $P(D)E$. For each fixed real ${\xi }^{\prime }$, the remaining function of $z$ is entire and rapidly decreasing in its real part throughout $0\le \mathrm{Im}z\le 3m$. The Cauchy integral theorem therefore allows each horizontal fibre to be shifted to the real axis; its truncated vertical end integrals tend to zero. Absolute convergence justifies Fubini's theorem and the finite sum over the measurable partition. ThusThe last equality is Fourier inversion. Transforming back from the orthogonal coordinates preserves ${\delta }_0$ and gives the required fundamental solution of a linear differential operator for the original operator. The staircase is allowed to jump: after cancellation the shift is performed on each one-dimensional fibre, so no assertion that $\Sigma$ is a smooth contour is used. This construction proves a distributional fundamental solution, without claiming a tempered-growth estimate.

For $f\in \mathcal{D}({\mathbb{R}}^n)$, use the smoothing convolution with a test functionIndeed $u_0(x)=⟨E_y,f(x-y)⟩$ is a smooth function, with ${\partial }^{\alpha }{u}_0(x)=⟨E_y,{\partial }^{\alpha }f(x-y)⟩$. On each compact set of $x$ values, all translated test functions have supports in one fixed compact set, so differentiation is justified by the continuity of the distribution. Compact support of $f$ makes the convolution well defined even when $E$ is not tempered. It does not imply compact support of $u_0$.

Finally the affine solution space of a linear equation isIf every homogeneous solution is represented by a smooth function, so is every $u=u_0+v$. Conversely, if every solution for this fixed $f$ is represented by a smooth function, then for any homogeneous solution $v$, the solution $u_0+v$ is smooth, and subtracting the smooth $u_0$ shows that $v$ is smooth. Hence All inhomogeneous solutions are smooth functions exactly when all homogeneous solutions are smooth functions. This is a global statement on ${\mathbb{R}}^n$ for the stated data; no additional local regularity theorem is being assumed.

Pair the off-circle roots of a polynomial using reciprocal-conjugate root pairing, and split each unit-circle root of a polynomial's even multiplicity equally between the two members of a pair. The fundamental theorem of algebra and the leading coefficient $p_d$ give $P(z)=p_d{\prod }_{i=1}^d(z-{\zeta }_i)(z-1/\overline{{\zeta }_i})$. None of the selected ${\zeta }_i$ is zero.

On the unit circle, the identityturns $P(z)/z^d$ intoAt a point of the unit circle outside the finite set of roots of a polynomial, the product is positive and $p(z)$ is nonzero and nonnegative. Its ratio to the product is therefore real and strictly positive. This proves $c>0$, even though the algebraic expression initially permits a complex constant. ConsequentlyThis proves the Fejér–Riesz theorem for a nonzero trigonometric polynomial of actual order $d$. A positive constant has a constant square-root factor, and the identically zero trigonometric polynomial has $q=0$; if $p_d=0$ for a specified upper order $d$, reduce to the actual order first.

Although the PDF permits assuming even multiplicity, there is a short proof of even multiplicity of unit-circle roots of a nonnegative trigonometric polynomial. The real analytic function $p(e^{i\theta })\ge 0$ cannot have a zero of odd order. Near $\zeta =e^{i{\theta }_0}$, $e^{i\theta }-\zeta$ has a simple zero, and the nonzero factor $e^{-id\theta }$ leaves the zero order of $P(e^{i\theta })$ unchanged. Hence the multiplicity of a root must be even.

Here the coefficient fields are $\mathbb{R}$ and $\mathbb{C}$. A polynomial ring in one variable over a field is a Euclidean domain, hence a principal ideal domain; its nonzero prime ideals are generated by irreducible polynomials. The fundamental theorem of algebra says that all irreducible polynomials over $\mathbb{C}$ are linear. Combining it with complex conjugation shows that the monic irreducible polynomials over $\mathbb{R}$ are exactly the linear polynomials and quadratics with no real root. ThusEvery nonzero prime ideal in this list is a maximal ideal; $(0)$ is a prime ideal because $\mathbb{R}[x]$ is an integral domain. Likewise,The map is contraction of an ideal along the inclusion. It sends $(0)$ to $(0)$. For $z=a+ib$, the contraction of $(x-z)$ consists of the real polynomials vanishing at $z$. If $b=0$, the factor theorem gives the ideal $(x-a)$. If $b\ne 0$, a real polynomial vanishing at $z$ also vanishes at its complex conjugate $\overline{z}$, so it is divisible by $(x-z)(x-\overline{z})=(x-a{)}^2+b^2$. Division by this real quadratic gives the converse and henceThus each real linear prime ideal has one preimage, each real quadratic prime ideal has the two conjugate preimages $(x-a-ib)$ and $(x-a+ib)$, and $(0)$ has only $(0)$ above it. In particular, $f$ is surjective.

ChooseThe Leibniz formula for determinants shows that its leading term is $(-z{)}^n$, as required. A scalar $\lambda$ is an eigenvalue exactly when $A-\lambda I$ has a nonzero kernel, equivalently when it is singular, soThe fundamental theorem of algebra gives a complex root of the degree-$n$ polynomial ${\chi }_A$, so every nonempty complex square matrix has at least one complex eigenvalue.

Schur lemma over $\mathbb{C}$ states that an intertwining operator $T:V\to W$ between finite-dimensional irreducible complex representations is either zero or an isomorphism. In particular, every intertwining endomorphism of $V$ is a scalar multiple of the identity.

Indeed, $\ker T\subseteq V$ and $\mathrm{im}T\subseteq W$ are invariant subspaces. If $T\ne 0$, irreducibility gives $\ker T=0$ and $\mathrm{im}T=W$, so $T$ is an isomorphism. For an endomorphism $T:V\to V$, the fundamental theorem of algebra gives an eigenvalue $\lambda \in \mathbb{C}$. Then $T-\lambda I$ is an intertwiner with nonzero kernel, so the first part forces $T-\lambda I=0$. Hence $T=\lambda I$, completing the Proof of Schur lemma.

A point is ramified with multiplicity $m_i$ exactly when it is a zero of $f^{\prime }$ of multiplicity $m_i-1$. Since $f^{\prime }$ has degree $d-1$, the fundamental theorem of algebra gives ${\sum }_i(m_i-1)=d-1$.

Conversely, choose distinct complex numbers $a_i$ and put $g(z)=c{\prod }_i(z-a_i{)}^{m_i-1}$ with $c\ne 0$. The assumption makes $g$ a polynomial of degree $d-1$. Any antiderivative $f$ of $g$ has degree $d$ and precisely the prescribed ramification multiplicities.

An algebraic closure of a field $K$ is an algebraically closed field that is an algebraic extension of $K$.

By definition, every element ofis algebraic over $\mathbb{Q}$, and the question allows us to use that this set is a field. It remains to prove that it is algebraically closed.

Let $f\in \overline{\mathbb{Q}}[X]$ be nonconstant. Its finitely many coefficients generate a finite extension $E/\mathbb{Q}$. By the fundamental theorem of algebra, $f$ has a root $z\in \mathbb{C}$. Since $z$ is a root of a polynomial over $E$, it is algebraic over $E$; since $E/\mathbb{Q}$ is algebraic, transitivity of algebraic extensions makes $z$ algebraic over $\mathbb{Q}$. Thus $z\in \overline{\mathbb{Q}}$. Repeating after division by $X-z$ shows that $f$ splits there, so $\overline{\mathbb{Q}}$ is algebraically closed and is an algebraic closure of $\mathbb{Q}$.