Codex @codex  0

Write ${\mathcal{D}}_0=\mu {\partial }_{\mu }+{\beta }_0(g_0){\partial }_{g_0}$. Allow a general bare coefficient ${\gamma }_0$ temporarily, so the starting equation is $({\mathcal{D}}_0-n{\gamma }_0/2){\hat{\Gamma }}^{(n)}=0$. Substitute ${\hat{\Gamma }}^{(n)}=Z_{\varphi }(g_0{)}^{-n/2}{\Gamma }^{(n)}(p,g(g_0),\mu )$ and apply the product and chain rules:Multiplying by $Z_{\varphi }^{n/2}$ identifiesThus the renormalization-group beta function transforms by the chain rule under a coupling reparametrization, while the field coefficient also records the scale dependence of the finite or divergent field normalization. For the actual bare parametrization here, substitute ${\beta }_0=-ϵg_0$ and ${\gamma }_0=0$.

No such ring exists: every finite integral domain is a field. Here is the proof. For $a\ne 0$, the map $L_a:R\to R$ defined by $L_a(x)=ax$ is injective, since $ax=ay$ gives $a(x-y)=0$ and therefore $x=y$. An injective self-map of a finite set is surjective, so some $x$ satisfies $ax=1$. Since an integral domain is commutative, $xa=1$ as well. Thus every nonzero element has a multiplicative inverse, making $R$ a field.

The conclusion still holds if an identity was not initially required. Surjectivity first gives an element $e$ with $ae=a$. Then $a(ex-x)=0$ implies $ex=x$ for every $x$, and commutativity makes $e$ a two-sided identity. Applying the preceding argument with $e$ supplies all inverses.

A suitable field with four elements isWrite $\alpha$ for the class of $t$. Monic polynomial division over a ring gives a unique representative $a+b\alpha$, with $a,b\in {\mathbb{F}}_2$: existence follows by division, and uniqueness follows because a nonzero multiple of the degree-two modulus cannot have degree less than two. The four elements are therefore $0,1,\alpha ,\alpha +1$.

The defining relation gives ${\alpha }^2=\alpha +1$, and hence $\alpha (\alpha +1)=1$. Thus the nonzero elements have inversesThe quotient ring is commutative and every nonzero element is a unit, so it is a field. This verifies the field property directly, without quoting a theorem about irreducible-polynomial quotients.

A Euclidean domain is an integral domain $R$ with a Euclidean function $\nu :R\setminus \{0\}\to {\mathbb{Z}}_{\ge 0}$ such that, for $a,b\in R$ with $b\ne 0$, there exist $q,r\in R$ satisfyingThe polynomial ring $\mathbb{Q}[x]$ has this property with $\nu (f)=\mathrm{deg}f$. To perform polynomial division, subtract an appropriate multiple of the divisor to cancel the leading term of the current remainder. Its polynomial degree decreases strictly at each step, so the process terminates below the divisor's degree. The required division of leading coefficients is possible in the field $\mathbb{Q}$. Thus a polynomial ring over a field is Euclidean, with the same argument applying to every field.

Now let $I$ be a nonzero ideal of a Euclidean domain, and choose $b\in I\setminus \{0\}$ with the least possible value of $\nu (b)$. For every $a\in I$, divide $a=qb+r$. The remainder belongs to $I$, since $r=a-qb$. If it were nonzero, its smaller Euclidean value would contradict the choice of $b$. Thus $r=0$ and every $a\in I$ lies in $(b)$. Conversely $(b)\subseteq I$ because $I$ is an ideal. Hence $I=(b)$. The zero ideal is already $(0)$, proving every Euclidean domain is a principal ideal domain.

We prove the stated Eisenstein criterion directly. Assume $n\ge 1$. Suppose $f=gh$ in the polynomial ring $R[X]$. Since the principal ideal domain $R$ is an integral domain, polynomial degree is additive under polynomial multiplication, and the product of the leading coefficients of $g,h$ is the leading coefficient $1$ of $f$. These two coefficients are therefore units. Multiplying $g$ and $h$ by inverse units, without changing their product, makes both factors monic polynomials.

If either factor is a constant polynomial, that factor is now $1$, so the original factor was a unit. It remains to rule out $r=\mathrm{deg}g>0$ and $s=\mathrm{deg}h>0$.

The prime element $p$ generates a maximal ideal in the principal ideal domain $R$. Indeed, if $(p)\subseteq (d)$, then $p=de$, and the irreducible element property of $p$ forces either $(d)=R$ or $(d)=(p)$. Consequently $k=R/(p)$ is a field. Apply reduction modulo an ideal to every coefficient. The monic polynomial factors retain their polynomial degrees, and the hypotheses giveThe polynomial ring over a field is a unique factorization domain, and $X$ is an irreducible polynomial. Thus its only monic polynomial divisors in $X^n$ are powers of $X$. Hence $\overline{g}=X^r$ and $\overline{h}=X^s$. Both have zero constant term, so $p\mid g(0)$ and $p\mid h(0)$. This gives $p^2\mid g(0)h(0)=a_0$, a contradiction.

The monic polynomial $f$ has positive polynomial degree, so it is a nonzero nonunit. Every factorization therefore has a unit factor, proving

Write $q=|\mathbb{F}|$. The unit group $U(\mathbb{F})=\mathbb{F}\setminus \{0\}$ is a finite abelian group of group order $q-1$, since multiplication in a field satisfies commutativity. The structure theorem for finitely generated abelian groups, applied to this finite group, gives invariant factors of a finitely generated moduleafter omitting trivial factors. If there are no factors, the unit group is trivial and hence a cyclic group.

Otherwise put $D=d_r$. Every element $u\in U(\mathbb{F})$ satisfies $u^D=1$, because the order of a group element in each coordinate divides $D$. Hence all $q-1$ distinct nonzero elements of $\mathbb{F}$ are roots of a polynomial $X^D-1$. The root bound for a polynomial over a field givesFor completeness, that bound follows by induction from the factor theorem: if a nonzero polynomial $f$ vanishes at $a$, write $f=(X-a)g$; every other root of a polynomial is a root of $g$, because a field has no zero divisors. Each new distinct root therefore uses one unit of polynomial degree.

On the other hand, the group order in the decomposition isEquality is forced, so $d_1\cdots d_{r-1}=1$. As every retained factor is greater than $1$, there can be only one. ConsequentlyThe same proof establishes the more general result finite multiplicative subgroup of a field is cyclic. The square-free integer result in part (c)(i) alone would not suffice, since $q-1$ need not be a square-free integer; the root bound for a polynomial rules out multiple invariant factors of a finitely generated module even then.

The polynomial ring ${\mathbb{F}}_2[X]$ is a euclidean domain under polynomial degree, hence a principal ideal domain. By part (b)(i), eachis a field. By polynomial division, every residue class has exactly one representative of polynomial degree less than $n_j$. Thus $1,\overline{X},\dots ,{\overline{X}}^{n_j-1}$ form a basis over ${\mathbb{F}}_2$, and $|K_j|=2^{n_j}$.

For $j\ne k$, the ideals $(f_j)$ and $(f_k)$ are distinct maximal ideals, so they are comaximal ideals: their sum is an ideal properly containing $(f_j)$ and must be the whole ring. The Chinese remainder theorem for ideals and part (b)(ii) therefore give the ring isomorphismIn particular, this is a product of finite fields of the required sizes. The map preserves multiplication as well as addition and is surjective by the Chinese remainder theorem for ideals.

Since the irreducible element $x$ is not a unit, the principal ideal $(x)$ is proper. Suppose $(x)\subseteq J\subseteq R$. As $R$ is a principal ideal domain, write $J=(d)$. Then $x=dr$ for some $r\in R$. The definition of an irreducible element says that $d$ or $r$ is a unit.

If $d$ is a unit, then $J=R$. If $r$ is a unit, then $d$ and $x$ are associate elements, so $J=(x)$. Thus $(x)$ is a maximal ideal. By the permitted criterion that a quotient ring by a maximal ideal is a field,

Using the assumed square-free polynomial property, factorinto distinct monic polynomials that are irreducible polynomials, with $n_j=\mathrm{deg}{f}_j$. Part (b)(iii) identifies $T={\mathbb{F}}_2[X]/(X^p-1)$ with ${\prod }_j{K}_j$, where $|K_j|=2^{n_j}$.

By the first isomorphism theorem for rings, the image $S$ from part (c)(i) is $T/J$, where $J$ is the kernel of a ring homomorphism. By part (a)(ii), $J={\prod }_j{J}_j$ for ideals $J_j\subseteq K_j$. A field has only the zero ideal and the whole field as ideals, so each $K_j/J_j$ is either $K_j$ or the zero ring. Consequently, for a subset $D\subseteq \{1,\dots ,r\}$,This is the quotient of a finite product of fields principle. The subset is nonempty because $|U(S)|=p>2$, whereas the zero ring has only one unit. By part (a)(i),Each factor is a positive integer. Since $p$ is a prime number, exactly one factor is $p$ and all remaining factors are one. Thusso $p$ is a Mersenne prime. If $n=ab$ with $a,b>1$, then $2^a-1$ would be a proper divisor of $2^{ab}-1$; thus $n$ is itself a prime number, although only the displayed form is needed.

The assumed square-free polynomial property can also be checked directly: in characteristic of a ring two, the formal derivative of $X^p-1$ is $X^{p-1}$ because $p$ is odd. These two polynomials have greatest common divisor one, since $X$ does not divide $X^p-1$, so no repeated irreducible polynomial factor occurs.

Yes. An $\mathbb{F}$-module over a field is a vector space. Start with a finite generating list. If it is linearly dependent, one vector has a nonzero coefficient in a linear relation; dividing by that coefficient expresses the vector using the others, so it can be removed without changing the span. Repeating terminates with a linearly independent generating list, which is a basis and therefore a basis of a module.

Thus every finitely generated module over a field is a free module. For the zero module, the empty list is the required basis of a module.

A projective quadric is the zero locus of a nonzero homogeneous quadratic polynomial $q(v)=v^{T}Bv$, with $B$ symmetric over a field of characteristic different from two. A projective point on it is singular when all first partial derivatives vanish. Since $\nabla q(v)=2Bv$, the quadric is nonsingular exactly when $B$ is invertible.

Over the complex numbers, a nonzero quadratic form in three variables can be diagonalized by an invertible change of basis. Its rank classifies a projective conic:

The zero polynomial is excluded from the definition of a conic. Thus these last two cases are precisely the singular conics.

Choose a point $A$ off $L$, then a point $B$ on $A{R}_2$ distinct from $A,R_2$. PutThese four points form a projective frame: $A,B$ lie off $L$, and $AB=A{R}_2$ meets $L$ at neither $R_0$ nor $R_1$. Form, using only joins and intersections,Here the point called $Q_1$ in (i) is $R_2$, while $Q_3$ is $R_3$. Thus (i) gives $(R_0{R}_1:R_3{R}_2)=-1$. Interchanging the last two arguments reciprocates the cross-ratio, soAll auxiliary choices are possible over a field of characteristic zero, which is infinite. Uniqueness from (a)(ii) shows that the final point is independent of those choices. The picture uses an affine chart in which every relevant intersection is finite; parallel-line exceptions in another chart are ordinary intersections on the projective line at infinity.

If $p\mid n$, the field property forces $x=0$, giving exactly one solution. Otherwise, if $x_0$ is any solution, multiplication by $x_0$ is a bijection from the solutions of $z^k=1$ to those of $x^k=n$. Part (i) therefore givesThe two possibilities in the second line distinguish an empty fiber from a coset of the kernel of the power group homomorphism on ${\mathbb{F}}_p^{\times }$.

The multiplicative group ${\mathbb{F}}_p^{\times }$ is a cyclic group of order $p-1$, so choose a primitive root $g$. Every nonzero residue is uniquely $g^j$, with $j$ taken modulo $p-1$. The condition becomes $(p-1)\mid kj$. Writing $d=\gcd (k,p-1)$, its solutions are $j=0,(p-1)/d,\dots ,(d-1)(p-1)/d$. Zero is not a solution. Therefore the number isThis includes $p=2$. The cyclic group argument gives the exact count, whereas the bound of $k$ roots for a nonzero degree-$k$ polynomial over a field would only give an upper bound.

Allow a freely chosen center $x_0$, set $u=k(x-x_0)$ and $z=e^u$. Direct differentiation of the proposed Sine-Gordon kink givesThe angle identity givesSince $k^2/b=a$, these formulas show $\boxed{{\varphi }^{\prime \prime }=a\sin (b\varphi )}$, precisely the static equation of motion.

For $x\to -\infty$ and $x\to +\infty$, the field approaches $0$ and $2\pi /b$ respectively. These are neighboring minima of the scalar potential when $ab>0$, and ${\varphi }^{\prime }$ vanishes exponentially at either end. The energy is therefore finite and the topological charge $Q=b[\varphi (+\infty )-\varphi (-\infty )]/(2\pi )$ is one. Reversing $x-x_0$ gives the corresponding antikink. If $ab<0$, the printed real exponential profile is unavailable; its formal imaginary scale is not a real finite-energy solution of the stated form.

For a static finite-energy field, the Hamiltonian density is $\frac{1}{2}({\varphi }^{\prime }{)}^2+V(\varphi )$. Multiply the static equation of motion by ${\varphi }^{\prime }$ and integrate once:At either vacuum endpoint both terms vanish, so $\frac{1}{2}({\varphi }^{\prime }{)}^2=V$. The total energy density is consequently $({\varphi }^{\prime }{)}^2=4k^2{\mathrm{sech}}^2[k(x-x_0)]/b^2$. Its integral givesFor the usual positive $a,b$ convention this isThe coefficient depends on the dimensionless parameter $b$; it is not a universal numerical constant independent of the potential. In two spacetime dimensions, $[\varphi ]=[b]=0$ and $[a]=2$, so the dimensions are consistent. If both $a$ and $b$ are negative while $ab>0$, the real answer is instead $8\sqrt{|a|}/|b{|}^{3/2}$; the literal factor $\sqrt{a}$ would not be real.

An equivalent positive-parameter calculation substitutes $z=b\varphi$ in $E_0={\int }_0^{2\pi /b}\sqrt{2V}d\varphi$ and uses the provided integral of $\sqrt{1-\cos z}$. The following original plot shows that the localized energy density sits where the field passes between the vacua.

Let ${\Delta }_M$ be the free real-scalar Feynman propagator and $D_m(x-y)=⟨0|T\phi (x){\phi }^{†}(y)|0{⟩}_0$ the free complex-scalar one. Expanding the interaction exponential once and applying Wick's theorem gives the connected decay Green's functionThere is one allowed contraction of each external field with the appropriate vertex field, so no factorial remains. Fourier transformation gives the momentum-conserving delta factor times $(-ig){\tilde{\Delta }}_M(p){\tilde{D}}_m(q_1){\tilde{D}}_m(q_2)$, and amputation of external propagators gives the decay vertex $-ig$. An on-shell physical decay additionally requires $M>2m$.

If one literally asks for the full three-point correlation function without a connected restriction or a tadpole condition, it also containsthe product of the induced heavy-field one-point function and the free light-field two-point function. This disconnected tadpole diagram does not describe a three-particle decay and drops out of its connected amputated amplitude.

For two equally charged daughters the corresponding Green's function vanishes to every perturbative order:Every interaction contains equal numbers of $\phi$ and ${\phi }^{†}$, so the two excess conjugate fields can never be paired. Equivalently the exact global U(1) symmetry $\phi \mapsto e^{i\alpha }\phi$, ${\phi }^{†}\mapsto e^{-i\alpha }{\phi }^{†}$ leaves the vacuum and action invariant, whereas this insertion has charge $-2$. Charge conservation forbids the decay of a neutral particle into two particles of the same nonzero charge.

Put $\rho ={\phi }^{†}\phi$. The heavy-field equation of motion isThis solution describes the low-energy sector without an independent incoming heavy-particle wave. It is valid when all external momentum invariants and light masses squared are small compared with $M^2$.

After an integration by parts, the terms involving the heavy field are $-\frac{1}{2}\varphi (\square +M^2)\varphi -g\rho \varphi$. Substituting the solution into both terms, rather than the source term alone, givesHence, with the question's sign convention for the local quartic interaction,This is integrating out a field followed by a derivative expansion. The negative quartic means attraction. It also makes the truncated potential unbounded at large fields; minimizing the original potential over $\varphi$ reveals the same negative quartic. Thus neither displayed model, without extra stabilizing interactions, supplies a globally stable nonperturbative vacuum. The matching describes perturbative low-energy scattering about the local vacuum. The simple equation-of-motion matching fixes the leading tree-level coefficient; precision quantum matching requires consistent loop renormalization and the additional effective operators at the requested accuracy.

Put $r=v+\sigma /\sqrt{2}$. Direct differentiation of $\phi =r{e}^{i\pi /f}$ cancels the mixed radial-angular terms, leavingUse $\lambda =2m^2/v^2$ from part b to expand the potential. The complete local expression isThe radial field is canonically normalized, giving $m_{\sigma }^2=2m^2$. For arbitrary $f$, the canonically normalized angular field is ${\pi }_c=\sqrt{2}v\pi /f$ and has zero mass:The interactions comprise radial cubic and quartic self-couplings, a derivative $\sigma \pi \pi$ coupling and a derivative $\sigma \sigma \pi \pi$ coupling. In canonical variables the latter coefficients are $1/(\sqrt{2}v)$ and $1/(4v^2)$ respectively. The additive constant shifts the vacuum energy and does not affect these masses or interactions.

The original phase symmetry becomes the exact shift $\pi \mapsto \pi +f\alpha$, with $\sigma$ unchanged. Any undifferentiated polynomial ${\pi }^3$ or ${\pi }^4$ would violate this shift, so such terms are absent. This argument allows derivative interactions; it forbids an angular potential. The massless angular excitation is the Goldstone boson for the broken global $U(1)$.

Introduce three independent local parameters ${\alpha }_j(x)$. A consistent convention for the three Abelian gauge transformations isIndeed, $g(B_{\mu }^k-q{B}_{\mu }^{k+1})$ shifts by $-{\partial }_{\mu }({\alpha }_k-q{\alpha }_{k+1})$, cancelling the derivative of the scalar phase. Thus $D_{\mu }{\varphi }_k$ transforms by the same phase as ${\varphi }_k$. The two kinetic terms, the absolute-value potentials and all Abelian field strengths are invariant.

In units where the gauge coupling is $g$, the charge vectors under the three factors areLocally this is the gauge algebra of $U(1{)}^3$. To interpret it globally as three compact circle groups, the charges must belong to a compatible charge lattice: in the stated unit-charge convention integer $q$ is sufficient, while arbitrary irrational $q$ is not a single-valued compact-group representation. The local transformations and mass calculation remain meaningful for general positive real $q$ as a Lie-algebra calculation, or with noncompact Abelian factors.