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Normalize the weight-four and weight-six Eisenstein series to have constant term one. Their Fourier expansions beginTheir absolutely convergent lattice sums transform by modular weights four and six under reindexing by the modular group, and their Fourier expansions prove holomorphy at its modular cusp. Thus they are elements of $M_4,M_6$.

We use the valence formula for the modular group: for a nonzero weight-$k$ form,The sum uses one point of each ordinary orbit. This is the argument principle on a truncated fundamental region: paired vertical integrals cancel, the circle pairing $f(-1/z)=z^{k}f(z)$ contributes the modular weight term, and local quotient coordinates at the two elliptic points of a modular curve modular weight their orders by $1/2,1/3$. All orders are nonnegative for a holomorphic modular form.

Define the modular discriminant byIt is a weight-twelve cusp form. Its order at infinity is one, exhausting the valence total $12/12=1$. Hence $\Delta$ has no zeros in $\mathbb{H}$. Therefore every weight-$k$ cusp form is divisible by $\Delta$, with quotient holomorphic on $\mathbb{H}$ and at the modular cusp:Negative modular weights vanish by the valence formula, and odd modular weights vanish because $-I$ acts by $(-1{)}^k$. Weight-zero forms are constant: they descend holomorphically to the compactified modular curve, and the maximum modulus principle applies. There are no weight-two forms: $f(i)=i^{2}f(i)=-f(i)$ forces a zero at $i$, whose valence contribution $1/2$ exceeds $2/12$.

Every nonnegative even modular weight other than two is $4a+6b$ with $a,b\ge 0$. If $k\equiv 0(\mathrm{mod}4)$, take $b=0$; if $k\equiv 2(\mathrm{mod}4)$ and $k\ge 6$, take $b=1$. For $f\in M_k$, choose such a monomial $h=E_4^a{E}_6^b$, whose constant term is one. Subtracting the constant term $c$ of $f$ gives $f-ch\in S_k$, soInduction on modular weight, together with $1728\Delta =E_4^3-E_6^2$, proves that every modular form is a polynomial in $E_4,E_6$.

For algebraic independence, first fix one modular weight $w$. All solutions of $4a+6b=w$ have the same residue of $a$ modulo three and the same parity of $b$. Consequently their monomials can be writtenfor suitable $a_0,b_0,L$. On an open set where the common factor and $E_6$ are nonzero, a linear relation would give a polynomial relation inThis is nonconstant. By the open mapping theorem, a polynomial vanishing on all its values must be the zero polynomial. Thus same-weight monomials are linearly independent.

Different modular weights cannot cancel either. Suppose ${\sum }_w{P}_w(E_4,E_6)=0$ with each $P_w$ weighted homogeneous. Transform by ${\gamma }_n=\left(\begin{array}{cc}1& 0\\ n& 1\end{array}\right)$. At any fixed $z\in \mathbb{H}$ this givesThe numbers $nz+1$ are infinitely many distinct values, so the polynomial in that number vanishes identically, and every $P_w(z)=0$. The same-weight independence then sets every coefficient to zero. This is the graded weights separate analytic polynomial relations argument. Hencewith algebraically independent generators. Multiplying the two Eisenstein series by their nonzero lattice-normalization constants does not alter either generation or independence.

Write $d\mu =dxdy/y^2$ and distinguish the all-pairs Eisenstein series in this problem from its primitive versionSeparating each nonzero integer pair into a positive integer multiple of a primitive pair givesThe factor two accounts for the two signs of a primitive pair. It must not be omitted in the Rankin–Selberg unfolding.

Because $f\overline{g}{y}^k$ and $d\mu$ are invariant, unfolding its integral against $E_0$ produces the strip $0\le x<1$, $y>0$:The $x$ integral of the product Fourier expansions is ${\sum }_{n\ge 1}a(n)\overline{b(n)}{e}^{-4\pi ny}$. The Gamma function integral then givesThese steps first follow by absolute convergence sufficiently far right. For $f=g$, positive unfolding and modular cusp decay show the square-coefficient series converges for $\mathrm{Re}s>k$; the Cauchy-Schwarz inequality gives the same absolute-convergence range for the mixed series.

Put $w=s-k+1$ and defineThis is the zeta-completed Rankin-Selberg coefficient series. Equivalently, solving this formula for $F(s)$ expresses the requested series directly in terms of the all-pairs Eisenstein integral.

The exponential decay of cusp forms at infinity dominates the polynomial growth in $y$ of the Eisenstein series and its parameter derivatives, locally uniformly away from its poles. Thus its stated continuation can be passed through the integral. The completed series is meromorphic on $\mathbb{C}$, with possible simple poles only at $s=k-1,k$, and its functional equation isIndeed $w\mapsto 1-w$ corresponds to $s\mapsto 2k-1-s$. If $⟨f,g⟩=0$, both pole residues vanish and the completion is entire.

The normalization of the poles can also be made explicit. The constant term of the given Eisenstein series isThe second term has residue one at $w=1$, since the pole of $\zeta (2w-1)$ has residue $1/2$; the functional equation gives residue minus one at $w=0$. Hence the residues of $\mathcal{R}$ at $k$ and $k-1$ are respectively $⟨f,g⟩$ and $-⟨f,g⟩$, the Petersson inner product.

It is important to separate this completed statement from the raw $F$. Inverting the completion gives meromorphic continuation of $F$, but zeros of $\zeta (2s-2k+2)$ in the denominator can give additional possible poles unless canceled by the integral. Nor does $F$ itself satisfy the unweighted symmetry $s\leftrightarrow 2k-1-s$. At the two displayed completion poles, for a nontrivial modular cusp modular weight,The latter is finite because the pole of $\Gamma (w)$ cancels that of the integral at $w=0$, using $\zeta (0)=-1/2$. In particular, taking $f=g\ne 0$ gives a raw-series pole at $k$ and regularity at $k-1$. The direct analogue of the stipulated two-pole Eisenstein properties is therefore the completed function $\mathcal{R}$, not the raw series with all completion factors suppressed.

Finally suppose the two $L$-series have Euler products. With first coefficients normalized to one, their coefficients are multiplicative, so $a(n)\overline{b(n)}$ is multiplicative too. Thereforeinitially in its absolute-convergence half-plane. For unnormalized forms multiply the product by $a(1)\overline{b(1)}$ and use normalized coefficients inside it.

For the usual degree-two Hecke products of Question 3, take roots ${\alpha }_p,{\beta }_p$ with sum $a(p)$ and product $p^{k-1}$, and ${\gamma }_p,{\delta }_p$ with sum $b(p)$ and the same product. The prime-power recurrence gives $a(p^r)=({\alpha }_p^{r+1}-{\beta }_p^{r+1})/({\alpha }_p-{\beta }_p)$ and the analogous expression for $b$. Multiplying and summing four geometric series proves the Euler factor of a coefficientwise product of Hecke eigenforms:Repeated roots are handled by continuity, or directly by the recurrence. Multiplication by $\zeta (2s-2k+2)$ cancels the local numerators and yields the degree-four convolution product. This also explains why the zeta factor belongs in the natural analytic completion.

Let $\mathrm{Par}$ be the set of finite weakly decreasing sequences of positive integers, together with the empty sequence; equivalently pad them indefinitely with zeros. The size is $|\lambda |={\sum }_i{\lambda }_i$, the length counts positive parts, and $\mathrm{Par}(n)$ consists of size-$n$ integer partitions. Their Young diagrams have cells $(i,j)$ with $1\le j\le {\lambda }_i$.

Diagram containment is the partial order $\mu \subseteq \lambda$ defined by ${\mu }_i\le {\lambda }_i$ for every $i$. Componentwise minimum and maximum give its meet and join, so this is Young's lattice. Its covers add one cell. On $\mathrm{Par}(n)$ the dominance order on partitions isIt is not diagram containment: distinct equal-size diagrams cannot contain one another. For the reverse lexicographic order on partitions, use the descending-dictionary convention: $\lambda {<}_R\mu$ when the first unequal pair of parts has ${\lambda }_i>{\mu }_i$. Thus for size four its increasing list is $(4),(3,1),(2,2),(2,1,1),(1,1,1,1)$. This is a total order opposite to increasing dictionary order on integer partitions, not the last-differing-exponent order on polynomial monomials.

Define the elementary symmetric functions and complete homogeneous symmetric functions byFor a partition, write $e_{\lambda }={\prod }_i{e}_{{\lambda }_i}$ and $h_{\lambda }={\prod }_i{h}_{{\lambda }_i}$. The monomial symmetric function $m_{\lambda }$ sums each monomial with exponent multiset $\lambda$ once.

Here is a proof of the Fundamental theorem of symmetric polynomials sufficient for the stable algebra. In $N$ variables, the lexicographically largest monomial of a symmetric homogeneous polynomial has exponents ${\lambda }_1\ge \cdots \ge {\lambda }_N\ge 0$. Otherwise exchanging an inverted pair of variables produces a larger monomial with the same coefficient. The polynomialhas that monomial as its largest one, with coefficient one. Subtract a suitable multiple and continue; there are only finitely many monomials of the fixed degree, so the procedure terminates and proves generation. Distinct products $e_1^{a_1}\cdots e_N^{a_N}$ have distinct largest exponents $(a_1+\cdots +a_N,a_2+\cdots +a_N,\dots ,a_N)$. The largest such monomial in any nonzero polynomial relation cannot cancel, proving algebraic independence. Stabilizing in each degree, where only $e_r$ with $r$ no greater than that degree can occur, gives

The two generating functions satisfySince the $e_r$ freely generate, there is a unique algebra homomorphism $\omega$ with $\omega (e_r)=h_r$. Apply it to the last identity:It follows that $\omega (h_r)=e_r$ and ${\omega }^2(e_r)=e_r$. Hence the symmetric-function involution is an involutory automorphism, and

For the forgotten coefficients, it is useful to derive the required duality explicitly. Put $p_r={\sum }_i{x}_i^r$. The logarithmic generating-function identities giveThe transition $p_r=r{h}_r+$ a polynomial in lower-index $h$'s is triangular and invertible over $\mathbb{Q}$, so the $p_r$ are also free generators and $p_{\lambda }$ is a basis. Define the Hall inner product of symmetric functions byExpand the same formal kernel in two ways:The last expansion is a reproducing kernel for this inner product. Comparing it with the $h,m$ expansion gives $⟨h_{\lambda },m_{\mu }⟩={\delta }_{\lambda \mu }$. One can see this without assuming any Schur orthogonality: pair either kernel with $h_{\nu }(y)$; the reproducing property gives $h_{\nu }(x)$, and uniqueness of its $h$ expansion proves the duality. The signs in $\omega (p_{\lambda })$ show that $\omega$ is an isometry; together with ${\omega }^2=1$ this makes it self-adjoint.

If $n=|\lambda |$ and $\ell =\ell (\lambda )$, define the forgotten symmetric function $f_{\lambda }=(-1{)}^{n-\ell }\omega (m_{\lambda })$. Its monomial coefficient is consequentlyIt remains to compute this coefficient rather than merely invoke duality. Expand the formal inverse $E(t)=H(-t{)}^{-1}$ as a geometric series in $H(-t)-1$:In $e_{\mu }={\prod }_j{e}_{{\mu }_j}$, choose a positive composition of each ${\mu }_j$. A term is $h_{\lambda }$ precisely when the concatenated list is a distinct rearrangement $\alpha$ of the parts of $\lambda$. Its boundaries at the end of each chosen composition are exactly the prescribed prefix sums of $\mu$. Conversely such a rearrangement uniquely determines the compositions by cutting at those prefix sums. Every one of its signs is $(-1{)}^{\sum {\mu }_j-\ell }=(-1{)}^{n-\ell }$, canceled by the prefactor in $a_{\lambda \mu }$. ThusIdentical parts are not artificially distinguished. For example $f_{(2,1)}=m_{(2,1)}+2m_{(3)}$: only the order $(2,1)$ refines $(2,1)$, while both orders refine the single block $(3)$.

In $N$ variables, pad $\lambda$ to length $N$ and set $\delta =(N-1,N-2,\dots ,0)$. Define the alternantThe last equality is the Vandermonde determinant with the indicated descending-exponent convention.

We use the Jacobi–Trudi identity proved above and give the determinant step explicitly. LetPut $b_i={\lambda }_i+N-i\ge 0$. The generating-function identityshows, by extracting the coefficient of $t^{b_i}$, thatWhen $\lambda$ is empty, $J_{ij}=h_{-i+j}$ is upper triangular with diagonal one, and the product matrix is $[x_k^{N-i}]$. Thus $\det B=a_{\delta }$. Taking determinants for general $\lambda$ gives $s_{\lambda }{a}_{\delta }=a_{\lambda +\delta }$ and proves the bialternant formulaThe equality before division is polynomial. Consequently the quotient extends at coincident variables as the Schur polynomial.

For the Fourier minors, choose a square submatrix of size $d\le p$, with distinct row residues $r_1,\dots ,r_d$ and column residues. Sort the columns into decreasing exponents $b_1>\cdots >b_d$ in $\{0,\dots ,p-1\}$; this changes a determinant only by sign. PutBecause the $b_i$ are distinct nonnegative integers, $b_i\ge d-i$, and $b_i-b_{i+1}\ge 1$. Hence $\lambda$ is a partition. With $x_j={\zeta }^{r_j}$, transposition of the submatrix and the bialternant formula giveThe $x_j$ are distinct, so the Vandermonde factor does not vanish.

To prove that the Schur factor cannot vanish, work in $\mathbb{Z}[\zeta ]$ and reduce modulo $1-\zeta$. Since the prime cyclotomic polynomial is ${\Phi }_p(t)=1+t+\cdots +t^{p-1}$ and ${\Phi }_p(1)=p$,All $x_j$ reduce to one, and the Schur factor reduces to $s_{\lambda }(1^d)$. We need its value, not an assumption that the specialization is nonzero. The Schur evaluation at all ones formula isFor completeness, substitute $x_j=e^{t{y}_j}$ with distinct $y_j$ into the bialternant. In the exponential expansions, the first nonzero determinant coefficient uses exactly the distinct powers $0,1,\dots ,d-1$ and has degree $d(d-1)/2$ in $t$. The determinant in the $y_j$ and the factorial factors are common to numerator and denominator. The remaining ratio is the Vandermonde product in $b_i$ divided by that in $d-i$, giving the formula as $t\to 0$. Its value is an integer because $s_{\lambda }$ has integer tableau coefficients.

Every positive difference $b_i-b_j$ and $j-i$ is strictly less than $p$. Thus each is a unit modulo $p$, and the displayed product has nonzero residue. A zero Schur factor would have zero residue, a contradiction. This proves the prime Fourier matrix minor theorem:

Finally use the one-row shape $\lambda =(r)$, for which $s_{(r)}=h_r$. Its numerator determinant has first-row exponents $N-1+r$ and subsequent exponents $N-2,\dots ,0$. Expanding along the first row gives cofactors equal, up to their signs, to the Vandermonde determinant in all variables except $x_k$. Sincethe cofactor sign $(-1{)}^{1+k}$ cancels the indicated sign in the quotient. ThereforeThis rational expression is first computed for pairwise distinct variables; its sum is the polynomial $h_r$, giving the continuous polynomial extension when variables coincide.

For genuine degrees three and two, the resultant is the Sylvester matrix determinantEquivalently, if $r_1,r_2,r_3$ are the roots of $p$, counted with multiplicity, then $R(p,q)=a^2{\prod }_{i}q(r_i)$. Thus $R(p,q)\ne 0$ means that the polynomials have no common root, or equivalently are coprime over $\mathbb{C}$. One can also see this from the matrix: its singularity gives a nonzero relation $Ap+Bq=0$ with $\mathrm{deg}A<2$, $\mathrm{deg}B<3$. If $p,q$ were coprime, $p\mid B$ would force $B=0$, then $A=0$, a contradiction. Conversely, a common factor supplies such a relation after dividing $p,q$ by that factor. When a leading coefficient is zero, use the resultant for the actual polynomial degrees rather than assuming the displayed fixed-degree criterion unchanged.

An algebraic addition theorem for a meromorphic function $f$ means a nonzero polynomial $P(U,V,W)$ withidentically where the values are finite. A local relation extends meromorphically to the other regular arguments.

For a polynomial $f$ of positive degree $d$, let $z_i$ be the $d$ roots of $f(z)-U$ and $w_j$ the $d$ roots of $f(w)-V$. FormIts coefficients are separately symmetric polynomials in the two sets of roots. The Fundamental theorem of symmetric polynomials and the fixed nonzero leading coefficient of $f$ show they are polynomials in $U,V$. The expression is monic of degree $d^2$ in $W$, so it is nonzero. If $U=f(z)$, $V=f(w)$, one factor vanishes at $W=f(z+w)$. This proves the addition relation, including at exceptional multiple-root values by polynomial identity. A constant polynomial instead has the relation $W-f(0)=0$.

For the elliptic case, let $E=\mathbb{C}/\Lambda$. A nonconstant elliptic function $f$ defines a finite morphism $E\to {\mathbb{P}}^1$, so the function field of $E$ is a finite algebraic extension of $\mathbb{C}(f)$. The elliptic function-field decomposition expresses it as $\mathbb{C}(\wp ,{\wp }^{\prime })$, withThe Weierstrass addition formula makes translation algebraic:Differentiating and using the cubic differential equation also makes ${\wp }^{\prime }(z+w)$ rational in the four separate values. Therefore $f(z+w)$ lies in a finite algebraic extension of $\mathbb{C}(f(z),f(w))$. Its algebraic equation, after clearing denominators, is the required polynomial relation. This sketches why every elliptic function, not only $\wp$, has an algebraic addition theorem. Constants are already covered.

For the final assertion, consider $f(z)=e^{e^z}$, an entire function. If $c$ is a period, then $e^{e^z(e^c-1)}=1$ for every $z$. The continuous exponent must be a constant integer multiple of $2\pi i$; its derivative forces $e^c=1$. Hence its periods are exactly $2\pi i\mathbb{Z}$, making it a simply periodic function.

Suppose it had an algebraic addition theorem $P$. There are only finitely many values $V_0$ for which $P(U,V_0,W)$ is identically zero: a nonzero coefficient polynomial in $V$ already bounds this exceptional set. Choose a positive irrational $\alpha$ with $e^{\alpha }$ outside this set, and set $c=\log \alpha$. Specializing $w=c$ gives a nonzero polynomial $Q(U,W)=P(U,e^{\alpha },W)$ and, with $t=e^z$,All exponents with nonzero coefficients are distinct, by irrationality of $\alpha$. Divide by the exponential with largest real exponent and let $t\to +\infty$; its nonzero coefficient would have to tend to zero. This contradiction proves that simple periodicity does not imply an algebraic addition theorem. It is the simply periodic entire function without an algebraic addition theorem example.