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Take the even primitive Dirichlet character ${\chi }_{12}$ whose values at $1,5,7,11$ modulo twelve are $1,-1,-1,1$. Its Gauss sum of a Dirichlet character is $\sqrt{12}$. The displayed series $A$ has $A(-1/\tau )=\sqrt{-i\tau }A(\tau )$ and $A(\tau +1)=e^{\pi i/12}A(\tau )$. Thus $A^{24}$ is the normalized weight-twelve modular discriminant. It has no interior zeros by the valence formula for the modular group. Its logarithmic derivative equals $E_2/24$, so integration gives the Dedekind eta function product. The odd character modulo four similarly produces a series $B$ with $B^8=A^{24}$ and leading term $q^{1/8}$; consequently $B=A^3$.

For nonnegative integer $k$ and a finite-index subgroup $\Gamma$, a nonzero modular form has order at infinity, measured in a genuine periodic cusp parameter, at most $k[S{L}_2(\mathbb{Z}):\Gamma ]/12$. The product of its slash translates over the left cosets is a nonzero level-one form; the translates along the infinity-cusp orbit contribute its full local order to this product. The valence formula for the modular group gives the bound. Initial coefficients through that bound therefore give an injective linear map.

The rational space of level-one cusp forms has a full lattice of forms with integral Fourier coefficients. It has finite rank by the valence formula for the modular group, and is preserved by the Fourier coefficients of a composite-index Hecke operator formula. Thus every Hecke operator has a monic integral characteristic polynomial on this lattice, and its eigenvalues are algebraic integers. The common eigenvalue field is totally real by the Petersson inner product self-adjointness.

Choose $4a_j+6b_j=k-12j$ with nonnegative exponents, for every $j$ with $k-12j\ge 0$ and $k-12j\ne 2$. The Eisenstein series and the modular discriminant give modular forms with integer coefficients and distinct first powers of $q$. The valence formula for the modular group shows that these form a basis. Successively subtracting their leading terms proves that integral coefficients up to the greatest such $j$ force all coefficients to be integral. This is more precise than merely spanning by arbitrary monomials in $E_4,E_6$.

The valence formula for the modular group shows that $\Delta =(E_4^3-E_6^2)/1728$ has no zeros in the half-plane. For every $a\in \mathbb{C}$, the weight-twelve form $E_4^3-a\Delta$ has total weighted zero order one. At $a=0$ its sole zero is the order-three zero at the cubic elliptic orbit; at $a=1728$ its sole zero is the order-two zero at the quadratic elliptic orbit. Otherwise it has one simple ordinary zero orbit. Thus each value of the Klein j-invariant occurs on exactly one modular orbit and classifies one homothety of complex lattices.

The modular discriminant has a simple zero at infinity and no zero in the upper half-plane, by the valence formula for the modular group. Multiplication by it identifies weight-$k$ modular forms with weight-$(k+12)$ cusp forms. Conversely dividing a cusp form by it preserves interior holomorphy and removes one order of vanishing at infinity.

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.

Put $q=e^{2\pi i\tau }$ and ${\sigma }_j(n)={\sum }_{d\mid n}{d}^j$. Separate the $m=0$ terms in the lattice Eisenstein series. Differentiating the cosecant partial-fraction identity $2k-2$ times gives, for $\Im z>0$,The initial identity is ${\sum }_n(z+n{)}^{-2}={\pi }^2{\csc }^2(\pi z)=-4{\pi }^2{\sum }_{r\ge 1}r{e}^{2\pi irz}$; differentiation accounts for the factorial and the sign. Pair $m$ and $-m$ and collect the products $mr$. Absolute convergence of the lattice sum for $k\ge 2$, and local uniform convergence of the resulting power series, justify these operations. Thus the requested Fourier expansions areHere $E_{2k}=G_{2k}/(2\zeta (2k))$, and the Bernoulli numbers have convention $B_2=1/6$. The second formula uses Euler's even-value identity for the Riemann zeta function, $2\zeta (2k)=(-1{)}^{k+1}{B}_{2k}(2\pi {)}^{2k}/(2k)!$. In particular $E_4=1+240\sum {\sigma }_3(n)q^n$ and $E_6=1-504\sum {\sigma }_5(n)q^n$.

We need the rational, rather than merely complex, version of the polynomial ring of level-one modular forms. Recall how the valence formula for the modular group supplies its proof. For a nonzero weight-$w$ modular form, with $\rho =e^{2\pi i/3}$,Integrate $f^{\prime }/f$ around a truncated standard fundamental domain of the modular group, indenting around boundary zeros. The paired vertical sides cancel by translation; pair the two halves of the circular edge using $f(-1/\tau )={\tau }^{w}f(\tau )$. The logarithmic derivative of ${\tau }^w$ supplies $w/12$, while the top edge supplies $v_{\infty }(f)$. The half-disc and third-disc at the two elliptic points account for the fractions $1/2,1/3$. The argument principle then gives the formula. All orders in it are nonnegative for a holomorphic modular form. It follows that negative weights have no forms, $M_0=\mathbb{C}$, and $M_2=0$: a weight-two form vanishes at $i$, contributing at least $1/2>2/12$.

Define ${\Delta }_0=(E_4^3-E_6^2)/1728$. It is a weight-twelve cusp form and its expansion starts with $q$. The valence formula for the modular group forces its order at infinity to be exactly one and leaves no interior zeros. Thus division by ${\Delta }_0$ identifies $S_w$ with $M_{w-12}$. Every nonnegative even $w\ne 2$ has a monomial $E_4^a{E}_6^b$ of weight $w$: solve $2a+3b=w/2$, taking $b=0$ when $w/2$ is even and $b=1$ when it is odd and at least three. Subtract the constant coefficient of a modular form times such a monomial and divide by ${\Delta }_0$. Induction on weight proves generation by $E_4,E_6$.

If the original Fourier coefficients are rational, every step preserves rationality: both generators and ${\Delta }_0$ have rational expansions, and formal division by ${\Delta }_0=q+O(q^2)$ does too. This proves the rational structure of level-one modular forms. Write $E_{2k}=\sum r_{a,b}{E}_4^a{E}_6^b$ with $r_{a,b}\in \mathbb{Q}$ and $4a+6b=2k$. Rescaling givesThe Bernoulli number formula above makes each zeta ratio rational: its powers of $\pi$ cancel precisely because the weights agree. This proves the required rational coefficients for the unnormalized lattice Eisenstein sums.

Finally $S_8=S_{10}=0$ because the quotient weights are negative, and $S_{14}=0$ because $M_2=0$. In each of these weights a form is determined by its constant coefficient. The indicated monomials have that coefficient equal to one, so

The product defining the modular discriminant converges locally uniformly for $|q|<1$ and has integral Fourier coefficients. Its logarithmic derivative isTherefore $D\Delta =E_2\Delta$. Comparing the coefficient of $q^n$ gives $n\tau (n)=\tau (n)-24{\sum }_{j=1}^{n-1}{\sigma }_1(j)\tau (n-j)$, orThis also fixes the initial value $\tau (1)=1$.

For the Ramanujan tau congruence modulo five, first justify the link between this product and $E_4,E_6$. Question 1 constructs ${\Delta }_0=(E_4^3-E_6^2)/1728\in S_{12}$ with first coefficient one. Its Serre derivative is in $S_{14}$, which vanishes by the valence formula for the modular group: a nonzero such cusp form would contribute at least $1+1/2>14/12$, since it vanishes at infinity and at $i$. Hence $D{\Delta }_0=E_2{\Delta }_0$. The product and ${\Delta }_0$ have the same differential equation and leading coefficient. Equivalently their quotient has derivative zero near $q=0$ and limit one; the identity theorem givesAlso $D_6{E}_6\in M_8=\mathbb{C}{E}_4^2$. Its constant coefficient is $-1/2$, so the Serre derivative gives $2D{E}_6=E_2{E}_6-E_4^2$.

Work now in the formal power series ring ${\mathbb{F}}_5[[q]]$. The integral expansions give $E_4\equiv 1$ and $E_2\equiv E_6(\mathrm{mod}5)$, because $d^5\equiv d(\mathrm{mod}5)$, $-24\equiv -504\equiv 1$, and $240\equiv 0$. The two identities reduce toThus $3\Delta \equiv -2D{E}_6\equiv 3D{E}_6$, and division by the nonzero element $3$ gives $\Delta \equiv D{E}_6$. Since the coefficient of $q^n$ in $D{E}_6$ is $-504n{\sigma }_5(n)$, we conclude

Extend $\chi$ by zero to integers not coprime to $N$, and write $g(\chi )={\sum }_{a\mathrm{mod}N}\chi (a)e^{2\pi ia/N}$ for its Gauss sum of a Dirichlet character. We use the analytic branch of $\sqrt{-i\tau }$ that is positive when $\tau$ is positive imaginary. With the convention $\hat{F}(t)={\int }_{\mathbb{R}}F(x)e^{-2\pi ixt}dx$, the Gaussian Fourier transform and its differentiated version areThe first formula follows by completing the square, first on the imaginary axis and then by analytic continuation; the second follows by differentiating the Fourier transform in $t$. All the functions are Schwartz functions on the real axis.

For completeness, the finite Fourier transform of a primitive Dirichlet character satisfiesFor a unit $r$ this is substitution by its inverse. If $p\mid (r,N)$, primitivity supplies a unit $u\equiv 1(\mathrm{mod}N/p)$ with $\chi (u)\ne 1$: otherwise the character would factor through modulus $N/p$. Multiplication by $u$ fixes the additive exponential but multiplies the sum by $\chi (u)$, so the sum vanishes. This proves the identity also on nonunits. Finite Parseval identity now gives $|g(\chi ){|}^2=N$, since exactly $\phi (N)$ frequencies have nonzero character values. In particular this Gauss sum of a Dirichlet character is nonzero.

Apply Poisson summation to each residue class: ${\sum }_{j}F(a+Nj)=N^{-1}{\sum }_r{e}^{2\pi ira/N}\hat{F}(r/N)$. Multiply by $\chi (a)$ and sum over $a$. The formulas above give the two character theta transformationsandIn the odd case the prefactor originally reads $g(\chi )/(N^2\tau \sqrt{-i\tau })$; using $\tau =i(-i\tau )$ explains the factor $i$. Opposite parity makes the corresponding series identically zero by pairing $n,-n$. This is why the weight-three-halves series is needed for odd Dirichlet characters.

To deduce the products from these transformations, use two specific primitive Dirichlet characters. The even character ${\chi }_{12}$ takes values $1,-1,-1,1$ on $1,5,7,11$ modulo twelve. It is primitive: its values do not factor through six or four, hence through no proper divisor of twelve. Its Gauss sum of a Dirichlet character is $2\sqrt{3}=\sqrt{12}$. PutThe even transformation gives $A(-1/\tau )=\sqrt{-i\tau }A(\tau )$; also $A(\tau +1)=e^{\pi i/12}A(\tau )$, since each nonzero term has $n^2\equiv 1(\mathrm{mod}24)$. Thus $A^{24}$ transforms as a weight-twelve modular form under the generators $S,T$ of the modular group. It is holomorphic and starts with $q$, so it is a normalized cusp form. The proof in Question 1 shows $S_{12}=\mathbb{C}{\Delta }_0$, giving $A^{24}={\Delta }_0$. In particular $A$ has no interior zeros, by the valence formula for the modular group.

The Serre derivative argument in Question 2 gives $D{\Delta }_0=E_2{\Delta }_0$. Hence $DA/A=E_2/24$. The analytic product $P(\tau )=q^{1/24}{\prod }_{n\ge 1}(1-q^n)$ is nonzero and has the same logarithmic derivative, since $DP/P=1/24-\sum {\sigma }_1(n)q^n$. Therefore $A/P$ is constant on the connected complex upper half-plane; its limit at infinity is one. This proves $A=P$, without assuming either of the product identities being deduced. Pair the terms $n,-n$ in $A$, choosing representatives $n=6r+1$. Since ${\chi }_{12}(6r+1)=(-1{)}^r$, we getThis is the Euler pentagonal identity, derived by the character theta proof of the eta product.

For the second product use the odd primitive character ${\chi }_4(1)=1$, ${\chi }_4(3)=-1$, with Gauss sum of a Dirichlet character $2i$. SetThe odd transformation gives $B(-1/\tau )=(-i\tau {)}^{3/2}B(\tau )$ and the series gives $B(\tau +1)=e^{\pi i/4}B(\tau )$. Thus $B^8$ is another normalized weight-twelve cusp form, and $B^8={\Delta }_0=A^{24}$. Because $A$ is nonzero, $B/A^3$ has eighth power one, so it is constant; its leading coefficient fixes it to one. Pairing positive and negative odd integers now givesSince $B=A^3=q^{1/8}{\prod }_{n\ge 1}(1-q^n{)}^3$, the Jacobi cubic product identity follows:All pairings and differentiations take place in locally uniformly convergent theta functions or products on $|q|<1$.

Let $\mathbb{H}$ be the upper half-plane and put $q=e^{2\pi i\tau }$. A weight-$k$ modular form for the modular group is a holomorphic function on $\mathbb{H}$ satisfyingwhose Fourier expansion at the modular cusp at infinity has only nonnegative powers of $q$. These functions form the vector space $M_k$. Its subspace $S_k$ consists of the cusp forms, whose constant Fourier coefficient is zero. For the full modular group every cusp is equivalent to infinity, so this is the complete cusp condition. The matrix $-I$ gives $M_k=0$ for odd $k$, and the valence formula for the modular group excludes nonzero forms of negative weight.

The normalized Eisenstein series start as $E_4=1+240q+O(q^2)$ and $E_6=1-504q+O(q^2)$. Consequently the modular discriminantis a nonzero cusp form of weight twelve, with order one at infinity. Its contribution there already exhausts the valence formula, so it has no zero in $\mathbb{H}$. Multiplication by $\Delta$ takes $M_k$ injectively into $S_{k+12}$. Conversely, dividing any member of $S_{k+12}$ by $\Delta$ is holomorphic in $\mathbb{H}$, has weight $k$, and has no negative power at infinity because its numerator vanishes there. Thus multiplication by the modular discriminant gives

The valence formula for the modular group shows that a weight-zero form minus its constant term must vanish: otherwise it would be a cusp form with positive order at infinity and total zero order zero. Hence $\dim M_0=1$. A weight-two form satisfies $f(i)=i^{2}f(i)=-f(i)$, so any nonzero one would contribute at least $1/2$ to the valence formula, exceeding $2/12$. Therefore $M_2=0$.

Every even integer $k\ge 4$ has a representation $k=4a+6b$ with $a,b\ge 0$. Indeed, if $k/2$ is even, take $b=0$; if it is odd, take $b=1$ and $a=(k-6)/4$. The form $R_k=E_4^a{E}_6^b$ has constant term one. Taking the constant Fourier coefficient therefore maps $M_k$ onto $\mathbb{C}$, with kernel $S_k=\Delta M_{k-12}$. Starting with the negative-weight vanishing and the two initial dimensions givesand hence the dimension of level-one modular forms isThis also proves finite dimensionality, rather than presupposing it in the recurrence.

For the generation assertion, if $f\in M_k$ has constant coefficient $a_0$, subtract $a_0{R}_k$. The difference is a cusp form, and so equals $\Delta h$ with $h\in M_{k-12}$. Induction on weight expresses $h$ as a polynomial in $E_4,E_6$; substituting the polynomial expression for $\Delta$ does the same for $f$. Constants start the induction, while weight two contributes nothing. The result is a weighted homogeneous polynomial, assigning weights four and six to its variables. In fact the monomials $E_4^a{E}_6^b$ with $4a+6b=k$ number exactly the displayed dimension: $b$ runs from zero to $\lfloor k/6\rfloor$ with parity $k/2$. Their spanning property therefore makes them a basis in each weight. This identifies the polynomial ring of level-one modular forms:

Use the normalized Eisenstein series $E_4=1+240q+O(q^2)$ and $E_6=1-504q+O(q^2)$, and putThe valence formula for the modular group applied to this nonzero weight-twelve cusp form gives weighted zero order one. Its order at infinity is already one, so $\Delta$ has no zeros in $\mathbb{H}$. Consequently the Klein j-invariant is holomorphic there and invariant under $S{L}_2(\mathbb{Z})$.

Set $\rho =e^{2\pi i/3}$. The stabilizer of $\rho$ forces $E_4(\rho )=0$, since the weight-four automorphy factor for $\left(\begin{array}{cc}0& -1\\ 1& 1\end{array}\right)$ is $(\rho +1{)}^4\ne 1$. The stabilizer of $i$ forces $E_6(i)=0$, since $i^6=-1$. The valence formula for the modular group giveswith no other zero orbits for either form. In particular $j(\rho )=0$ and $j(i)=1728$.

For any $A\in \mathbb{C}$, the weight-twelve formis nonzero, with constant Fourier coefficient one and hence order zero at infinity. Its total weighted zero order is one. If $A=0$, the triple zero at $\rho$ accounts for this entire order. If $A=1728$, then $h_A=E_6^2$, and the double zero at $i$ accounts for it. For all other $A$, neither elliptic orbit is a zero: $E_4(\rho )=0$ and $\Delta (\rho )\ne 0$ exclude $\rho$, while $E_6(i)=0$ gives $E_4(i{)}^3=1728\Delta (i)$ and excludes $i$. Therefore there is precisely one ordinary zero orbit, of multiplicity one. Since $h_A(\tau )=0$ is exactly $j(\tau )=A$, every complex number occurs at exactly one modular orbit. ThusThis proves Klein j-invariant classifies complex lattice homothety using the weighted zeros, including the two exceptional stabilizers.

Use $\Theta =(2\pi i{)}^{-1}d/d\tau$. For $\gamma =\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$, differentiating the weight-$k$ law givesThe transformations of $E_2$ under the generators $S,T$, and the chain rule for their compositions, giveEquivalently, one can check just $S,T$ in the next cancellation, since they generate the group. Substituting these formulas intocancels the extra term and gives $(D_{k}f)(\gamma \tau )=(c\tau +d{)}^{k+2}{D}_{k}f(\tau )$. Both constituents are holomorphic on $\mathbb{H}$ and have Fourier expansions of a modular form with nonnegative exponents, so $D_{k}f$ is holomorphic at the modular cusp. ThusThis operator is the Serre derivative.

If $f={\sum }_{n\ge 0}{a}_n{q}^n$, then its derivative has constant coefficient zero, while $E_2$ has constant coefficient one. ThereforeFor $k\ne 0$ this vanishes exactly when $a_0(f)=0$, proving the claimed cusp form equivalence in the intended positive-weight setting. At weight zero the printed equivalence is false: $f=1$ is a noncuspidal weight-zero modular form, but $D_{0}f=0$ belongs to $S_2$. Indeed all weight-zero forms are constants by compactness of $X(1)$. Negative-weight level-one modular forms are zero by the valence formula for the modular group, so there is no further nonzero exception. The transformation and holomorphy conclusion for $D_{k}f$ remains true at weight zero.

For the full modular group $\Gamma =S{L}_2(\mathbb{Z})$ and integer weight $k$, a modular form is a holomorphic function on the upper half-plane satisfyingand holomorphic at the cusp at infinity. Explicitly, periodicity under $\tau \mapsto \tau +1$ gives a Fourier expansion of a modular form in $q=e^{2\pi i\tau }$, and holomorphy at the cusp means $f={\sum }_{n\ge 0}{a}_n{q}^n$. A cusp form has $a_0=0$. All cusps are equivalent for this group. The matrix $-I$ forces $f=(-1{)}^{k}f$, so every odd-weight modular form is zero.

There is a corresponding qualification to the displayed Eisenstein series formula: the Bernoulli-normalized formula is for even integers $k\ge 4$. For odd integers $k>2$, pairing $(m,n)$ with $(-m,-n)$ gives $G_k=0$, while $B_k=0$, so the printed expression divided by $B_k$ is undefined. We prove the asserted nonzero formula in its even-weight range.

On each compact set in the upper half-plane, the real-linear map $(m,n)\mapsto m\tau +n$ has a uniformly bounded inverse. Hence $|m\tau +n|\ge C(m^2+n^2{)}^{1/2}$ there. For $k>2$, lattice convergence gives locally uniform absolute convergence of $G_k$, and therefore holomorphy. If $\gamma =\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)\in \Gamma$, thenThe map $(m,n)\mapsto (ma+nc,mb+nd)$ permutes the nonzero integer pairs. Reindexing proves the weight-$k$ transformation law.

To compute the Fourier expansion of a normalized Eisenstein series, start from the cotangent partial-fraction expansionFor $\mathrm{Im}z>0$, the cotangent Fourier expansion is $\pi \cot (\pi z)=-\pi i-2\pi i{\sum }_{r\ge 1}{e}^{2\pi irz}$. Differentiating both normally convergent expressions $k-1$ times givesFor even $k$, the terms with $m=0$ sum to $2\zeta (k)$, and the terms with $m<0$ duplicate those with $m>0$. Substituting $z=m\tau$ in the last identity and collecting $N=mr$ givesThe rearrangement is absolutely convergent for every $|q|<1$, since the divisor function grows at most polynomially.

For completeness, the normalization follows directly from the definition of the Bernoulli numbers. Near zero, the cotangent partial-fraction expansion givesOn the other hand $\pi \cot (\pi z)=\pi i+2\pi i/(e^{2\pi iz}-1)$. Substitute $t=2\pi iz$ in the generating function of the Bernoulli numbers. The term $B_1=-1/2$ cancels $\pi i$, and the higher odd terms vanish because $t/(e^t-1)+t/2$ is even. Comparison of Laurent coefficients proves the Euler evaluation of even zeta valuesThusThis Fourier series also proves holomorphy at infinity, completing the modular form verification.

For the Ramanujan congruence modulo 691, the standard facts we use are that the modular discriminant is the weight-twelve cusp form $\Delta =q{\prod }_{n\ge 1}(1-q^n{)}^{24}={\sum }_{n\ge 1}\tau (n)q^n$, and that $S_{12}=\mathbb{C}\Delta$. The latter also follows from the valence formula for the modular group: a nonzero weight-twelve modular form has total weighted zero order one, so a cusp form whose coefficient of $q$ vanishes would have cusp order at least two and must be zero. The product makes the Ramanujan tau function integer-valued.

Write $E_k=G_k/(2\zeta (k))$. Using $B_4=-1/30$ and the supplied $B_{12}=-691/2730$, we haveBoth $E_{12}$ and $E_4^3$ have weight twelve and constant term one. Their difference is a cusp form, so is a multiple of $\Delta$. Its coefficient of $q$ is $65520/691-720=-432000/691$, which determines that multiple:The coefficients of $E_4^3$ are integers. Comparing the positive-index Fourier coefficients and reducing modulo $691$ therefore gives $65520{\sigma }_{11}(n)\equiv -432000\tau (n)$. Both numerical constants are congruent to $566$ modulo $691$, and $\gcd (566,691)=1$. Cancellation proves

The Hecke operators organize modular forms simultaneously by their transformation law and their Fourier coefficients. Work with $\Gamma =S{L}_2(\mathbb{Z})$, $M_k$ its space of weight-$k$ modular forms, and $S_k$ its cusp form subspace, with even $k\ge 4$. Odd weights give zero spaces. The normalization below makes the first coefficient of $T_{n}f$ equal to the $n$th coefficient of $f$, which is the key link between operators and arithmetic.

For $A=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$ with positive determinant, the determinant-normalized slash operator for modular forms isThe automorphy factor identity proves $(f{|}_{k}A){|}_{k}B=f{|}_k(AB)$. Positive scalar matrices act trivially. Let ${\mathcal{A}}_n$ be the integral matrices of determinant $n>0$. Left multiplication by $\Gamma$ reduces their first column, by Bezout identity, to $(a,0)$ with $a>0$; the determinant then gives $d=n/a>0$. A row shear makes $0\le b<d$. These are unique determinant-n matrix representatives for Hecke operators, so defineRight multiplication by any element of $\Gamma$ permutes these left cosets. The slash operator for modular forms identity therefore proves that $T_{n}f$ has the same modular transformation law as $f$. Each summand is holomorphic on the upper half-plane. Its cusp behavior follows explicitly from the coefficient calculation below, so $T_n$ preserves both $M_k$ and $S_k$.

Insert $f={\sum }_{r\ge 0}{a}_r{q}^r$ in the sum. The finite sum over $b$ is zero unless $d\mid r$, in which case it is $d$. Writing $r=d\ell$ leaves the term $a_{d\ell }{q}^{a\ell }$ with multiplier $n^{k-1}{d}^{1-k}=a^{k-1}$. Thus the Fourier coefficients of a composite-index Hecke operator areFor $m=0$ use $\gcd (0,n)=n$. In particular the constant coefficient becomes ${\sigma }_{k-1}(n)a_0$, so cusp forms remain cusp forms, andfor a prime number $p$. The absence of negative powers proves cusp holomorphy. Notice that no primality assumption is needed in the composite-index formula.

The Hecke multiplication relations can be derived directly, rather than assumed from the coset description. On Fourier series define auxiliary maps $U_p(\sum a_m{q}^m)=\sum a_{pm}{q}^m$ and $V_{p}f(q)=f(q^p)$. They need not separately preserve $M_k$, but $U_p{V}_p=1$. With $c=p^{k-1}$, the coefficient formula givesMultiplying on the left by $U_p+c{V}_p$, the term $j=0$ in the $U_p$ sum contributes $U_p^{r+1}$; the terms with $j\ge 1$ become $c^j{V}_p^{j-1}{U}_p^{r-j}$. Together with the $c{V}_p$ sum this is exactly $T_{p^{r+1}}+c{T}_{p^{r-1}}$. Hence the Hecke prime-power recurrence from q-shift operators isFor coprime $m,n$, substituting the coefficient formula twice gives a sum over $a\mid \gcd (m,r)$ and $b\mid \gcd (n,r)$ with coefficient $(ab{)}^{k-1}{a}_{mnr/(ab{)}^2}(f)$. The divisors $a,b$ correspond uniquely to a divisor $ab\mid \gcd (mn,r)$, so $T_m{T}_n=T_{mn}$. The prime-power recurrence, by induction on the smaller exponent, givesCombining the coprime and prime-power formulas provesIn particular the Hecke operators commute, and the modular Hecke algebra is generated by the prime operators.