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Start from the integral triangular basis of level-one modular forms. Proceed downwards in $j$. Set $g_{d-1}=F_{d-1}$; once $g_{j+1},\dots ,g_{d-1}$ are constructed, subtract from $F_j$ its coefficients at $q^{j+1},\dots ,q^{d-1}$ times these respective forms. The already constructed rows have a single nonzero coefficient among those initial positions, so this removes every unwanted initial coefficient without changing the coefficient one at $q^j$. All subtractions use integers. The resulting integral echelon basis of level-one modular forms therefore hasIts first $d$ coefficient vectors are the standard coordinate vectors, hence it is a basis. If another form had the same first $d$ coefficients as $g_j$, their difference would have those coefficients all zero; the triangular basis proves that difference vanishes. This proves uniqueness of every $g_j$, not merely uniqueness up to basis change.

For any $f\in M_k$, its expansion in this basis is $f={\sum }_{j=0}^{d-1}{a}_j(f)g_j$, since its first $d$ coefficients determine the basis coordinates. Comparing subsequent coefficients givesThe coefficients of $f$ need not be integral for this last identity. When needed at earlier indices, extend the notation by $c_n(j)={\delta }_{nj}$ for $0\le n<d$. This agrees with the actual coefficients of the basis forms and is the convention needed for the next part's all-index wording.

Write $\Gamma =S{L}_2(\mathbb{Z})$, $\mathbb{H}=\{\tau :\mathrm{Im}\tau >0\}$ and $q=e^{2\pi i\tau }$. The space $M_k$ consists of holomorphic functions on the complex upper half-plane satisfyingand holomorphic at the cusp: their Fourier expansion of a modular form is $f={\sum }_{n\ge 0}{a}_n{q}^n$. The subspace $S_k$ of cusp forms has $a_0=0$. Since all cusps are equivalent at level one, this condition treats every cusp. The element $-I$ forces odd-weight forms to vanish.

Use the permitted properties of the normalized Eisenstein series and modular discriminant:They have integer Fourier coefficients and weights $4,6,12$. For even $k\ge 0$, the assumed dimension formula gives $d=\lfloor k/12\rfloor +1$ except when $k\equiv 2(\mathrm{mod}12)$, when $d=\lfloor k/12\rfloor$. For $0\le j<d$, the remaining weight $k-12j$ is nonnegative and is not two. Choose $b_j\in \{0,1\}$ and $a_j\ge 0$ with $4a_j+6b_j=k-12j$. Every allowable remaining weight admits such a choice. The formsgive an integral triangular basis of level-one modular forms: their distinct first exponents prove independence, and their number $d$ proves that they span.

Now take $f$ with integral coefficients. Subtract its constant coefficient times $F_0$. The remainder still has integral coefficients and has no constant term. Subtract its coefficient at $q$ times $F_1$, then its current coefficient at $q^2$ times $F_2$, and continue. Every subtraction uses an integer. After $d$ steps the remainder's first $d$ coefficients vanish; triangularity and the basis property force that remainder to be zero. ConsequentlyThis proves integral polynomial generation without dividing by $1728$. The relation $E_4^3-E_6^2=1728\Delta$ explains why the additional integral generator is useful, although over $\mathbb{C}$ two generators suffice. For $k=0$ the basis is the constant one; when $M_k=0$, including weight two and odd weights, the claim is vacuous.

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.

The analytic reason for simultaneous diagonalization is the Petersson inner productwhich is positive definite on $S_k$ and convergent because of exponential cusp decay. Here is the adjoint argument. For a matrix $A$, set $H_A=\Gamma \cap A^{-1}\Gamma A$. The left cosets in $\Gamma A\Gamma$ are represented by $A\gamma$ with $\gamma \in H_A\backslash \Gamma$. Unfolding this finite coset sum in the Petersson inner product givesUnder $z=A\tau$, hyperbolic area is unchanged and the determinant factors in the slash operator for modular forms giveThe integration group becomes $A{H}_A{A}^{-1}=\Gamma \cap A\Gamma A^{-1}$. Folding this finite cover back to $\Gamma \backslash \mathbb{H}$ identifies the adjoint with the correspondence for $A^{-1}$. For $A\in {\mathcal{A}}_n$, the matrix $\mathrm{adj}(A)=n{A}^{-1}$ is again integral with determinant $n$; scalar matrices act trivially, and adjugation permutes the double cosets making up ${\mathcal{A}}_n$. The real normalizing factor $n^{k/2-1}$ is unchanged. Thus this proves that Hecke operators are self-adjoint for the Petersson inner product:The valence formula for the modular group makes $S_k$ a finite-dimensional vector space. Commuting self-adjoint operators on a finite-dimensional inner product space have a common orthogonal eigenbasis: diagonalize one operator, use commutativity to preserve each of its eigenspaces, and repeat within those spaces. Therefore $S_k$ has a basis of Hecke eigenforms, with real Hecke operator eigenvalues.

If $f$ is a common Hecke eigenform, say $T_{n}f={\lambda }_{n}f$, its first Fourier coefficient cannot vanish: $a_1=0$ would imply $a_n=a_1(T_{n}f)={\lambda }_n{a}_1=0$ for every $n$. Normalize $a_1=1$. Then the Hecke operator eigenvalue ${\lambda }_n$ is precisely the Fourier coefficient $a_n$. The Hecke multiplication relations becomeConsequently the coefficients are multiplicative at coprime indices, and their local generating function satisfiesThis gives the Euler product of a Hecke eigenformin a half-plane of absolute convergence. The Fourier coefficient bound for a cusp form already supplies such a half-plane; Question 4 improves it to $\mathrm{Re}s>(k+1)/2$.

The L-function of a cusp form also has an analytic description independent of the Hecke eigenform condition. Termwise Mellin transform in an initial right half-plane givesAt infinity the cusp form decays exponentially. Modular inversion gives $f(i/y)=(iy{)}^{k}f(iy)$, so at zero it is $O(y^{-k}{e}^{-2\pi /y})$. The integral therefore converges locally uniformly for every $s$, continuing $\Lambda$ to an entire function. Changing $y$ to $1/y$ provesFor a normalized Hecke eigenform, the Euler product of a Hecke eigenform and this functional equation describe the same completed L-function of a cusp form.

There is also an integral arithmetic structure. One standard consequence of the level-one dimension formula and the valence formula for the modular group is the integral triangular basis of level-one modular forms: if $k=12m+s$ with $s\in \{0,4,6,8,10,14\}$, choose $F_s=1,E_4,E_6,E_4^2,E_4{E}_6,E_4^2{E}_6$ respectively. Thenform a basis of $M_k$, and the terms $1\le j\le m$ form a basis of $S_k$. Their first nonzero terms are $q^j$ with coefficient one and all coefficients are integers. Successive elimination of coefficients of $q,q^2,\dots ,q^m$ shows that these cusp basis elements generate exactly the lattice of rational cusp forms with integral Fourier coefficients. The composite-index coefficient formula preserves this lattice, so each $T_n$ has an integral matrix on it. Its characteristic polynomial is monic with integral coefficients. This proves that Hecke eigenvalues are algebraic integers; self-adjointness also makes every root of that characteristic polynomial real. The field generated by the eigenvalues is a finite totally real number field: a finite set of operators spans the finite-dimensional rational modular Hecke algebra, and every conjugate of each eigenvalue remains a root of its real-rooted characteristic polynomial.

Two basic examples connect the operator theory to familiar modular forms. The normalized Eisenstein series $E_k$ has coefficients proportional to ${\sigma }_{k-1}(n)$. At a prime number its coefficient formula useswhere the second term is zero unless $p\mid m$. This follows by writing $m=p^{r}u$ with $p\nmid u$ and summing a finite geometric series. It proves $T_p{E}_k=(1+p^{k-1})E_k$, and the Hecke multiplication relations give $T_n{E}_k={\sigma }_{k-1}(n)E_k$. Meanwhile $S_{12}=\mathbb{C}\Delta$ implies $T_n\Delta =\tau (n)\Delta$, because $a_1(\Delta )=1$. Thus the Ramanujan tau function is multiplicative and satisfies $\tau (p^{r+1})=\tau (p)\tau (p^r)-p^{11}\tau (p^{r-1})$. For example, $\tau (2)=-24$ gives $\tau (4)=(-24{)}^2-2^{11}=-1472$. This illustrates how Hecke operators turn analytic Fourier coefficients into a coherent arithmetic system.

The normalized Eisenstein series satisfy $E_4=1+240{\sum }_{m\ge 1}{\sigma }_3(m)q^m$ and $E_6=1-504{\sum }_{m\ge 1}{\sigma }_5(m)q^m$. Their weights are four and six. The valence formula for the modular group makes the normalized weight-twelve cusp form $\Delta$ have one simple cusp zero and no zero in $\mathbb{H}$. Comparing its first coefficient gives $\Delta =(E_4^3-E_6^2)/1728$.

Odd weights vanish because $-I$ acts by $(-1{)}^k$. Weight two vanishes by the valence formula for the modular group: the forced zero at $i$ alone contributes $1/2>2/12$. For every remaining nonnegative even weight, except two, choose $a,b\ge 0$ with $4a+6b=k$. Subtract the constant coefficient of $f$ times $E_4^a{E}_6^b$. The remainder is a cusp form; division by $\Delta$ produces a holomorphic modular form of weight $k-12$. The cusp and interior orders justify this division. Induction on the weight provesThe integral triangular basis of level-one modular forms is $F_j={\Delta }^j{E}_4^{a_j}{E}_6^{b_j}$, with $4a_j+6b_j=k-12j\ge 0$ and $k-12j\ne 2$. The induction just given shows these span, and their distinct leading terms $q^j$ show linear independence. Every coefficient is integral by the hypothesis on $\Delta$ and the displayed Eisenstein expansions. In $f={\sum }_j{c}_j{F}_j$, compare coefficients successively: $c_0=a_0$, then $c_1$ is $a_1$ minus an integer combination of already integral coefficients, and similarly for every $j$. The largest permitted $j$ is at most $\lfloor (k+1)/12\rfloor$, so the stipulated initial coefficients make every $c_j$ integral. Therefore all Fourier coefficients of $f$ are integers.