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The preceding dimension of level-one modular forms gives $\dim M_{16}=2$ and $\dim S_{16}=\dim M_4=1$. Thus $E_4\Delta$, which begins with $q$, spans $S_{16}$. A normalized weight-sixteen Eisenstein series and $E_4^4$ both have constant term one, so their difference is a multiple of $E_4\Delta$.

There is a normalization error in the printed auxiliary expansion: for the usual normalized $E_{16}$, the coefficient multiplier is $16320/3617$, rather than $255/(8\cdot 3617)$. Indeed, the Fourier expansion of a normalized Eisenstein series gives $-32/B_{16}$, and the Bernoulli number $B_{16}=-3617/510$ gives $16320/3617$. The printed multiplier is smaller by a factor of $512$. With the correct normalization, the first Fourier coefficients yieldComparing all positive-index coefficients therefore gives the exact identityThe Fourier coefficients of $E_4^4$ are integers, as are those of $E_4\Delta$: $E_4$ has its integral divisor-sum expansion and the modular discriminant has the product $q{\prod }_{r\ge 1}(1-q^r{)}^{24}$. Also $3456000+16320=960\cdot 3617$, and $\gcd (16320,3617)=1$. Reducing the identity modulo $3617$ and cancelling the invertible multiplier proves the weight-sixteen Eisenstein congruenceThe desired congruence is valid; it is the auxiliary normalization in the PDF that requires correction.

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:

Subtract the constant term of a modular form times a same-weight monomial in $E_4,E_6$. The remainder is a cusp form, hence divisible by the modular discriminant. Induction on weight and $1728\Delta =E_4^3-E_6^2$ express every form as a weighted homogeneous polynomial in these two Eisenstein series. The number of monomials of a given weight equals the dimension of level-one modular forms, so they form a basis and the graded ring has no polynomial relation.