= Solution
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 $\mathbb H$ satisfying
$$
f\!\left(\frac{a\tau+b}{c\tau+d}\right)=(c\tau+d)^kf(\tau)\qquad\begin{pmatrix}a&b\\c&d\end{pmatrix}\in SL_2(\mathbb Z),
$$
whose <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 discriminant>
$$
\Delta=\frac{E_4^3-E_6^2}{1728}=q+O(q^2)
$$
is 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
$$
\boxed{M_k\xrightarrow{\;f\mapsto\Delta f\;}S_{k+12}\text{ an isomorphism}.}
$$
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 $f(i)=i^2f(i)=-f(i)$, so any nonzero one would contribute at least $1/2$ to the valence formula, exceeding $2/12$. Therefore $M_2=0$.
Every even integer $k\geq4$ has a representation $k=4a+6b$ with $a,b\geq0$. Indeed, if $k/2$ is even, take $b=0$; if it is odd, take $b=1$ and $a=(k-6)/4$. The form $R_k=E_4^aE_6^b$ 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 gives
$$
\dim M_k=1+\dim M_{k-12}\quad(k\geq4\text{ even}),
$$
and hence the <dimension of level-one modular forms> is
$$
\boxed{\dim M_k=\begin{cases}\lfloor k/12\rfloor,&k\equiv2\pmod{12},\\\lfloor k/12\rfloor+1,&k\not\equiv2\pmod{12},\end{cases}\qquad k\geq0\text{ even}.}
$$
This 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 $a_0R_k$. 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 $E_4^aE_6^b$ 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>:
$$
\boxed{M_*(SL_2(\mathbb Z))=\mathbb C[E_4,E_6].}
$$
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