= Solution
Normalize the weight-four and weight-six <Eisenstein series> to have constant term one. Their <Fourier expansions> begin
$$
E_4=1+240q+O(q^2),\qquad E_6=1-504q+O(q^2),\qquad q=e^{2\pi iz}.
$$
Their 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,
$$
v_\infty(f)+\tfrac12v_i(f)+\tfrac13v_\rho(f)+\sum_{z\ne i,\rho}v_z(f)=\frac{k}{12}.
$$
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 $f(-1/z)=z^kf(z)$ 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> by
$$
\Delta=\frac{E_4^3-E_6^2}{1728}=q+O(q^2).
$$
It 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>:
$$
\boxed{S_k=\Delta M_{k-12}.}
$$
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: $f(i)=i^2f(i)=-f(i)$ 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 $a,b\geq0$. If $k\equiv0\pmod4$, take $b=0$; if $k\equiv2\pmod4$ and $k\geq6$, take $b=1$. For $f\in M_k$, choose such a monomial $h=E_4^aE_6^b$, whose constant term is one. Subtracting the constant term $c$ of $f$ gives $f-ch\in S_k$, so
$$
f=cE_4^aE_6^b+\Delta g,\qquad g\in M_{k-12}.
$$
Induction 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 written
$$
E_4^{a_0}E_6^{b_0}(E_4^3)^j(E_6^2)^{L-j},\qquad j=0,\ldots,L,
$$
for 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 in
$$
R(z)=\frac{E_4(z)^3}{E_6(z)^2}=1+1728q+O(q^2).
$$
This 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 $\gamma_n=\begin{pmatrix}1&0\\n&1\end{pmatrix}$. At any fixed $z\in\mathbb H$ this gives
$$
\sum_w(nz+1)^wP_w(E_4(z),E_6(z))=0\qquad(n\in\mathbb Z).
$$
The 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. Hence
$$
\boxed{M_*(SL_2(\mathbb Z))\cong\mathbb C[E_4,E_6],\qquad \deg E_4=4,\ \deg E_6=6,}
$$
with \b[algebraically independent generators]. Multiplying the two <Eisenstein series> by their nonzero lattice-normalization constants does not alter either generation or independence.
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