Does $L^2((0, 1])$ contain $(L^\infty((0, 1]))^2$ as $L^\infty((0, 1])$-modules? Does $L^2((0, 1])$ contain $(L^\infty((0, 1]))^2$ as $L^\infty((0, 1])$-modules?
Here $L^\infty((0, 1])$ is just treated as a ring.
 A: The question seems to be asking whether there exists an injective $L^\infty([0,1])$-module map $(L^\infty([0,1]))^2\to L^2([0,1])$. The answer is "NO".
In fact, I claim that every finitely generated submodule of $L^2([0,1])$ is actually generated by a single element and it is isomorphic to $L^\infty([0,1])\cdot\chi_A$ for some measurable $A\subset [0,1]$.
This follows from the three claims below.
Claim 1: For every $f\in L^2([0,1])$ there exists $\phi_f\in L^\infty[0,1]$ such that $\phi_f f=|f|$.
Proof: For $t$ such that $f(t)=0$ define $\phi_f(t)=0$.
Otherwise define $\phi_f(t)=\overline{f(t)}/|f(t)|$. 
Claim 2: Every finitely generated submodule of $L^2([0,1])$ is generated by a single element.
Proof: For the module generated by $f_1,\ldots,f_n$ consider the element $f=\sum \phi_{f_i}f$.
Claim 3: The module generated by a single element $f$ is isomorphic to $L^\infty([0,1])\cdot\chi_{\{t\mid f(t) \neq 0\}}$.
Proof: The kernel of the map $L^\infty([0,1])$ to $L^2([0,1])$ given by $1\mapsto f$ is clearly $L^\infty([0,1])\cdot\chi_{\{t\mid f(t)= 0\}}$.
Corollary: There is no injective $L^\infty([0,1])$-module map $\phi: (L^\infty([0,1]))^2\to L^2([0,1])$.
Proof: Since the image of $\phi$ is finitely generated it could be injected further into $L^\infty([0,1])$, thus we obtain a map $(L^\infty([0,1]))^2\to L^\infty([0,1])$. We conclude by the fact that for every commutative unital ring $R$, an $R$-module map $\psi:R^2\to R$ cannot be injective: either $\psi=0$ or the nonzero element $(\psi(0,1),-\psi(1,0))$ is in its kernel.
Remark: Note that it follows that for every $n\geq 2$, there is no injective $L^\infty([0,1])$-module map $(L^\infty([0,1]))^n\to L^2([0,1])$.
