Let $R=\mathbb{\mathbb{C}}[x_1^{\pm1}, \cdots, x_n^{\pm 1}]$, $M$ a free $R$-module which at the same time is an $R$-algebra (meaning it contains the identity). Let $N$ be an $R$-subalgebra. Is $N$ a free $R$-module?
Note that if we drop the algebra condition, that is, if we only assume that $M$ is a free $R$-module and $N$ a submodule of $M$, then $N$ is not necessarily a free $R$-module. For example if $M=R=\mathbb{C}[x_1^{\pm 1}, x_2^{\pm 1}]$, then the ideal $N=(x_1, x_2)$ is not free over $R$.