Graded version of Quillen and Suslin's theorem (graded projectives of graded-free and finite rank modules are graded-free) I'm trying to solve the following exercice about graded $R$-modules, when $R=\mathbb{R}[x_1,\ldots,x_n]$ is a polynomial ring :

Let $M$ be a graded-free $R$-module of finite rank. Show that any graded summand $N$ of $M$ is graded-free.

Note that everything should be understood in the graded sense, eg graded-free means free with a homogeneous basis.
I haven't got far: let ${m_k}$ be a homogeneous basis of $M$, and $p\colon M\to N$ the projection map (graded homogeneous, as $N$ is a graded summand). Then $\{p(m_k)\}$ is a homogeneous generating set of $N$, and we can extract a minimal generating set $\{p(m_{k_i})\}$. We must then show that if $\sum_i \lambda_{k_i} p(m_{k_i})=0$, $\lambda_{k_i}=0$ for all $i$. I guess I should use that $R$ is a polynomial ring, but I don't know where.
Any hint?
 A: Thanks to @lhl73's comment, I found a reference which shows this result using Nakayama's lemma. I showed it in a difference way again using this lemma, which I describe here in case it helps anyone (any confirmation that this is indeed correct is welcome).
Polynomial rings have the special feature that their lowest grading space is a field $k$. For $R=k[x_1,\ldots,x_n]$, any $R$-module $M$ is in a particular a $k$-module, that is a vector space, and we can often use linear algebra to solve questions about $R$-module. Nakayama's lemma tells us how. We denote $R_+ = \oplus_{i>0} R_i$ and $[m]$ the projection of $m\in M$ in $M/R_+ M$.

(one version of the) Graded Nakayama's lemma. Let $M$ be a finitely generated $R$-module, and $m_1,\ldots,m_r\in M$ homogeneous elements such that $[m_1],\ldots,[m_r]$ is a basis of the $k$-vector space $M/R_+ M$. Then $m_1,\ldots,m_r$ is a minimal generating set of $M$.

Let $M=N\oplus P$ and $m_1,\ldots,m_k$ such a minimal generating set (it always exists). As $[m_1],\ldots,[m_r]$ is a basis of $N/R_+ N\oplus P/R_+ P$, we can modify it into the union of a basis of $N/R_+ N$ and a basis $P/R_+ P$. Such a linear change can directly be applied to the elements $m_1,\ldots,m_k$, so that we now assume we have elements $n_1,\ldots,n_s$ and $p_1,\ldots,p_t$ defining a minimal generating set (ie a basis as $M$ is free) such that $[n_1],\ldots,[n_s]\in N/R_+ N$ and $[p_1],\ldots,[p_t]\in P/R_+ P$.
Let $\pi_N$ and $\pi_P$ be the projections. $\{\pi_N(n_1),\ldots,\pi_N(n_s),\pi_P(p_1),\ldots,\pi_P(p_t)\}$ is still a basis of $M$ by Nakayama's lemma, as the values of the projections onto $M/R_+ M$ are left unchanged. This shows that $\{\pi_N(n_1),\ldots,\pi_N(n_s)\}$ is a basis of $N$, and hence $N$ is free.
