Generating connected module over a connected $K$-algebra I have an algebraic question that I cannot solve. It is extracted from Adams and Margolis' paper on modules over the Steenrod Algebra. Here is the problem :

Let $K$ be a commutative ring with unit, $R$ a connected $K$-algebra (not commutative in my case), i.e., graded as $R \cong K \oplus R_1 \oplus R_2 \oplus \cdots$, and $M$ be a connected $R$-module, i.e., only positively graded (no condition on $M_0$, but I am okay to assume $M_0 \cong K$, only as a $K$-module).
Consider elements $\{m_i\} \in M$ such that $\{ 1 \otimes m_i \} \in K \otimes_R M$ form a $K$-basis.

(Note that $K \otimes_R M \cong M / I(R)M$ where $I(R)$ is the augmentation ideal for the canonical augmentation here, so another way to see it is that $\{ \overline{m_i} \}$ form a $K$-basis)
Then they claim that "the following facts are well known and easily proved:"


*

*The $\{ m_i \}$ $R$-generates $M$.

*If $M$ is $R$-free, then the $\{ m_i \}$ are an $R$-basis of $M$.


Thank you for your help.

*

*If $(m_i)$ are homogeneous, then use Nakayama Lemma, graded version. 

*Since $M$ is free it has a basis $(x_j)_{j\in J}$ and this remains so in $K\otimes_RM$, therefore $|J|=|I|$. (For the sake of simplicity let's assume $J=I$.) Now let $f:M\to M$ given by $f(x_i)=m_i$. Then $f$ is surjective, and using this result $f$ is bijective, so $(m_i)$ is a basis. (Here I need $M$ finitely generated and I don't have a proof if remove this condition.)
