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Let $R$ be a graded ring and $M$ a graded$R$-module. $M$ is called a graded free module if $M$ has a Basis consisting of homogeneous elements

Now in my lecture note there is the following Theorem without a proof.

$M$ is a graded free $R$-module if and only if there exists homogeneous elements $e_{\lambda} \in M$ ($\lambda \in \Lambda$), such that $\{e_{\lambda},\lambda \in \Lambda$} is a linearly Independent generating set for $M$.In this case we have

\begin{equation} M = \bigoplus\limits_{\lambda \in \Lambda}R_{\lambda} \cong \bigoplus\limits_{\lambda \in \Lambda}R(-d_{\lambda})~~~~~~~~~\text{(1)} \end{equation} where $R_{\lambda}$ is a copy of $R$ and $d_{\lambda} = \text{deg}~ e_{\lambda}$ .

I don't understand why the relationship (1) holds. Can someone explain it please?

  • The set ${e_\lambda}{\lambda\in \Lambda}$ is your basis, and $R\lambda = R e_\lambda$. Does that help? – Billy Jul 10 '18 at 13:42
  • but $e_{\lambda}$ is an element of $M$. Why is $R_{\lambda} = R e_{\lambda}$? –  Jul 10 '18 at 13:45
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    Yes, $Re_\lambda$ is the $R$-submodule of $M$ generated by $e_\lambda$. The phrase "$R_\lambda$ is a copy of $R$" just means "$R_\lambda$ is isomorphic to $R$ as an $R$-module". – Billy Jul 10 '18 at 13:49
  • ok...thanks. How does the isomorphism looks like? Is it just the $R$-module homomorphism defined by mapping $e_{\lambda}$ to $1$? –  Jul 10 '18 at 13:52
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    Yes, that's it. – Billy Jul 10 '18 at 14:21
  • love you...thanks (: –  Jul 10 '18 at 14:24

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