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Below we have a definition of a graded $k$-algebra where $k$ is a field. I have a few questions. First, looking it up, there seems to be some ambiguity as to what $R$ is a direct sum of the $R_n$ as. Groups? $k$-modules? Is there anything from context that could tell me? Secondly, why is $R_0$ a subalgebra. If the $R_n$ are submodules, then for $r\in R_n$, $kr\in R_n$, and therefore its not hard to see that if $R$ is an integral domain, $k\in R_0$, but I can't see why this should work if $R$ is not an integral domain. Thanks in advance.

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    $\begingroup$ Direct sum of $k$-vector subspaces. Since $R_0R_0\subseteq R_0$ we have that $R_0$ is a subring of $R$. And by assumption, $R_0$ is a $k$-vector subspace so $R_0$ is a $k$-subalgebra of $R$. For the general case when $R$ is not necessarily a $k$-algebra, see the beginning of math.unl.edu/~tmarley1/905notes.pdf $\endgroup$
    – Michael
    May 10 at 0:41
  • $\begingroup$ I assume your rings contain a $1$. Is this the case? $\endgroup$
    – Michael
    May 10 at 0:45
  • $\begingroup$ Yes, but why is it in $R_0$? $\endgroup$
    – Thomas Anton
    May 10 at 0:54
  • $\begingroup$ Remark 1.1 of the link I gave provides a nice proof that $1\in R_0$. $\endgroup$
    – Michael
    May 10 at 0:58
  • $\begingroup$ @MichaelMorrow please consider writing an answer so that this question can be marked as dealt with. $\endgroup$
    – KReiser
    May 10 at 1:29

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The phrase "where the subspaces..." indicates that we want to think of this as a direct sum of $k$-vector subspaces. The general definition of graded ring uses a direct sum of abelian groups, see https://math.unl.edu/~tmarley1/905notes.pdf for more details.

One can show without difficulty (see Remark 1.1 of the aforementioned link) that $1\in R_0$. Furthermore, since $R_0R_0\subseteq R_{0+0}=R_0$ we have that $R_0$ is a subring of $R$. Finally, combining this with the fact that $R_0$ is a $k$-subspace of $R$ we obtain that $R_0$ is indeed a $k$-subalgebra of $R$.

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