# Definition of a Graded Algebra and $R_0$

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.

• 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 May 10 at 0:41
• I assume your rings contain a $1$. Is this the case? May 10 at 0:45
• Yes, but why is it in $R_0$? May 10 at 0:54
• Remark 1.1 of the link I gave provides a nice proof that $1\in R_0$. May 10 at 0:58
• @MichaelMorrow please consider writing an answer so that this question can be marked as dealt with. May 10 at 1:29

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 $$R$$.