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.

enter image description here

  • 1
    $\begingroup$ 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 $\endgroup$
    – morrowmh
    May 10 at 0:41
  • $\begingroup$ I assume your rings contain a $1$. Is this the case? $\endgroup$
    – morrowmh
    May 10 at 0:45
  • $\begingroup$ Yes, but why is it in $R_0$? $\endgroup$ May 10 at 0:54
  • $\begingroup$ Remark 1.1 of the link I gave provides a nice proof that $1\in R_0$. $\endgroup$
    – morrowmh
    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

1 Answer 1


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 $R$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.