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So, there are two types of definitions of graded rings (I will consider only commutative rings) that I have seen:

1) A ring $R$ is called a graded ring if $R$ has a direct sum decomposition $R = \bigoplus_{n \in \mathbb{Z}} R_n$, where for all $m,n \in \mathbb{Z}, R_mR_n \subset R_{m+n}$.

2) A ring $R$ is called a graded ring if $R$ has a direct sum decomposition $R = \bigoplus_{n \in \mathbb{Z}} R_n$, where for all $m,n \in \mathbb{Z}, R_mR_n \subset R_{m+n}$, and $R_0$ is a subring of $R$, i.e., $1 \in R_0$.

In the second definition, is the additional condition that $R_0$ is a subring, i.e., basically the condition that $1 \in R_0$, redundant?

J W
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Rankeya
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    Well, if $1\in R_k$, then $R_n\subseteq R_{n+k}$ for any $n$. But $R_n\cap R_{n+k}=0$ (if $k\ne 0$), so $R_n=0$ for all $n$ and $R=0$ (then $1\in R_0$). So the answer is yes. –  Nov 04 '11 at 23:08
  • So, Atiyah-Macdonald has this construction: For an ideal $\alpha \subset A$ (where A is a ring), you define the graded ring $A^{*} = A \bigoplus \alpha \bigoplus \alpha^{2} \bigoplus ...$. Here it does not seem that $R_n \cap R_{n+k} = 0$ for $k \neq 0$. Am I missing something here? – Rankeya Nov 05 '11 at 00:58
  • This is a conceptual error on my part, and if you can explain where I am wrong, it will be very helpful to me. While I know the internal direct sum notation makes sense only for subgroups with trivial intersection, I am not sure how to make sense of Atiyah-Macdonald's construction of the graded ring above (which the C-ring project calls the blowup algebra), because each consecutive homogeneous part is contained in the one before. – Rankeya Nov 05 '11 at 01:06
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    For the graded ring you are thinking of, by definition, the graded components are considered not to intersect. A formal way of doing this is to define $A^* = A \oplus t\alpha \oplus t^2 \alpha^2 \oplus \dots$, viewed as a subring of $A[t]$. – Michael Joyce Nov 05 '11 at 01:11

1 Answers1

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Yes it is redundant, you just need the definition 1). This is because:

if $ R_mR_n \subset R_{m+n}$ then $ R_0R_0 \subset R_{0}$ , thus $R_{0}$ is subring. Second we have $1=\sum x_n$ where there is only a finite number of non-zero $x_n$. Also note that $x_m = 1\cdot x_m=\sum x_nx_m$. By comparing degree we see that $x_m=x_0x_m$ and $x_0= 1 \cdot x_0=\sum x_n \cdot x_0=\sum x_n=1$, therefore $1 \in R_0$.

user26857
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Lonely Penguin
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    I don't think I would write $1 \times x_n = \sum_n x_n \times x_n$, but rather $x_m = 1 \times x_m = \sum_n x_n \times x_m$. Otherwise it's a bit confusing! – Patrick Da Silva Dec 22 '14 at 11:22