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I've just read that if $R=R_0\oplus R_1 \oplus \dots$ is a graded ring and $f\in R$ then there's a unique decomposition of $f$ as $f=f_0+\dots+f_n$ with $f_i\in R_i$. I can't see immediately why in general this would have to be a finite sum! Could someone possibly enlighten me? I've got a feeling I'm being stupid, so apologies if it's completely obvious!

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Basically, the direct sum is defined to be the subset of the direct product with all but finitely many of the terms equal to zero. Without a topology there isn't any way to define what it would mean for an infinite sum to converge. – Mike B Aug 2 '12 at 21:35
up vote 3 down vote accepted

It's because the graded ring is a direct sum of the $R_i$'s, not a direct product. Every element of a direct sum has only finitely many non-zero summands by definition.

PS: I always try to keep examples in mind when learning a definition. In the case of graded rings, the first I think of is $k[x_1,\ldots, x_n]$ graded by degree, where $k$ is a field. Here the idea that only finite sums are allowed is exactly what we should expect from experience.

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