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In the definition of graded ring (https://en.wikipedia.org/wiki/Graded_ring), there is a statement $R_iR_j\subset R_{i+j}$.

Just to clarify how is $R_iR_j$ defined? I can think of two possibilities:

1) $R_iR_j=\{r_ir_j\mid r_i\in R_i, r_j\in R_j\}$

2) $R_iR_j=\{\sum r_ir_j\mid r_i\in R_i, r_j\in R_j\}$ (finite/infinite sum?)

Or maybe it is another definition?

Thanks for the clarification.

Side question: What is the name of such a product $R_i R_j$?

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    It doesnt really matter since $R_{i+i}$ is closed under addition. – 2017-02-07
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    As Rene said, it isn't so important. In the end, what they are getting at is 1). The statement $R_iR_j \subseteq R_{i+j}$ means that a product of something of degree $i$ and something of degree $j$ has degree $i+j$. On a side note, you cannot have an infinite sum in a ring. There are objects called power series rings that look as if there are infinite sums, but that is an illusion. – 2017-02-07
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    @ReneSchipperus : I don't understand your comment. How can you always factors $\sum r_i r_j$ as $ab$ with $a \in R_i, b \in R_j$ ? – 2017-02-07
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    @N.H. I believe that Rene was pointing out either definition being contained in $R_{i+j}$ would imply that the other is also contained. – 2017-02-07
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    Ok but the OP is talking about the definition of $R_iR_j$. I believe the correct definition is as sum of $a_ib_j$ where the $a_i \in R_i$ and $b_j \in R_j$ and not simply the set of all elements $a_ib_j$, maybe it is obvious that such sums factors as a product with an element of $R_i$ and an element of $R_j$ but I don't see why this is true. – 2017-02-07
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    @N.H. I would agree with you on that being the definition, perhaps I should have mentioned that in my above comment. The two certainly aren't equal as sets. As for the purposes of understanding what the statement means, I like 1) better. – 2017-02-07
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    Thanks for the insightful comments. – 2017-02-07
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    @yoyostein : sure ! If you want to have a good intuition for it the best example is the ring of polynomials with several variables, where the degree of a monomial $x_1^{d_1}\dots x_n^{d_n}$ is simply $d_1 + \dots + d_n$. – 2017-02-07

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