Question about definition of graded ring In the course of algebraic geometry I follow, the professor has introduced the notion of 'graded ring' to decompose polynomial ring $k[x_0, \ldots, x_n]$ as follows:
$$k[x_0, \ldots, x_n] = \oplus_{d \geq 0}S_d$$
where he denoted $S_d$ the set of all homogeneous polynomials of degree $d$.
The definition of graded ring he gave confuses me a little bit. This was the definition:
A ring $S$ is graded if $S = \oplus_{d \geq 0}S_d$ for abelian groups $S_d$ such that $S_d \cdot S_e \subseteq S_{d+e}$. 
The part which confuses me is the last condition: It makes sense to me if the $S_d$ are subgroups of $S$ (in that case I see elements of $S_d \cdot S_e$ as $x_dx_e$, with $x_d \in S_d$ and $x_e \in S_e$). But what if the $S_i$ are not subgroups and we have that $S$ is actually isomorphic to such a direct product? How is $S_d \cdot S_e$ defined in this case?
 A: If we have an isomorphism
$$ \varphi : S \to \coprod_{d \geq 0} S_d $$
(note that, while $\oplus$ is more traditional than $\amalg$ here, I find it a poor choice of notation for reasons beyond the scope of this post)
Then, if the canonical insertions are notated as
$$ i_n : S_n \to \coprod_{d \geq 0} S_d $$
then, for each $n$ we intend to identify the three groups:


*

*$S_n$

*$i_n(S_n) \subseteq \coprod_{d \geq 0} S_d$

*$\varphi^{-1}(i_n(S_n)) \subseteq S $


Notation is intended to be flexible: we should be prepared to switch back and forth between any of these three groups on demand.
In particular, we should be prepared to multiply an element of $S_d$ with an element of $S_e$ by moving them both over to $S$.
So, $S_d \cdot S_e \subseteq S_{d+e}$ should be read as meaning exactly the same thing as
$$ \varphi^{-1}(i_d(S_d)) \cdot \varphi^{-1}(i_e(S_e)) \subseteq \varphi^{-1}(i_{d+e}(S_{d+e}))$$

For those allergic to this ambiguity, however, there is a rescue: you can always strictify such a scenario. Define the groups
$$S_d' = \varphi^{-1}(i_d(S_d)) \subseteq S$$
Then, the family of maps
$$ s_d = \varphi^{-1} \circ i_d : S_d \to S'_d $$
defines an isomorphism $s$ between any choice of coproduct
$$ s : \coprod_{d \geq 0} S_d \to \coprod_{d \geq 0} S_d'$$
and thus
$$ s \circ \varphi : S \to \coprod_{d \geq 0} S_d' $$
is an isomorphism. In particular, this means $S$ itself is a coproduct of the $S_d'$, and so we can choose the coproduct so that 
$$ S = \coprod_{d \geq 0} S_d' $$
$$ s \circ \varphi = 1_S$$
and so we've made $S$ into a coproduct of subgroups.

However, I strongly suggest getting used to working coherently with passing back and forth between the various identified objects. In my opinion, advances in category theory, homological algebra, and homotopical algebra very strongly suggest that the notion of equality is too strong; that one should only care about things up to isomorphism, or even just up to equivalence.
A: I don't think I fully understand your question, but I may know where your confusion lies. The decomposition of a graded ring $S_{\bullet}=\oplus_{d\ge0}S_d$ is a decomposition of abelian groups, not of rings.
For instance, if $A,B$ are rings, then we'd usually define $A\oplus B$ to be the ring where multiplication is $(a,b)\cdot(a',b')=(aa',bb')$. This is not the multiplication we are talking about in $S_{\bullet}$.
In our case, the $S_i$ are always additive subgroups, but they are not going to be subrings in general (by definition, the only one closed under multiplication is $S_0$).
Look at the original example of $S_{\bullet}=k[x_0,\dots,x_n]$. Then $S_d$ is the abelian group generated by the monomials of degree $d$, so for instance $S_0=k$, $S_1=\langle x_0,\dots,x_n\rangle$, $S_2=\langle x_ix_j:1\le i,j\le n\rangle$, etc. (again, these are abelian groups, not rings/ideals). These are all additive subgroups of $S_{\bullet}$ but they are not closed under multiplication for $i>0$. But you can immediately see that $S_d\cdot S_e\subseteq S_{d+e}$.
Let me know if you still have some confusions. If you can be more precise with your questions, I can try to update my answer and help out any way I can.
A: If $S:= \oplus S_i$, then each $S_i$ is a subgroup of $S$. For example, consider the group homomorphism $(x_1, \dots, ) \mapsto (x_1, \dots)$, whose image is $S_1$.
If $S \cong \prod S_d$ as an abelian group, then $S \neq \oplus S_d$, since there is no decomposition $(1,1,1, \dots)=( x_1, \dots) \in \oplus S_d$, since by definition there can only be finitely many nonzero terms. Hence, this would just not be a grading. 
The multiplication condition is just making sure that the secondary operation (multiplication) also respects the grading structure of $S_d$ in a way compatible with "polynomial" considerations. 
