Defining ring structure for the Rees algebra of an ideal For a commutative ring $R$ and an ideal $I$, Aluffi III.5.17 defines the object $\text{Rees}_R(I)$ to be the direct sum $R \oplus I \oplus I^2 \oplus I^3 \oplus \dots$. Then he asks to define a ring structure on this direct sum and prove that if $a \in R$ is a non-zero-divisor, then $\text{Rees}_R((a)) \cong R[x]$ as an $R$-algebra.
There's a similar question, but I'm a bit stuck before that: namely, what shall be the ring structure?
Assuming that the direct sum in the definition is a direct sum of the corresponding abelian groups (what else could it be?), the only multiplication that I could come up with is $(r_0, r_1, \dots) \cdot (s_0, s_1, \dots) = (r_0 s_0, r_1 s_0 + r_0 s_1, \dots)$ (more generally, $k$th position in the product is equal to $\sum_{i +j = k}r_i s_j$).
This definition seems to produce a valid multiplication, and, moreover, the second part of the exercise follows by considering $\varphi : R[x] \rightarrow \text{Rees}_R((a))$, $\varphi(r) = (r, 0, \dots)$ for $r \in R$, $\varphi(x) = (0, a, 0, \dots)$ and defining $\varphi$ on the rest of the domain via the homomorphism condition.
But is this multiplication the intended one? It looks quite artificial to me.
 A: The ring structure is essentially unique. Write elements as formal sums
$$
(a_0,a_1,a_2,\dotsc)=\sum_n a_n
$$
with $a_n\in I^n$ (and all zero, but a finite number). Since you want the product to be distributive with respect to addition, and
$$
\Bigl(\sum_n a_n\Bigr)+\Bigl(\sum_n b_n\Bigr)=\sum_n (a_n+b_n)
$$
you have to define the product of $a_m$ by $b_n$, where $a_m\in I^m$ and $b_n\in I^n$: since $a_mb_n\in I^{m+n}$, the choice is obvious.
The verification of the ring laws is a bit tedious, but simple.
For the second part, define $e_0=(1,0,0,\dotsc)\in\operatorname{Rees}_R((a))$, $e_1=(0,a,0,\dotsc)$ and note that $(e_1)^k=e_k$, the element having $a^k$ at the $k$-th position and $0$ elsewhere.
The subring generated by $e_0$ is clearly isomorphic to $R$ and the action of $R$ on $\operatorname{Rees}_R((a))$ as module is the same as the action of the subring. You can so define uniquely a ring homomorphism $\varphi\colon R[x]\to\operatorname{Rees}_R((a))$ by $\varphi(r)=re_0$ and $\varphi(x)=e_1$. Explicitly,
$$
\varphi(r_0+r_1x+\dots+r_nx^n)=r_0e_0+r_1e_1+\dots+r_ne_n
$$
This homomorphism is surjective (prove it). Now compute the kernel.
