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.
Best Answer
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.