Let $R$ be a noetherian commutative ring. Why is $R\cong R_1 \times \cdots \times R_n$ with $R_i$ connected?
I know that one can prove this result using topological arguments on $\mathrm{Spec}(R)$, but since I'm not familiar with Zariski's topology, I would prefer a "purely algebraic" proof.
Using the fact that $R \cong aR \times (1-a)R$ for every idempotent element $a$ of $R$ (which is also central since $R$ is commutative), I already tried writing $R$ as $R\cong \prod b_1 b_2 \cdot \cdot \cdot b_n R$ where $b_k$ is either $a_k$ or $1-a_k$ with $a_k$ an idempotent element. But:
- How do I know that $R$ admits a finite number of idempotent elements? I tried using the fact that $R$ is noetherian so a chain of ideals $(a_1,\dotsc,a_n)$ with each $a_i$ idempotent has a largest element but I did not manage to prove that $a_{n+1}$ cannot be written as $\sum _{i=1} ^n \lambda_i a_i$ with $\lambda _i \in A$.
- How do I know that if $b_k$ is either $a_k$ or $1-a_k$ with $a_k$ those idempotent elements, $b_1\cdot \cdot \cdot b_n R$ is connected?
Could anyone help me with those two problems if my idea works, and if not give some hints on how to prove this result ?
Let $R$ be a commutative Noetherian ring.
Proof: If $e_0,e_1,e_2,\dotsc$ are infinitely many pairwise orthogonal idempotents in $R$, then since $R$ is Noetherian we have $\langle e_0,\dotsc,e_s \rangle = \langle e_0,\dotsc,e_{s+1}\rangle$ for some $s \geq 0$. Hence, $e_{s+1} = \sum_{i=0}^{s} u_i \cdot e_i$ for some $u_i \in R$. Multiply this equation with $e_{s+1}$. Then we get $e_{s+1}=0$, a contradiction. $\checkmark$
Proof: by induction on the number of pairwise orthogonal idempotents $\neq 0$. If that number is $0$, we have $1=0$ in $R$, so $R=0$ is the empty product and we are done. If that number is $1$, $R$ is connected and we are done. Otherwise, there is some idempotent $e \in R$ with $e \neq 0,1$. We have $R \cong eR \times (1-e)R$. Both factors are commutative Noetherian rings, and their number of pairwise orthogonal idempotents $\neq 0$ is less than that of $R$: This is because for every sequence of non-zero pairwise orthogonal idempotent elements $x_1,\dotsc,x_n \in eR$ we get such a sequence $(1-e),x_1,\dotsc,x_n$ in $R$. By induction hypothesis, $eR$ and $(1-e)R$ are both finite products of connected Noetherian commutative rings. Hence, the claim follows for $R$ as well. $\checkmark$