Still another question about an exercise in Lenstra's Galois Theory for Schemes.
The purpose of Exercise 1.18 is to demonstrate that a compact, Hausdorff topological ring $R$ (with unity) is a profinite ring. Lenstra breaks the argument up into four pieces, and I'm fairly confident that I've resolved the last three, and a portion of the first part; however there is still one aspect of it that I can't seem to establish (which seems to me like it should be the easiest out of all of them, but nevertheless…). The statement is:
(a) For an open neighborhood $U$ of 0 in $R$, let $V = \{x \in R : RxR\subset U\}$. Prove that $V$ is a neighborhood of 0 in $R$. If moreover $U$ is an additive subgroup of $R$, prove that $V$ is an open two-sided ideal of $R$.
Proving that U being an additive subgroup implies that $V$ is an open two-sided ideal is straightforward, but I can't seem to show that $V$ is a neighborhood of 0. I'm certain that the argument is short and simple, but for some reason it alludes me. Any suggestions?
Let $U$ be any neighborhood of $0$. For every pair $(a,b)\in R\times R$, there is a neighborhood $W_{a,b}$ of this pair and there is a neighborhood $V_{a,b}$ of $0$ such that, whenever $(a',b')\in W_{a,b}$ and $x\in V_{a,b}$, then $a'xb'\in U$ (because multiplication is continuous and $a0b=0$). By compactness, finitely many of the $W_{a,b}$'s cover $R\times R$. The intersection of the finitely many corresponding $V_{a,b}$'s will be included in your $V$, so the latter is a neighborhood of $0$.