# $p^2=p\in \bar{I}$, I ideal of Banach algebra $\Rightarrow p\in I$

Let $I\subset A$ be a ideal of a Banach algebra $A$. Assume $p\in \overline I$ and $p^2=p$.

Show: $p\in I$.

Can someone give me a little hint how to solve this, please?

• Being complete is the only hypothesis you have on $A$ ? How is its norm ? Dec 23, 2014 at 10:55
• Well, I have that $A$ is a Banach algebra. What do you mean by "How is its norm?" ? Dec 23, 2014 at 11:12
• There are Banach algebras over anything, over any Banach ring in fact, like $\mathbf{Z}$ or $\mathbf{Z}_p$ ($p$-adic integers) or $\mathbf{Q}(( T ))$, or, why not, over $\mathbf{F}_p$. This wikipedia's article is not accurate at all. Dec 23, 2014 at 11:45
• Okay, sorry. Since I haven't heard of these algebras so far, I thought my language is clear. Is it at least precise enough now? Dec 23, 2014 at 11:50
• Yeah, no worries ! Now it's clear. Dec 23, 2014 at 11:50

I assume that $A$ is unital, if not, we can pass to the unitisation.
Suppose that $(t_n)_{n=1}^\infty$ is a sequence in $I$ that converges to an idempotent $p$. Since $pAp$ is a Banach algebra with identity $p$ and $pt_np \to p^3=p$, there exists $n$ so that $pt_np$ is invertible in $pAp$. Consequently, $p = (psp)(pt_np)$ for some $s\in A$. This proves that $p\in I$.