Show if pAp is a factor given a minimal projection

Question

Let $A$ be a Von Neumann algebra on a hilbert space $H$.

If $p$ is a minimal projection in the center of $A$, show that $pAp$ is a factor on $pH$.

My attempt :

Since $p$ is minimal, $pAp=\mathbb{C}p$.

So the center of $pAp$ will be $ \mathbb{C}p\bigcap(\mathbb{C}p)'$

But I am not even convinced if this $(\mathbb{C}p)'$ makes sense since I think it is not a von neumann algebra.

Any help would be appreciated.

Answer 1

The equality $pAp=\mathbb Cp$ would occur if $p$ was a minimal projection in $A$, which is not the case here. The hypothesis is that $p$ is minimal in the centre of $A$.

Note that $Z(A)=Z(A')$. The centre of $pAp=Ap$ is, since $p\in A\cap A'$, $$ Ap\cap(Ap)'=Ap\cap A'p=(A\cap A')p=Z(A)p=\mathbb Cp, $$ since $p$ is minimal in $Z(A)$. So $pAp=Ap$ is a factor.

Answer 2

Firstly, $A'''=A'$ in any scenario, so $(\mathbb Cp)'$ is a von Neumann algebra. Note that $\mathbb Cp\cap(\mathbb Cp)'=\mathbb Cp$ is just the linear span of its unit, hence $\mathbb Cp$ is a factor.