Let $B$ be a commutative C$^*$-algebra and let $M_n$ denote the algebra of $n\times n$ complex matrices. Let $f$ be a state on the tensor product of $B$ and $M_n$, $B\otimes M_n$.
How can I show that if $f$ is pure then it must be a product state? I encountered this in Stormer's book on completely positive maps (proof of Th.1.2.4) and have been stuck on it. Please help.
The argument goes as follows. You take $b\in B\otimes\mathbb C1$, with $0\leq b\leq 1$. For any $a\geq0$ in $B\otimes M_k$, $ab=a^{1/2}ba^{1/2}$; this shows that $0\leq ab\leq a$. Then $$\tag{1} 0\leq \rho(ab)\leq\rho(a),\ \ \ a\geq0,\ \ a\in B\otimes M_k. $$ Now define states $f_b$ and $g_b$ on $B\otimes M_k$ by $$ f_b(x)=\frac1{\rho(b)}\,\rho(bx),\ \ \ g_b(x)=\frac1{\rho(1-b)}\,\rho((1-b)x). $$ Note that these are positive by $(1)$. Now we can write the convex combination $$ \rho=\rho(b)\,f_b+\rho(1-b)\,g_b. $$ As $\rho$ is pure (this means that it is extreme), we have $\rho=f_b$. That is, $$ \rho(x)=\frac1{\rho(b)}\,\rho(bx),\ \ x\in B\otimes M_k, $$ or $\rho(bx)=\rho(b)\rho(x)$ for all $x\in B\otimes M_k$. Now you finish by letting $\omega$ the state on $B$ given by $\omega(x)=\rho(x\otimes 1)$, and $\eta$ the state on $M_k$ given by $\eta(y)=\rho(1\otimes y)$. Then, for any $x\in B$ and $y\in M_k$, $$ \rho(x\otimes y)=\rho((x\otimes 1)(1\otimes y))=\rho(x\otimes1)\rho(1\otimes y)=\omega(x)\eta(y). $$