[proofplan] We prove comparability by invoking the [central comparison theorem for projections](/theorems/9282) and then using the defining feature of a factor: its only central projections are $0$ and $1$. The central comparison theorem splits $M$ by a central projection $z$ so that $zp$ is subequivalent to $zq$ on one central summand and $(1-z)q$ is subequivalent to $(1-z)p$ on the complementary summand. In a factor, this central projection has only the two possible values $0$ and $1$, and each value gives exactly one of the desired alternatives. [/proofplan] [step:Apply central comparison to split the projections by a central projection] By the [Central Comparison Theorem][citetheorem:9282] applied to the von Neumann algebra $M$ and the projections $p,q\in M$, there exists a central projection $z\in Z(M)$ such that \begin{align*} zp \precsim zq \end{align*} and \begin{align*} (1-z)q \precsim (1-z)p. \end{align*} Here the subequivalences are Murray-von Neumann subequivalences inside $M$. [guided] The central comparison theorem is designed for exactly this situation: it compares two projections after cutting the algebra by a central projection. We may apply it because $M$ is a von Neumann algebra and $p,q\in M$ are projections. Its conclusion gives a projection $z\in Z(M)$ satisfying two comparison relations: \begin{align*} zp \precsim zq \end{align*} and \begin{align*} (1-z)q \precsim (1-z)p. \end{align*} The point is that the theorem compares $p$ and $q$ separately on the two central summands $zM$ and $(1-z)M$. In a general von Neumann algebra both summands may be nonzero, so the result is a split comparison. In a factor there are no nontrivial central projections, so one of these two summands must disappear. [/guided] [/step] [step:Use the factor condition to force the central projection to be trivial] Since $M$ is a factor, its center satisfies $Z(M)=\mathbb C1$. Equivalently, by the [[Factor Criterion by Central Projections](/theorems/9284)][citetheorem:9284], the only central projections in $M$ are $0$ and $1$. Since $z\in Z(M)$ is a projection, either $z=0$ or $z=1$. If $z=1$, then the first comparison relation becomes \begin{align*} p=1p \precsim 1q=q, \end{align*} so $p\precsim q$. If $z=0$, then the second comparison relation becomes \begin{align*} q=(1-0)q \precsim (1-0)p=p, \end{align*} so $q\precsim p$. [/step] [step:Conclude the Murray-von Neumann comparability] The two possible values of the central projection $z$ exhaust all cases. Therefore either $p\precsim q$ or $q\precsim p$, which proves the comparison theorem for projections in the factor $M$. [/step]