Corollary 3

Let $p$ be a prime number and let K be a perfect field of characteristic $p$. Let A be a central simple algebra of finite degree over K. Then $p$ does not divide the index of A.

Let us prove that the Brauer group of K does not contain any element of order $p$. Every Galois extension M of K of finite degree is a perfect field (V, §7, No. 1, p. 36, Proposition 2). Hence taking the $p$-th power is an automorphism of the group $M^*$. It follows that multiplication by $p$ is an automorphism of the group $H^2$(Gal(M$/K$)$,M^*$) that is isomorphic to Br(M$/K$).

Consequently, the order of [A] is prime to $p$ and, by Corollary 2, its index is not divisible by $p$.