Corollary 2

Let K be a commutative field, and let A be a central simple algebra of finite degree over K. Let $p$ be a prime number. If $p$ divides the index of A, then it divides the exponent of A.

Let us suppose that the prime number $p$ does not divide the exponent of A and prove that it does not divide its index. It suffices to prove this result in the case when A is a field. Let L be a Galois extension of K of finite degree that splits A. Let G be the Galois group of L over K, and let H be a Sylow $p$-subgroup of G (I, §6, No. 6, p. 78). We denote by $L'= L^H$ the subfield of L of invariants of H. The exponent of $A_{(L')}$ divides that of A and is therefore prime to $p$. By Theorem 6, we have $[L : L'][A_{(L')}] = 0$ in the Brauer group of $L'$. It follows that $[A_{(L')}] = 0$ and that the field $L'$ splits A. We then apply Corollary 2 of VIII, p. 283; it follows that the index of A divides $[L': K]$ and is therefore not divisible by $p$.