See part (1) here. We will assume that is a field and is a finite dimensional central simple -algebra of degree

**Lemma 3**. Let be a Galois splitting field of with a -algebra isomorphism Then for all

*Proof*. Recall that always have a Galois splitting field (see the corollary in this post). Let and put Clearly but we want to prove that Let By Lemma 1 in part (1), there exists a -algebra isomorphism such that By Lemma 2 in part (1), and so i.e. Thus for all and all So because is Galois, and thus

**Lemma 4**. Let and be splitting fields of with algebra isomorphisms and Then for all

*Proof*. Fix a Galois splitting field of and a -algebra isomorphism Let and put

By Lemma 3, Let

Then, in order to prove the lemma, we only need to show that Let where is a maximal ideal of So is a field. Define the -algebra homomorphisms and by and for all and By Lemma 1 in part (1), there exist -algebra isomorphisms

such that and By Lemma 2, and so We also have because Hence and therefore because is injective.

**Definition**. Let be a splitting field of with a -algebra isomorphism For let The monic polynomial is called the **reduced characteristic polynomial** of

**Theorem**. Let Then and does not depend on or

*Proof. *By Lemma 2 in part (1), does not depend on By Lemma 4, and does not depend on