Throughout this post, is a ring with

**Theorem** (Jacobson). If for some integer and all then is commutative.

In fact in Jacobson’s theorem, doesn’t have to be fixed and could depend on i.e. Jacobson’s theorem states that if for every there exists an integer such that then is commutative. But I’m not going to prove that here.

In this post, we’re going to prove Jacobson’s theorem. We have already proved the theorem for many values of (see here and here) and we didn’t need to have we didn’t need that much ring theory either. But to prove the theorem for any we need a little bit more ring theory.

**Lemma**. If Jacobson’s theorem holds for division rings, then it holds for all rings with

**Proof**. Let be a ring with such that for some integer and all Then clearly is reduced, i.e. has no non-zero nilpotent element. Let be the set of minimal prime ideals of

By the structure theorem for reduced rings, is a subring of the ring where is a domain. Clearly for all and all But then, since each is a domain, we get or i.e. each is a division ring. Therefore, by our hypothesis, each is commutative and hence which is a subring of is commutative too.

**Example**. Show that if for all then is commutative.

**Solution**. By the lemma, we may assume that is a division ring. Then

gives or Suppose that is not commutative and choose a non-central element Then are also non-central and so which gives contradiction!

**Remark 1**. Let be a division ring with the center If there exist an integer and such that for all then is a finite field. This is obvious because the polynomial has only a finite number of roots in and we have assumed that every element of is a root of that polynomial.

**Remark 2**. Let be a domain and suppose that is algebraic over some central subfield Then is a division ring and if then is a finite dimensional division -algebra.

**Proof**. Let So for some integer and We may assume that Then and so is invertible, i.e. is a division ring.

Since is a subring of it is a domain and algebraic over and so it is a division ring by what we just proved. Also, since for some integer we have and so

**Proof of the Theorem**. By the above lemma, we may assume that is a division ring.

Let be the center of By Remark 1, is finite. Since is a division ring, it is left primitive. Since every element of is a root of the non-zero polynomial is a polynomial identity ring.

Hence, by the Kaplansky-Amtsur theorem, and so is finite because is finite. Thus, by the Wedderburn’s little theorem, is a field.

**Remark 3**. If is finite and satisfies where is an integer which is a function of then proving that is commutative is quite easy. In fact, we show that is a finite direct product of finite fields. First note that since is finite, it’s artinian and so its Jacobson radical is nilpotent hence zero because is reduced. Thus, by the Artin-Wedderburn theorem, for some division rings Since is finite, each is finite and hence, by the Wedderburn’s little theorem, each is a finite field. Finally, since is reduced, for all So is a finite direct product of fields hence commutative.