We will assume that is a commutative ring with 1. Also, given an integer we will denote by the ring of polynomials in non-commuting variables and with coefficients in

**Definition.** A -algebra is called a** PI-algebra** if there exists an integer and a non-zero polynomial

such that the coefficient of at least one of the monomials of the highest degree in is 1 and for all Then the smallest possible degree of such is called the **PI-degree** of

**Remark 1**. If an algebra satisfies a polynomial then obviously every subring of and homomorphic image of will satisfy too.

**Remark 2**. It is known and easy to prove that if an algebra satisfies a non-zero polynomial of degree then will also satisfy a non-zero multi-linear polynomial of degree at most

**Example 1.** Every commutative algebra is PI because it satisfies the polynomial

**Example 2.** Let be a commutative ring and Then for any we have Thus by Cayley-Hamilton theorem for some Therefore commutes with any element of So we’ve proved that is PI because it satisfies the polynomial

**Remark 2**. Let be a commutative ring and suppose that satisfies a non-zero polynomial of degree Then, by Remark 1, will also satisfy a non-zero multi-linear polynomial

for some and Renaming the variables, if necessary, we may assume that Then for some Therefore for all Thus which is a contradiction. So does not satisfy any non-zero polynomial of degree

**Theorem.** Let be an algebra. Suppose is left primitive, a faithful simple left module and If satisfies a polynomial of degree then and

*Proof***.** Suppose Then there exists some such that either or is a homomorphic image of some subring of In either case, by remark 1, and hence satisfies . Thus by remark 2: , which is nonsense.

Great!

Thank you