Posts Tagged ‘PI ring’

PI rings; definition and basic examples

Posted: February 20, 2024 in Noncommutative Ring Theory Notes, Polynomial Identity Rings
Tags: , , ,
0

Rings in this post may or may not have identity. As always, the ring of n \times n matrices with entries from a ring R is denoted by M_n(R).

PI rings are arguably the most important generalization of commutative rings. This post is the first part of a series of posts about this fascinating class of rings.

Let C be a commutative ring with identity, and let C\langle x_1, \ldots ,x_n\rangle be the ring of polynomials in noncommuting indeterminates x_1, \ldots ,x_n and with coefficients in C. We will assume that each x_i commutes with every element of C. If n=1, then C\langle x_1, \ldots ,x_n\rangle is just the ordinary commutative polynomial ring C[x].
A monomial is an element of C\langle x_1, \ldots ,x_n\rangle which is of the form y_1y_2 \cdots y_k, where y_i \in \{x_1, \cdots , x_n\} for all i. The degree of a monomial y_1y_2 \cdots y_k is defined to be k. For example, x_1x_2x_1^3x_5^2 is a monomial of degree 7. So an element of f \in C\langle x_1, \ldots ,x_n\rangle is a C-linear combination of monomials, and we say that f is monic if the coefficient of at least one of the monomials of the highest degree in f is 1. For example, x_1^2+x_2-3x_2x_3x_2+2x_4^3 is not monic because none of the monomials of the highest degree, i.e. x_2x_3x_2 and x_4^3, have coefficient 1, but x_1^2+x_2-3x_2x_3x_2+x_4^3 is monic.

Definition 1. A ring R is called a polynomial identity ring, or PI ring for short, if there exists a positive integer n and a monic polynomial f \in \mathbb{Z}\langle x_1, \ldots ,x_n\rangle such that f(r_1, \cdots , r_n)=0 for all r_i \in R. We then say that R satisfies f or f is an identity of R.

Definition 2. Let C be a commutative ring with identity, and let R be a C-algebra. If, in Definition 1, we replace \mathbb{Z} with C, we will get the definition of a PI algebra. So R is called a PI algebra if there exists a positive integer n and a monic polynomial f \in C\langle x_1, \ldots ,x_n\rangle such that f(r_1, \cdots , r_n)=0 for all r_i \in R. Note that since every ring is a \mathbb{Z}-algebra, every PI ring is a PI algebra.

Example 1. Every commutative ring R is a PI ring.

Proof. Since R is commutative, r_1r_2-r_2r_1=0, for all r_1,r_2 \in R, and therefore R satisfies the monic polynomial f=x_1x_2-x_2x_1. \ \Box

Remark 1. A PI ring could satisfy many polynomials. For example a finite field of order q satisfies the polynomial in Example 1, because it is commutative, and it also satisfies the polynomial x^q-x. Another example is Boolean rings; they satisfy the polynomial in Example 1, because they are commutative, and they also satisfy the polynomial x^2-x.

Example 2. If C is a commutative ring with identity, then R:=M_2(C) is a PI ring.

Proof. Let A_1,A_2 \in R. Then \text{tr}(A_1A_2 - A_2A_1)=0 and so, by Cayley-Hamilton,

(A_1A_2 - A_2A_1)^2 = cI_2,

for some c \in C. Thus (A_1A_2-A_2A_1)^2 commutes with every element of R, i.e.,

(A_1A_2-A_2A_1)^2A_3-A_3(A_1A_2-A_2A_1)^2=0

for all A_1,A_2,A_3 \in R. Thus R satisfies the monic polynomial

f=(x_1x_2-x_2x_1)^2x_3 - x_3(x_1x_2 - x_2x_1)^2. \ \Box

Example 3. The division ring of real quaternions \mathbb{H} is a PI ring.

Proof. Recall that \mathbb{H}=\mathbb{R}+\mathbb{R}\bold{i}+\mathbb{R}\bold{j}+\mathbb{R}\bold{k}, where \bold{i}^2=\bold{j}^2=-1, \ \bold{ij}=-\bold{ji}=\bold{k}. Let

x=a+b\bold{i}+c\bold{j}+d\bold{k} \in \mathbb{H}, \ \ \ \ \ a,b,c,d \in \mathbb{R}.

It is easy to see that x^2-2ax+a^2+b^2+c^2+d^2=0. Thus, since a,b,c,d are in the center of \mathbb{H}, we get that y(x^2-2ax)=(x^2-2ax)y for all y \in \mathbb{H}. So yx^2-x^2y=2a(yx-xy) and hence, since a is central, (yx-xy)(yx^2-x^2y)=(yx^2-x^2y)(yx-xy) for all x,y \in \mathbb{H}. Therefore \mathbb{H} satisfies the monic polynomial f=(x_1x_2-x_2x_1)(x_1x_2^2-x_2^2x_1)-(x_1x_2^2-x_2^2x_1)(x_1x_2-x_2x_1). \ \Box

Remark 2. If R is a PI ring with identity, then R could satisfy a polynomial with a nonzero constant. For example, the ring \mathbb{Z}/n\mathbb{Z} satisfies the polynomial in Example 1, and it also satisfies the polynomial f=(x-1)(x-2) \cdots (x-n)+n. Now suppose that R has no identity, and f \in \mathbb{Z}\langle x_1, \ldots ,x_n\rangle has a nonzero constant m. So f=g + m, where g \in \in \mathbb{Z}\langle x_1, \ldots ,x_n\rangle has a zero constant. Now, what is f(r_1, \cdots , r_n), \ r_i \in R? Well, It is not defined because it is supposed to be g(r_1, \cdots , r_n)+m1_R but 1_R does not exist. So if R has no identity, all identities of R must have zero constants.

Example 4. Any subring or homomorphic image of a PI ring is a PI ring.

Proof. If R satisfies f, then obviously any subring of R satisfies f too. If S is a homomorphic image of R, then S \cong R/I for some ideal I of R. Now, it is clear that for any f(x_1, \cdots , x_n) \in \mathbb{Z}\langle x_1, \ldots ,x_n\rangle and any r_1, \cdots , r_n \in R, we have f(r_1+I, \cdots , r_n+I)=f(r_1, \cdots , r_n) + I. Hence if R satisfies f, then R/I satisfies f too. \ \Box

Example 5. If I is a nilpotent ideal of a ring R, and R/I is a PI ring, then R is a PI ring.

Proof. So I^m=(0) for some positive integer m. Now suppose that R/I satisfies a monic polynomial f(x_1, \cdots , x_n) \in \mathbb{Z}\langle x_1, \ldots ,x_n\rangle. Then

I=0_{R/I}=f(r_1+I, \cdots , r_n+I)=f(r_1, \cdots , r_n) + I

for all r_i \in R, and so f(r_1, \cdots , r_n) \in I. Thus (f(r_1, \cdots , r_n))^m \in I^m=(0) and hence R satisfies the monic polynomial f^m. \ \Box

Example 6. A finite direct product of PI rings is a PI ring.

Proof. By induction, we only need to prove that for a direct product two PI rings. So let R_1, R_2 be PI rings that, respectively, satisfy f,g \in \mathbb{Z}\langle x_1, \ldots ,x_n\rangle, and let R:=R_1 \times R_2. It is clear that for any polynomial h \in \mathbb{Z}\langle x_1, \ldots ,x_n\rangle and (r_i,s_i) \in R, \ 1 \le i \le n, we have

h((r_1,s_1), \cdots , (r_n,s_n))=(h(r_1, \cdots , r_n),h(s_1, \cdots , s_n)).

Therefore

f((r_1,s_1), \cdots , (r_n,s_n))g((r_1,s_1), \cdots , (r_n,s_n))=(f(r_1, \cdots , r_n)g(r_1, \cdots , r_n), f(s_1, \cdots , s_n)g(s_1, \cdots , s_n))

=(0_{R_1},0_{R_2})=0_R,

because f(r_1, \cdots , r_n)=0_{R_1} and g(s_1, \cdots , s_n)=0_{R_2}. So R satisfies the monic polynomial fg. \ \Box

Exercise. Let R be a ring with the center C. Suppose that for every r \in R, there exist a,b,c \in C such that r^3+ar^2+br+c=0. Show that R is a PI ring.
Hint. See the proof of Example 3.

Note. The reference for this post is Section 1, Chapter 13 of the book Noncommutative Noetherian Rings by McConnell and Robson. Example 3 was added by me.