Exploration 10: Noncommutative rings

April 15, 2024.

Questions:

TODO

Domains

In our travels we’ve built up a hefty classification of integral domains: PIDs, UFDs, fields, Euclidean domains, and more. Here is a complete classification:

Now is the time to get into noncommutative rings. We begin with the ring at the bottom, a domain: a noncommutative integral domain.

In this section, we describe left and right versions of zero divisors.

The definition of domain is not exactly “no zero divisors”, though that is a sufficient definition. A more precise definition is to say that a domain has no left zero divisors and no right zero divisors. This is to distinguish elements $a$ where some nonzero $b$ exists so that $ab=0$ (making $a$ a left zero divisor and $b$ a right zero divisor) and elements $a$ where some nonzero $b$ exists so that $ba=0$ (making $b$ a left zero divisor and $a$ a right zero divisor). Our original notion of zero divisor in a commutative ring is a two-sided zero divisor, which is an element that is both a left and right zero divisor.

The distinction becomes important because if a ring has right zero divisors and no left zero divisors, for instance, then you partially get the cancellation property of integral domains, because you can cancel factors on the left: $ab=ac$ implies $b=c$ .

A domain is then a ring (not necessarily commutative) with no nonzero left or right zero divisors.

In this section, we describe left and right versions of ideals.

What happens to ideals in the noncommutative scenario? Let’s again go through how we defined an ideal, and spot where commutativity was used.

Therefore: for noncommutative rings, a left ideal can be defined as one that absorbs multiplication on the left.

We need to show that $x[a]=[xa]$ for all $x\in [x]$ . This is the same as proving that $xa\sim xb$ iff $a\sim b$ . Let’s see how this works out: $\begin{aligned} &xa\sim xb\\ \iff& xb\in [xa]\\ \iff& xb-xa\in H\\ \iff& x(b-a)\in H\\ \iff& b-a\in H\\ \iff& b\in[a]\\ \iff& a\sim b \end{aligned}$ Note that the cancellation step $x(b-a)\in H\iff b-a\in H$ requires $xH=H$ . This imposes $xH=H$ as a requirement for multiplication in $R/H$ to be well-defined.

Similarly, a right ideal absorbs multiplication on the right, and our original notion of ideal is a two-sided ideal, which are both left and right ideals.

To quotient a ring with an ideal it must be both a left and right ideal, so there are no changes needed for our understanding of quotient rings.

In this section, we measure the degree of commutativity of a ring.

Just like with groups, we can see how commutative a ring is by studying elements that commute with all other elements. Just like with groups, the set of all such elements is called the center $Z(R)$ of the ring $R$ .

Theorem: Forming a polynomial ring

R[x]

from a ring

R

preserves its center

Z(R)

. That is,

Z(R)[x]

is in

Z(R[x])

The center of a ring $Z(R)$ is all elements that (multiplicatively) commute with every element in $R$ .
We must show that an arbitrary element $\sum_i z_ix^i\in Z(R)[x]$ commutes with every element in $R[x]$ .
Let $r$ be an arbitrary element of $R[x]$ . Since each term $z_ix^i$ commutes with everything, we can show that $r(\sum_i z_ix^i)=\sum_i rz_ix^i=\sum_i z_ix^ir=(\sum_i z_ix^i)r$
Therefore an arbitrary element $\sum_i z_ix^i\in Z(R)[x]$ commutes with all of $R[x]$ . Therefore $Z(R)[x]\subseteq Z(R[x])$ .
The converse, $Z(R[x])\subseteq Z(R)[x]$ is similarly straightforward; if an element $\sum_i r_ix^i$ is in the center of $R[x]$ , then we need to show that $r_i$ commutes with an arbitrary element $s\in R$ . $\begin{aligned} s(\sum_i r_ix^i)&=(\sum_i r_ix^i)s\\ \sum_i sr_ix^i&=\sum_i r_ix^is\\ \sum_i sr_ix^i&=\sum_i r_isx^i\\ sr_i&=r_is \end{aligned}$ implying each coefficient $r_i$ is in $Z(R)$ , meaning the polynomial $\sum_i r_ix^i$ is in $Z(R)[x]$ .

Corollary: If $R$ is a commutative ring, so is $R[x]$ .

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