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Abstract algebra A Kursobjekt for Math 331

Aaron Greicius

Section 2.2 Subrings, units, zero divisors

We will now discuss various types of special ring elements, starting with units.
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Definition 2.2.1. Units and multiplicative inverses.

Let \(a\) be an element of the nontrivial ring \(R\text{.}\) A multiplicative inverse of \(a\) is an element \(b\in R\) satisfying \(ab=ba=1\text{.}\) We denote \(b=a^{-1}\) in this case. The element \(a\) is a unit (or is invertible) if it has a multiplicative inverse. We denote by \(R^*\) the set of all units of \(R\text{:}\) i.e.,
\begin{align} R^* \amp =\{a\in R\mid a \text{ is invertible}\}\text{.}\tag{2.2.1} \end{align}
It follows directly from the ring axioms that \(R^*\) is a group with respect to the ring multiplication operation.
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As we have intimated on multiple occasions, the ring \(M_n(R)\text{,}\) for \(R\) an arbitrary commutative ring, behaves very much like the ring \(M_n(\R)\text{.}\) Let’s make official some of these similarities, postponing a proof until later in the course. At some point we will be able to use our ring theory to prove a universal identity result allowing us to transfer certain familiar properties of familiar rings “defined over \(\R\) or \(\C\)” (meaning to be revealed later) and import these to analogous rings defined over an arbitrary commutative ring \(\R\text{.}\)
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Proof.

Specimen 22. General linear group.

Let \(R\) be a nontrivial commutative ring, and let \(n\) be a positive integer. We denote by \(\GL_n(R)\) the group of units of \(\ML_n(R)\text{:}\) i.e.,
\begin{align*} \GL_n(R) \amp =\{A\in \ML_n(R)\mid A \text{ is invertible}\}\\ \amp = \{A\in \ML_n(R)\mid \det A\in R^*\}\text{.} \end{align*}
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Example 2.2.3. Units in product rings.

  1. Let \(R\) be a product ring \(R=\prod_{i\in I}R_i\text{.}\) Prove: \(R^*=\prod_{i\in I}R_i^*\text{.}\)
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  2. \(S=\mathcal{F}(X,R)\) for some nonempty set \(X\text{.}\) Prove:
    \begin{align*} S^* \amp = \{f\in \mathcal{F}(X,R)\mid f(x)\in R^* \text{ for all } x\in X\} \text{.} \end{align*}
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Solution.
Left to the reader.
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Definition 2.2.4. Division rings and fields.

A nontrivial ring \(R\) is a division ring if every nonzero element of \(R\) is a unit: i.e., \(R^*=R- \{0\}\text{.}\) A field is a commutative division ring.
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Example 2.2.5. Division rings.

Determine whether the given ring is a division ring, and whether it is a field.
  1. \(\displaystyle \Z\)
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  2. \(\displaystyle \Q\)
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  3. \(\displaystyle \R\)
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  4. \(\displaystyle \C\)
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  5. \(\Z/n\Z\) for \(n\geq 2\) a positive integer. The answer depends on \(n\text{.}\)
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  6. \(\Q[\sqrt{d}]\text{,}\) where \(d\in \Q\) does not have a square root in \(\Q\text{.}\)
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  7. \(\ML_n(R)\text{,}\) for \(R\) a commutative ring and \(n\geq 2\text{.}\)
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Solution.
We mainly leave this to the reader, restricting our comments to the rings \(\Z/n\Z\text{,}\) \(\Q(\sqrt{d})\text{,}\) and \(\ML_n(R)\) .
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We showed previously that
\begin{align} (\Z/n\Z)^* \amp =\{[a]_n\in \Z/n/Z\mid \gcd(a,n)=1\}\text{,}\tag{2.2.3} \end{align}
from whence it easily follows that \(\Z/n\Z\) is a field if and only if \(n=p\) is a prime integer.
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We claim \(\Q[\sqrt[d]]\) is a field. Assume \(\alpha=a+b\sqrt{q}\) is a nonzero element of \(\Q[\sqrt{d}]\text{.}\) This is true if and only if \(a\ne 0\) or \(b\ne 0\text{.}\) (If \(\alpha=0\) and \(a\) or \(b\) is nonzero, then we could write \(\sqrt{q}=-a/b\in \Q\) or \(\sqrt{q}=-a/b\in \Q\)). Contradiction.) Let \(\beta=a-b\sqrt{q}\text{.}\) We have
\begin{align} \alpha\beta=a^2-b^2q \amp \text{,}\tag{2.2.4} \end{align}
and this is nonzero, since otherwise we would have \(q=a^2/b^2\text{,}\) contradicting the fact that \(q\) is not a square in \(\Q\text{.}\) Thus \(\alpha\) has inverse
\begin{align*} \alpha^{-1}\amp =\frac{1}{a^2-b^2q}(a-b\sqrt{q})\in \Q[\sqrt{d}] \amp \text{.} \end{align*}
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Lastly, from Theorem 2.2.2, we know that \(A\in M_n(R)\) is invertible if and only if \(\det A\in R^*\text{.}\) Assuming \(n\geq 2\text{,}\) it is easy to construct a nonzero matrix \(A\in M_n(R)\) with \(\det A=0\text{.}\) Thus \(M_n(R)\) is not a division ring.
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Notation 2.2.6.

Having shown that \(\Z/p\Z\) is a field for \(p\) a prime integer, we will write \(\F_p=\Z/p\Z\) from now on.
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Definition 2.2.7. Zero divisors and integral domains.

Let \(R\) be a ring. A nonzero element \(a\in R\) is a zero divisor if there exists a nonzero element \(b\in R\) satisfying \(ab=0\) or \(ba=0\text{.}\)
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A ring \(R\) is an integral domain if it is (i) commutative, and (ii) contains no zero divisors.
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Example 2.2.8. Zero divisors.

For each of the rings in Example 2.2.5, determine all zero divisors and determine whether the ring is an integral domain.
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Solution.
We leave most of this to the reader.
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One interesting observation is that \(\Z/n\Z\) has zero divisors if and only if \(n\) is not prime. It follows that \(\Z/n\Z\) is a field if and only if \(\Z/n\Z\) is an integral domain! This interesting fact is generalized in Theorem 2.2.10 below.
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We are tempted to say that \(A\in M_n(R)\) is a zero divisor if and only if \(\det A\) is either zero or a zero divisor of \(R\text{.}\) Indeed, this is the case! However, even with Theorem 2.2.2 at our disposal, one of the implications involved is somewhat involved. The easy direction is the forward one: if \(A\) is a a zero divisor, then there is a nonzero matrix \(B\) such that \(AB=0\text{;}\) multiplying on the left by the adjoint matrix \(A^\dagger\text{,}\) we see that \((\det A)IB=(\det A)B=0\text{.}\) Since \(B\ne 0\text{,}\) it follows that \(\det A\) is either zero or a zero divisor in \(R\text{.}\)
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Although you might guess that this same argument could be employed to prove the reverse implication, we are foiled by the fact that \(A^\dagger\) can be the zero matrix even when \(\det A\) is nonzero! (This is true even in the familiar setting \(R=\R\text{.}\)) The reader is invited to check out this StackExchange post, which provides an answer as well as some useful references.
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Proof.

Definition 2.2.11. Nilpotents and idempotents.

An element \(a\) of the ring \(R\) is nilpotent if \(a^n=0\) for some positive integer \(n\text{;}\) it is idempotent if \(a^2=a\text{.}\)
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