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Modern Algebra: Groups, Rings, Modules, Fields

Sam Macdonald

Section 8.2 Units, Zerodivisors, Domains

Subsection Absolute Units

“There is more power in unity than division.”
―Emmanuel Cleaver

Definition 8.17. Unit.

An element \(u\) of a unital ring \(R\) with \(1\neq 0\) is called a unit there exists \(v\in R\) such that \(uv=1\) and \(vu=1\text{.}\) If such a \(v\) exists, it is unique, it is called the inverse of \(u\) and denoted by \(u^{-1}\text{.}\)

Definition 8.18. Group of Units.

The set of units of a non-trivial unital ring \(R\) is denoted \(R^\times\text{.}\) This forms a group \((R^\times,\cdot)\) with respect to multiplication.
Finally, we have a formal definition for groups of units.

Exercise 8.19. Sweet, Sweet Rigor.

The group of units is indeed a group \((R^\times,\cdot)\) with respect to multiplication.
Well, that was validating.
 1 
Not as validating as it will be when we prove that matrix multiplication is associative, finally allowing us to complete the proof that \(\GL_n(F)\) is indeed a group. Soon...
Months in the making. But enough vindication: we move forward. It’s all we’ve ever known.

Remark 8.21.

Theorem 8.20 is often the standard way of proving that something is a field.
If you’ll remember all the way back to Example 1.16, you’ll see that we verified Theorem 8.20 for \(\Q,\R,\) and \(\C\text{.}\) This also explains why \(\Z^\times\) and \(\Z/n^\times\) had a different structure: they weren’t fields.

Example 8.22. \(\Z/p\) is a Field!

In Exercise 1.19 we showed that \(\Z/p^\times=\Z/p\sm\{0\}\text{.}\) Thus \(\Z/p\) is a field for all primes \(p\text{.}\)

Exercise 8.23. Matrices and Units.

For any field \(F\) we have \(\operatorname{M}_{n}(F)^\times = \operatorname{GL}_n(F)\text{.}\)

Remark 8.24.

Units in matrix rings are called invertible matrices, but they aren’t very useful and we’ll probably never see them again.
 2 
That was a joke. I’m very funny.

Exercise 8.25. Gaussian Units.

Let \(\Z[i] = \{a + bi \mid a,b \in \Z\}\) be the ring of Gaussian integers. Define a function
\begin{equation*} N:\Z[i]\to\Z, N(a+bi)=a^2+b^2. \end{equation*}
  1. Prove that \(N(\a\b)=N(\a)N(\b)\) for any \(\a,\b\in \Z[i]\text{.}\)
  2. Use the function \(N\) to determine the units of \(\Z[i]\text{.}\)
Solution.
  1. Let \(\alpha,\beta\in\Z[i]\) and behold
    \begin{equation*} \begin{aligned} N(\alpha\beta)&=N((a+bi)(c+di))\\ &=N(ac-bd+adi+bci)\\ &=(ac-bd)^2+(ad+bc)^2\\ &=((ac)^2-2acbd+(bd)^2)+((ad)^2+2abcd+(bc^2))\\ &=a^2c^2+b^2d^2+a^2d^2+b^2c^2\\ &=(a^2c^2+a^2d^2)+(b^2c^2+b^2d^2)\\ &=a^2(c^2+d^2)+b^2(c^2+d^2)\\ N(\alpha)N(\beta)&=(a^2+b^2)(c^2+d^2). \end{aligned} \end{equation*}
    So there’s that done.
  2. Suppose \(\alpha\beta=1\text{.}\) Then by Part (1) we know
    \begin{equation*} N(\alpha)N(\beta)=N(\alpha\beta)=N(1)=1, \end{equation*}
    so \((a^2+b^2)(c^2+d^2)=1\text{,}\) with \(a,b,c,d\in\Z\text{.}\) Thus we have \(a^2+b^2=1=c^2+d^2\text{.}\) So the units of \(\Z[i]\) are \(\{\pm 1,\pm i\}\text{.}\)

Remark 8.26.

The function \(N\) in Exercise 8.25 is some incredibly spicy foreshadowing of something called a norm function, which we’ll see more of in Section 10.1.

Subsection A Zero Divided Cannot Stand

“The only way on Earth to multiply happiness is to divide it.”
―Paul Scherrer
We might not be able to divide by zero, but sometimes we can get the next best thing.

Definition 8.27. Zerodivisor.

A zerodivisor in a ring \(R\) is an element \(x\ne 0\) such that \(xy = 0\) or \(yx=0\) for some \(y \ne 0\text{.}\)
 3 
We once again jump the gun on the lack-of-hyphenation situation. Language evolves at its own rate, but who has time for that?
However, as it turns out, most of the time we are trying to avoid the next best thing.

Definition 8.28. Integral Domain.

A unital ring \(R\) is an integral domain (often shortened to domain) if \(1 \ne 0\text{,}\) \(R\) is commutative, and \(R\) has no zerodivisors.

Remark 8.29.

Note that by Exercise 8.4 saying that \(1\neq0\) is equivalent to saying that \(R\) is nontrivial.

Example 8.30. \(\Z\) is an Integral Domain.

The ring of integers \(\Z\) is an integral domain. Find two nonzero integers that multiply to \(0\text{.}\) I dare you.
 4 
This is, in our humble opinion, how all conjectures should be proven moving forward.
Unlike in groups, which have a notion of Cancellation, rings do not necessarily come equipped with this property. One of the main niceties of integral domains is that cancellation is indeed possible (and in many cases encouraged).
Rather than jumping in and proving too many examples directly, let’s be a little more efficient and discover a whole class of examples in one fell swoop. Well, actually two fell swoops, since we’ll need a quick lemma first.

Proof.

Suppose that \(a\) is both a zerodivisor and a unit. Then there exists \(b\neq 0\) such that \(ab=0\) or \(ba=0\text{.}\) Multiplying either of these equations by \(a^{-1}\) gives \(b=0\text{,}\) a contradiction.
All right. Now it’s one fell swoop.

Proof.

  1. Let \(F\) be a field. By Theorem 8.20 we know that every nonzero element of \(F\) is a unit. Thus \(F\) has no zerodivisors by Lemma 8.32.
  2. Let \(x\in R\text{,}\) and consider the set \(\{x^n|n\in\N\}\text{.}\) As \(R\) is finite there must exist \(n<m\in\N\) such that \(x^n=x^m\text{.}\) Since \(R\) is an integral domain Lemma 8.31 allows us to cancel an \(x^n\) from both sides, yielding \(e=x^{m-n}\text{,}\) and thus \(x\cdot x^{m-n-1}=e\text{,}\) making \(x\) a unit of \(R\text{.}\) Thus \(R\) is a field by Theorem 8.20.

Qual Watch.

Proving Part (2) of Theorem 8.33 was Part (a) of [cross-reference to target(s) "jan-2022-4" missing or not unique] on the [cross-reference to target(s) "jan-2022" missing or not unique] qualifying exam.

Exercise 8.34. Finite Zerdivisors and Units.

  1. Prove that if \(R\) is finite then every element is either a unit or a zerodivisor.
  2. Give an example of a ring \(R\) and an element \(a\in R\) which is neither a unit nor a zerodivisor.

Definition 8.35. Nilpotent.

An element \(a\) of a ring \(R\) is called nilpotent if \(a^n = 0\) for some integer \(n \geq 1\text{.}\)

Definition 8.37. Idempotent.

Let \(R\) be a ring with identity. An element \(e\in R\) is called idempotent if \(e^2=e\text{.}\)

Exercise 8.38. Potent Idempotents.

Let \(R\) be a ring with identity.
  1. Give an example of a ring which has an idempotent other than \(0\) or \(1\text{.}\)
  2. Prove that if \(e\) is idempotent, so is \(1-e\text{.}\)
  3. Suppose \(R\) is commutative and \(e\) an idempotent. Let \(Re:=\{re\mid r\in R\}\text{.}\) Prove that \(Re\) is a commutative ring (with identity \(e\)).
  4. Prove that the only idempotents in an integral domain are \(0\) and \(1\text{.}\)
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