Invertible Elements and Zero Divisors

This post was machine-translated from the Korean original by Marvin (via Kimi). It may contain errors or awkward phrasing — the Korean original is the source of truth.

In this post we organize the two most fundamental classes of elements in the multiplicative structure of a ring: the unit, which possesses a multiplicative inverse, and the zero divisor, which has a partner multiplying to \(0\). Both concepts have already appeared implicitly in many places. An integral domain was defined as a commutative ring with no zero divisors (§Field of Fractions, ⁋Definition 5), and a field was a division ring in which every nonzero element is a unit (§Field of Fractions, ⁋Definition 3). Here we formally define the unit, verify that its collection forms a group, show that units and zero divisors are mutually exclusive, and prove that in a finite commutative ring every element that is not a zero divisor is automatically a unit. As an immediate corollary we obtain that every finite integral domain is a field.

Unless otherwise stated, \(A\) denotes a ring with identity \(1\neq 0\); commutativity is assumed only where explicitly needed.

Units

Definition 1 An element \(u\in A\) of a ring \(A\) is called a unit if there exists an element \(v\) in \(A\) satisfying \(uv=vu=1\). Such a \(v\) is unique when it exists, and we denote it by the inverse \(u^{-1}\) of \(u\). The set of all units of \(A\) is written \(A^\times\).

The uniqueness of the inverse follows immediately from the associativity of multiplication. For instance, if \(vu=1\) and \(uw=1\), then

\[v=v\cdot 1=v(uw)=(vu)w=1\cdot w=w\]

so the left and right inverses agree, and hence the inverse of an element possessing inverses on both sides is unique.

Moreover, \(A^\times\) is closed under multiplication. Closure under the empty product—that is, \(1\in A^\times\)—is clear from \(1\cdot 1=1\). If \(u,u'\in A^\times\), then

\[(uu')(u'^{-1}u^{-1})=u(u'u'^{-1})u^{-1}=uu^{-1}=1\]

and similarly \((u'^{-1}u^{-1})(uu')=1\), so \(uu'\in A^\times\) and its inverse is \(u'^{-1}u^{-1}\). Also, if \(u\in A^\times\), then \(u^{-1}\) has \(u\) as its inverse, so \(u^{-1}\in A^\times\). Therefore \(A^\times\) is a group under the multiplication of \(A\), called the unit group of \(A\).

Example 2 In the ring \(\mathbb{Z}\), the only integers \(u,v\) satisfying \(uv=1\) are \(u=v=1\) or \(u=v=-1\), so \(\mathbb{Z}^\times=\{1,-1\}\).

In any division ring \(A\), every nonzero element has an inverse by definition, so \(A^\times=A\setminus\{0\}\) (§Field of Fractions, ⁋Definition 3). In particular, for a field \(\mathbb{K}\), the set \(\mathbb{K}^\times=\mathbb{K}\setminus\{0\}\) is a commutative group under multiplication.

One subtlety is that a unit in a ring need not remain a unit in a given subring. For example, \(2\in\mathbb{Q}\) lies in \(\mathbb{Q}^\times\), yet in \(\mathbb{Z}\) there is no integer \(v\) with \(2v=1\), so \(2\not\in\mathbb{Z}^\times\).

Zero Divisors and Regular Elements

In the multiplicative structure, the opposite extreme from a unit may be said to be \(0\). Extending this, we examine elements that multiply to \(0\), namely zero divisors. (§Field of Fractions, ⁋Definition 5) Let us first refine the definition a little further.

Definition 3 For an element \(a\in A\) of a ring \(A\), we define the following.

1. \(a\) is a left zero divisor if there exists a nonzero element \(b\neq 0\) such that \(ab=0\).
2. Likewise, if \(ba=0\) for some \(b\neq 0\), we call \(a\) a right zero divisor.
3. An element that is not a zero divisor is called a regular element or a non-zero-divisor.

§Field of Fractions, ⁋Definition 5 does not distinguish between left and right zero divisors, and encompasses both notions. In particular, in a commutative ring this distinction vanishes, so omitting the direction causes no confusion.

By definition, \(0\) itself is a zero divisor: whenever \(A\neq 0\), multiplying \(0\) by any nonzero element (for instance \(1\)) yields \(0\). The contrapositive shows that every regular element must be nonzero.

Our interest lies in the relationship between regular elements and units. One direction always holds in an arbitrary ring.

Proposition 4 Every unit in a ring \(A\) is a regular element.

Proof

Suppose, for contradiction, that \(u\in A^\times\) and \(ub=0\) for some \(b\in A\). Multiplying both sides on the left by \(u^{-1}\) gives

\[b=1\cdot b=(u^{-1}u)b=u^{-1}(ub)=u^{-1}\cdot 0=0\]

so \(b=0\), a contradiction. A similar argument applies assuming \(bu=0\), and therefore \(u\) is a regular element.

Proposition 4 shows that every unit is regular, but the converse fails in general. For example, in \(\mathbb{Z}\) the element \(2\) is easily seen to be regular, yet it is not a unit. (Example 2)

However, if the ring is finite, the converse does hold. This rests essentially on the definition of a finite set (§Natural Numbers and Infinite Sets, ⁋Definition 1): a function from a finite set to itself is automatically bijective as soon as it is either injective or surjective.

Theorem 5 In a finite ring \(A\), an element is a regular element if and only if it is a unit.

Proof

By Proposition 4 every unit is regular, so it suffices to show that a regular element is a unit. Let \(a\in A\) be regular, and consider the left multiplication map

\[\lambda_a:A\rightarrow A;\qquad x\mapsto ax\]

If \(\lambda_a(x)=\lambda_a(y)\), then \(a(x-y)=0\), and since \(a\) is regular we have \(x-y=0\), i.e. \(x=y\). Thus \(\lambda_a\) is injective. But \(A\) is finite, and an injective self-map of a finite set is surjective; hence \(\lambda_a\) is surjective. Therefore there exists \(v\in A\) with \(\lambda_a(v)=1\), which means \(av=1\). In the same way, surjectivity of the right multiplication map yields a left inverse for \(a\), and the argument immediately following Definition 1 shows that these two inverses must coincide.

The most important corollary of this theorem concerns integral domains.

Corollary 6 Every finite integral domain is a field.

Proof

By definition any finite integral domain \(A\) is commutative and satisfies \(0\neq 1\). (§Field of Fractions, ⁋Definition 5) To show further that \(A\) is a field, we must verify that every nonzero element is a unit. Since an integral domain has no zero divisors other than \(0\), any nonzero element \(a\) is regular and hence a unit by Theorem 5.

Examples

Now let us examine some examples illustrating the results above.

Example 7 Consider the ring \(\mathbb{Z}/n\mathbb{Z}\) defined for \(n\geq 1\). An element \(a+n\mathbb{Z}\) of this ring is a unit precisely when there exists some \(x+n\mathbb{Z}\) such that

\[(a+n\mathbb{Z})(x+n\mathbb{Z})=1+n\mathbb{Z}\]

that is, when there exist integers \(x,k\) satisfying \(ax-kn=1\). By [Number Theory] §Euclidean Algorithm and Bézout’s Identity, ⁋Theorem 3, the existence of such \(x,k\) is equivalent to \(\gcd(a,n)=1\), i.e. to \(a\) being coprime to \(n\). Consequently,

\[(\mathbb{Z}/n\mathbb{Z})^\times=\{a+n\mathbb{Z}:\gcd(a,n)=1\}\]

and the order of this group is \(\varphi(n)\), the number of integers between \(1\) and \(n\) coprime to \(n\) ([Number Theory] §Euler’s Theorem and the Phi Function, ⁋Definition 1).

On the other hand, if \(\gcd(a,n)=d>1\) and \(a+n\mathbb{Z}\neq 0+n\mathbb{Z}\), then \(a+n\mathbb{Z}\) is a zero divisor. Indeed, \(n/d+n\mathbb{Z}\neq 0+n\mathbb{Z}\) and

\[(a+n\mathbb{Z})(n/d+n\mathbb{Z})=a\cdot(n/d)+n\mathbb{Z}=(a/d)n+n\mathbb{Z}=0+n\mathbb{Z}\]

Hence every nonzero element of \(\mathbb{Z}/n\mathbb{Z}\) is either a unit or a zero divisor, and this classification again illustrates Theorem 5.

In particular, when \(n=p\) is prime, the integers \(1,\ldots,p-1\) are all coprime to \(p\), so \((\mathbb{Z}/p\mathbb{Z})^\times=\mathbb{Z}/p\mathbb{Z}\setminus\{0+p\mathbb{Z}\}\) and \(\mathbb{Z}/p\mathbb{Z}\) is a finite integral domain with no zero divisors. By Corollary 6 it is a field, namely the prime field \(\mathbb{F}_p\) with \(p\) elements (§Fields).

Meanwhile, the unit group of a product ring is determined componentwise, because multiplication in a product ring is computed componentwise.

Proposition 8 For the product \(A=A_1\times\cdots\times A_n\) of rings \(A_1,\ldots,A_n\), an element \((a_1,\ldots,a_n)\) is a unit of \(A\) if and only if each \(a_i\) is a unit of \(A_i\). That is, as groups,

\[A^\times=A_1^\times\times\cdots\times A_n^\times\]

holds.

Proof

Multiplication in \(A\) is componentwise and the identity is \((1,\ldots,1)\). If \(a=(a_1,\ldots,a_n)\) is a unit, then there exists \(b=(b_1,\ldots,b_n)\) with \(ab=ba=(1,\ldots,1)\), which means \(a_ib_i=b_ia_i=1\) in each component; thus each \(a_i\) is a unit and \(b_i=a_i^{-1}\).

Conversely, if each \(a_i\) is a unit, then \(b=(a_1^{-1},\ldots,a_n^{-1})\) satisfies \(ab=ba=(1,\ldots,1)\), so \(a\) is a unit. Therefore \(a\in A^\times\) is equivalent to each \(a_i\in A_i^\times\), and the map

\[A^\times\longrightarrow A_1^\times\times\cdots\times A_n^\times,\qquad (a_1,\ldots,a_n)\longmapsto(a_1,\ldots,a_n)\]

is a well-defined bijection by the above equivalence; since multiplication is componentwise, it is a group homomorphism, hence an isomorphism.

On the other hand, for a matrix ring the unit group becomes the general linear group.

Example 9 Consider the ring \(\Mat_n(R)\) of \(n\times n\) matrices with entries in a ring \(R\). By definition, a unit of \(\Mat_n(R)\) is a matrix possessing a two-sided inverse under multiplication, i.e. an invertible matrix. The set of all such matrices is called the general linear group and denoted \(\GL(n;R)\). Thus

\[\Mat_n(R)^\times=\GL(n;R)\]

(§Matrices, ⁋Definition 1). It is known that when \(R\) is a commutative ring, a matrix \(M\in \Mat_n(R)\) is invertible if and only if its determinant \(\det M\) lies in \(R^\times\). (§Determinants, ⁋Corollary 3) Hence in this case

\[\GL(n;R)=\{M\in \Mat_n(R):\det M\in R^\times\}\]

For example, if \(R=\mathbb{Z}\), then \(\mathbb{Z}^\times=\{1,-1\}\), so \(\GL(n;\mathbb{Z})\) consists of the integer matrices with determinant \(\pm 1\).

When \(n\geq 2\), the ring \(\Mat_n(R)\) contains nontrivial zero divisors, so the distinction between units and zero divisors is meaningful. For instance, when \(n=2\) the matrix units

\[E_{11}=\begin{pmatrix}1&0\\0&0\end{pmatrix},\qquad E_{12}=\begin{pmatrix}0&1\\0&0\end{pmatrix}\]

satisfy \(E_{12}E_{11}=0\) while \(E_{12}\neq 0\) and \(E_{11}\neq 0\); thus both are zero divisors and hence not invertible by Proposition 4.

---

References

[AM] M. F. Atiyah and I. G. Macdonald, Introduction to commutative algebra, Addison–Wesley, 1969.

[DF] D. S. Dummit and R. M. Foote, Abstract algebra, 3rd ed., Wiley, 2004.

[Lam] T. Y. Lam, A first course in noncommutative rings, 2nd ed., Graduate Texts in Mathematics 131, Springer, 2001.