Commutative Rings, Identity, and Units

The group of units $R^\times$ is far from a curiosity; it is a working tool across mathematics and beyond. In elementary number theory, $(\mathbb{Z}/n\mathbb{Z})^\times$ is the group underlying Euler's theorem and Fermat's little theorem, and its structure (cyclic when $n$ is $1,2,4,p^k,2p^k$ for odd prime $p$, and a direct product otherwise) is the engine behind primality tests and discrete-logarithm-based cryptography such as Diffie–Hellman key exchange and RSA, where the security relies on the difficulty of certain computations inside $(\mathbb{Z}/n\mathbb{Z})^\times$. In algebraic number theory, units of rings of integers in number fields (generalizing $\mathbb{Z}[i]^\times = \{\pm 1, \pm i\}$) are governed by Dirichlet's unit theorem, which computes the rank of the unit group from the field's embeddings into $\mathbb{R}$ and $\mathbb{C}$. In linear algebra and representation theory, the units of the ring of $n \times n$ matrices over a field form the general linear group $GL_n$, a non-abelian example showing the theorem's proof deliberately avoided assuming commutativity. In commutative algebra and algebraic geometry, whether a local ring's maximal ideal consists exactly of the non-units is a defining structural fact used constantly when studying local behavior of schemes. In computer algebra systems, detecting units is the first step in implementing division algorithms, since an element is "safely invertible" for computation exactly when it lies in $R^\times$. Even in coding theory, unit groups of finite rings such as $\mathbb{Z}/n\mathbb{Z}$ or polynomial quotient rings determine which linear codes admit certain symmetries.

:::quiz[What is the group of units $\mathbb{Z}[i]^\times$ of the Gaussian integers, and what is its order?]

• $\{1,-1\}$, order 2
• $\{1,-1,i,-i\}$, order 4 ✓
• All nonzero Gaussian integers, infinite order
• $\{1\}$, order 1

Ask the tutor: How does the structure of $(\mathbb{Z}/n\mathbb{Z})^\times$ change as $n$ ranges over primes, prime powers, and products of distinct primes?

Think of a ring as having two layers of structure stacked on the same underlying set: an addition layer, which is always a full abelian group (every element has an additive inverse, subtraction always works), and a multiplication layer, which is usually not a group. Multiplication is only required to be associative and to distribute over addition — there is no promise that you can "divide." The unit condition isolates exactly the elements for which division by that element makes sense. A useful mental picture is a two-lane number line for $\mathbb{Z}$: every integer has an additive inverse (its negative), so additively the whole line is symmetric and complete as a group. But multiplicatively, only $1$ and $-1$ "reach back" to $1$ when multiplied by another integer; every other integer's reciprocal ($1/2, 1/3, \dots$) escapes the ring entirely. The units are the tiny surviving island of multiplicative invertibility inside a much larger ring. For $\mathbb{Z}/n\mathbb{Z}$, the picture changes: invertibility becomes a question about the greatest common divisor with $n$. An element $\bar a$ is a unit exactly when $\gcd(a,n) = 1$, because Bézout's identity produces integers $x, y$ with $ax + ny = 1$, and reducing mod $n$ gives $\bar a \bar x = \bar 1$. This is why $(\mathbb{Z}/n\mathbb{Z})^\times$ has exactly $\varphi(n)$ elements, where $\varphi$ is Euler's totient function.

This corollary is the reason $R^\times$ is often described as a subset of $R \setminus \{0\}$: it can never contain the zero element, though it need not equal all of $R\setminus\{0\}$ — that stronger property is exactly what will define a field.

Example 1: The group of units of $\mathbb{Z}/12\mathbb{Z}$. Let $R = \mathbb{Z}/12\mathbb{Z}$, the ring of integers modulo $12$, with the usual addition and multiplication of residue classes. This is a commutative ring with identity $\bar 1$. Claim: $\bar a \in R^\times$ if and only if $\gcd(a, 12) = 1$. Proof of claim. Suppose $\gcd(a,12) = 1$. By Bézout's identity there exist integers $x, y$ with

Reducing modulo $12$ (and noting $12 y \equiv 0$), we get $\bar a \bar x = \bar 1$ in $R$, so $\bar x$ is a two-sided inverse of $\bar a$ (the ring is commutative, so $\bar a \bar x = \bar x \bar a$). Hence $\bar a \in R^\times$. Conversely, suppose $\bar a \in R^\times$, so there is $\bar x \in R$ with $a x \equiv 1 \pmod{12}$, i.e. $ax - 12k = 1$ for some integer $k$. Any common divisor $d$ of $a$ and $12$ divides the left side $ax - 12k$, hence divides $1$, forcing $d = \pm 1$. So $\gcd(a,12) = 1$. $\square$ Now we compute directly. The residues $a \in \{0,1,\dots,11\}$ with $\gcd(a,12)=1$ are:

(Check: $\gcd(2,12)=2$, $\gcd(3,12)=3$, $\gcd(4,12)=4$, $\gcd(5,12)=1$, $\gcd(6,12)=6$, $\gcd(7,12)=1$, $\gcd(8,12)=4$, $\gcd(9,12)=3$, $\gcd(10,12)=2$, $\gcd(11,12)=1$.) So $R^\times = \{\bar 1, \bar 5, \bar 7, \bar{11}\}$, and $|R^\times| = 4 = \varphi(12)$ where $\varphi(12) = 12(1-\tfrac12)(1-\tfrac13) = 4$. Let us verify the group structure explicitly by computing the multiplication table modulo $12$:

So every non-identity element squares to $\bar 1$: $\bar 5^2 = \bar 7^2 = \bar{11}^2 = \bar 1$. This means $R^\times$ is a group of order $4$ in which every element has order dividing $2$; such a group is necessarily isomorphic to the Klein four-group $\mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}$ (it cannot be cyclic of order $4$, since a cyclic group of order 4 has an element of order 4, but here the maximal order of a non-identity element is $2$). This is a genuinely useful fact: $(\mathbb{Z}/12\mathbb{Z})^\times \cong \mathbb{Z}/2\mathbb{Z}\times\mathbb{Z}/2\mathbb{Z}$, not cyclic — illustrating that the group of units need not be cyclic even though $\mathbb{Z}/n\mathbb{Z}$ itself (additively) always is.

Example 2: The group of units of the Gaussian integers $\mathbb{Z}[i]$. Let $R = \mathbb{Z}[i] = \{a + bi : a, b \in \mathbb{Z}\} \subset \mathbb{C}$, with addition and multiplication inherited from $\mathbb{C}$. This is a commutative ring with identity $1 = 1 + 0i$; one checks directly that sums and products of Gaussian integers are again Gaussian integers (closure), associativity and distributivity are inherited from $\mathbb{C}$, and $(R,+)$ is an abelian group (it is a subgroup of $(\mathbb{C},+)$, being closed under addition and negation with integer coordinates). Claim: $R^\times = \{1, -1, i, -i\}$. Proof of claim. Define the norm $N: \mathbb{Z}[i] \to \mathbb{Z}_{\geq 0}$ by

The norm is multiplicative: for $z, w \in \mathbb{Z}[i]$,

using that complex conjugation is multiplicative ($\overline{zw} = \bar z \bar w$) and that complex multiplication is commutative and associative. Suppose $u \in \mathbb{Z}[i]^\times$, with inverse $v \in \mathbb{Z}[i]$, so $uv = 1$. Applying $N$ and multiplicativity,

Both $N(u)$ and $N(v)$ are non-negative integers (as $a^2+b^2$ for integers $a,b$), and their product is $1$, so we must have $N(u) = N(v) = 1$. Now we find all $u = a+bi \in \mathbb{Z}[i]$ with $N(u) = a^2+b^2 = 1$, $a,b \in \mathbb{Z}$. Since $a^2, b^2 \geq 0$ are integers summing to $1$, one of them is $0$ and the other is $1$. This forces $(a,b) \in \{(1,0),(-1,0),(0,1),(0,-1)\}$, i.e.

So $N(u) = 1$ implies $u \in \{1,-1,i,-i\}$, showing $\mathbb{Z}[i]^\times \subseteq \{1,-1,i,-i\}$. For the reverse inclusion we exhibit explicit inverses: $1\cdot 1=1$; $(-1)(-1)=1$; $i \cdot (-i) = -i^2 = 1$; $(-i)\cdot i = 1$. So each of $1,-1,i,-i$ has an inverse in $\mathbb{Z}[i]$ (in fact each is its own inverse or paired with its negative), confirming all four lie in $\mathbb{Z}[i]^\times$. Hence $\mathbb{Z}[i]^\times = \{1,-1,i,-i\}$, a cyclic group of order $4$ generated by $i$, since $i^1 = i$, $i^2=-1$, $i^3=-i$, $i^4=1$. Note the contrast with Example 1: here the unit group of order $4$ is cyclic ($\cong \mathbb{Z}/4\mathbb{Z}$), while there it was the Klein four-group — a reminder that $|R^\times|=4$ does not determine the isomorphism type of $R^\times$ without more information about $R$.