Commutative Ring Zoo

Recall that $\mathbb Z$ is a ring that is commutative, and contains a multiplicative identity $1$ . Commutative rings have many rich sub-structures, and the integers actually help us uncover some of these results.

Firstly, we note that given integers $m,n$ , if $m \neq 0$ and $n \neq 0$ , then $mn \neq 0$ . This property doesn’t always hold; in the commutative ring $\mathbb Z_6$ , where elements take of the form $[x] \equiv x + 6\mathbb Z$ and whose identity element is $[0]$ , we have $[2] \neq 0$ and $[3] \neq 0$ but

$[2] \cdot [3] = [2 \cdot 3] = [6] = [0].$

We will revisit these rings later on. But for now, we can, and should, appreciate the specialness of multiplying nonzero numbers.

Henceforth, let $R$ be a commutative ring.

Definition 1. We say that $R$ is an integral domain (taking reference from the integers), if $R^{*} := R \backslash \{0\}$ is closed under multiplication. Equivalently, if $x,y \in R$ , then

$xy \neq 0 \quad \Rightarrow \quad x \neq 0\quad \lor \quad y \neq 0.$

Theorem 1. Let $I \subseteq R$ be an ideal. Then $R/I$ forms an integral domain if and only if $R\backslash I$ is closed under multiplication, i.e. for any $x,y \in R$ ,

$x \notin I\quad \wedge \quad y \notin I \quad \Rightarrow \quad x \cdot y \notin I.$

Proof. $(\Rightarrow)$ Suppose $R/I$ forms an integral domain. Fix $x, y \in R$ and suppose $x\notin I$ and $y \notin I$ . Since $R/I$ forms an integral domain,

$[x] \neq [0],\quad [y] \neq [0] \quad \Rightarrow \quad [x \cdot y] = [x] \cdot [y] \neq [0] = I.$

Therefore, $x \cdot y \notin I$ , as required.

$(\Leftarrow)$ Suppose instead that $R\backslash I$ is closed under multiplication. Fix nonzero $[x], [y] \in R/I$ . Then $x \in R \backslash I$ and $y \in R \backslash I$ , so that

$x \cdot y \in R \backslash I \quad \Rightarrow \quad [x] \cdot [y] = [x \cdot y] \neq [0].$

Therefore, $R/I$ forms an integral domain.

Definition 2. We call an ideal $I \subseteq R$ prime if $R \backslash I$ is closed under multiplication.

Equivalently, if $x,y \in R$ , then

$xy \notin I \quad \Rightarrow \quad x \notin I\quad \lor \quad y \notin I.$

Equivalently again, if $x,y \in R$ , then

$xy \in I \quad \Rightarrow \quad x \in I\quad \lor \quad y \in I.$

By Theorem 1, $R/I$ forms an integral domain if and only if $I \subseteq R$ is a prime ideal.

Example 1. By Definition 2, $R \cong R/\{0\}$ is an integral domain if and only if $\{0\} \subseteq R$ is a prime ideal.

Example 2. For any number $n \in \mathbb N$ , $n \mathbb Z \subseteq \mathbb Z$ is a prime ideal if and only if $n$ is prime. By Theorem 1, $\mathbb Z_p = \mathbb Z/p\mathbb Z$ forms an integral domain if and only if $p$ is prime.

Proof. Fix $x, y \in \mathbb Z$ .

$(\Rightarrow)$ Suppose $n$ is prime. By Euclid’s lemma,

$\begin{aligned} xy \in n \mathbb Z \quad & \Rightarrow \quad n \mid xy \\ &\Rightarrow \quad n \mid x \quad \lor \quad n \mid y \\ &\Rightarrow \quad x \in n \mathbb Z\quad \lor \quad y \in n\mathbb Z, \end{aligned}$

so that $n \mathbb Z \subseteq \mathbb Z$ is prime.

$(\Leftarrow)$ Suppose $n$ is not prime. Then there exists two positive integers $1 < r,s < n$ such that $r \cdot s \in n \mathbb Z$ . By construction,

$\begin{aligned} r \nmid n \mathbb Z \quad &\Rightarrow \quad r \in \mathbb Z \backslash n\mathbb Z, \\ s \nmid n \mathbb Z \quad &\Rightarrow \quad s \in \mathbb Z \backslash n\mathbb Z. \end{aligned}$

Hence, $r,s \in \mathbb Z \backslash n\mathbb Z$ but $r \cdot s \in n \mathbb Z$ . Hence, $\mathbb Z \backslash n\mathbb Z$ is not closed under multiplication.

Lemma 1. Let $p$ be prime. Suppose $p \mathbb Z \subseteq I \subseteq \mathbb Z$ as ideals. Then $I = p\mathbb Z$ or $I = \mathbb Z$ .

Proof. Suppose $I \neq p\mathbb Z$ . Fix $q \in I \backslash p\mathbb Z$ . Then $p \nmid q$ implies that $\gcd(p, q) = 1$ , so that

$p\mathbb Z + q \mathbb Z = \mathbb Z.$

If $q \in I$ , then $q \mathbb Z \subseteq I$ . In particular,

$\mathbb Z \subseteq p\mathbb Z + q \mathbb Z \subseteq I + I \subseteq I \subseteq \mathbb Z.$

Therefore, $I = \mathbb Z$ .

Definition 3. We call an ideal $I \subsetneq R$ maximal if given the ideals $I \subseteq J \subseteq R$ , $J = I$ or $J = R$ .

Suppose furthermore that $R$ has a multiplicative identity $1$ .

Theorem 2. Let $I \subseteq R$ be an ideal. Then $I$ is maximal if and only if every nonzero element in $R/I$ has a multiplicative inverse. In this case, we call $R/I$ a field.

Proof. $(\Rightarrow)$ Fix a nonzero element $[x] \in R/I$ . Since $[x] \neq [0] = I$ , we have $x \notin I$ . Define the set

$J := \{z + xy : z \in I, y \in R\} \supsetneq I.$

It is not hard to check that $J \subseteq R$ as an ideal. Since $I$ is maximal, $J = R$ . In particular, there exist $z \in J$ and $y \in R$ such that

$u+ xy = 1 \quad \Rightarrow \quad xy \in 1 + I.$

Since $R$ is commutative, so is $R/I$ , and

$[x] \cdot [y] = [y] \cdot [x] = [1].$

$(\Leftarrow)$ Fix any ideal $J \subseteq R$ such that $I \subsetneq J$ . Fix $x \in J \backslash I$ . Since $R/I$ is a field, there exists $y \in R\backslash I$ such that

$[xy] = [x] \cdot [y] = [1].$

Hence, $xy - 1 \in J$ implies that

$1 = xy - (xy - 1) \in J.$

Hence,

$R = \langle 1 \rangle \subseteq J \subseteq R,$

implying that $I$ is maximal, as required.

Corollary 1. A maximal ideal $I \subseteq R$ is always prime.

Proof. Fix an ideal $I \subseteq R$ . Suppose $I$ is maximal. By Theorem 2, $R/I$ is a field. By definition of a field, $R/I$ is an integral domain. By Theorem 1 and Definition 2, $I$ is prime.

By Lemma 1, proper ideals of $\mathbb Z$ are maximal if and only if they are prime. This observation hints the specialness of $\mathbb Z$ that deems it a ring par excellence. One crucial observation is that $\mathbb Z$ contains a division or a Euclidean algorithm: given integers $a, b$ and $a \neq 0$ , there exist unique integers $q, r$ such that

$b = aq + r\quad \text{and} \quad (r = 0\quad \text{or}\quad |r| < |a|).$

We generalise this idea into that of a Euclidean domain.

Definition 4. We call an integral domain $R$ with multiplicative identity $1$ a Euclidean domain if there exists a non-negative function (called a norm) $f : R \to \mathbb N_0$ such that for any $a, b \in R$ and $a \neq 0$ , there exist unique $q, r \in R$ such that

$b = aq + r\quad \text{and} \quad (r = 0\quad \text{or}\quad f(r) < f(a)).$

In the case $R = \mathbb Z$ and $f( \cdot ) = | \cdot |: \mathbb Z \to \mathbb N_0$ , we recover the usual Euclidean algorithm.

A crucial example of a non-integers Euclidean domain would be the ring of polynomials with real coefficients. We will explore this ring in greater detail in the future.

Theorem 3. Let $R$ be a Euclidean domain. For ideal $I \subseteq R$ , there exists $x \in R$ such that $I = \langle x \rangle$ . In this case, we call $R$ a principal ideal domain, abbreviated PID.

Proof. Denote the norm in $R$ by $f$ . Fix an ideal $I \subseteq R$ . If $I = \{0\}$ , then $I = \langle 0 \rangle$ . If $I = R$ , then $I = \langle 1 \rangle$ . Suppose $\{0\} \subsetneq I \subsetneq R$ (i.e. $I$ is nontrivial).

By the well-ordering principle, the non-empty set $f(I) \subseteq \mathbb N_0$ has a minimum element $f(x)$ for some $x \in I$ , so that $\langle x \rangle \subseteq I$ . We claim that $(\supseteq)$ holds. Fix $y \in I$ . Since $R$ is a Euclidean domain, there exist unique $q,r \in R$ such that

$y = qx + r\quad \text{and} \quad (r = 0\quad \text{or}\quad f(r) < f(x)).$

Since $f(x)$ is minimal, we must have $f(r) \geq f(x)$ , so that $r = 0$ . Therefore, $y = qx \in \langle x \rangle$ , establishing $I \subseteq \langle x \rangle$ .

Definition 5. A ring $R$ possesses the cancellation law if for any $x,y,z \in R$ with $z \neq 0$ ,

$xz = yz \quad \Rightarrow \quad x=y.$

Lemma 2. A commutative ring is an integral domain if and only if it possesses the cancellation law.

Proof. $(\Rightarrow)$ If $xz = yz$ , then $(x-y) z = 0$ . Hence, $x-y = 0$ or $z = 0$ . Since the latter is impossible, we must have $x-y = 0 \Rightarrow x= y$ .

$(\Leftarrow)$ Suppose $xy = yx = 0$ . If $x = 0$ , we are done. Otherwise, $x \neq 0$ , and by the cancellation law, $y = 0$ .

Theorem 4. Let $R$ be a PID. For any nontrivial ideal $I \subseteq R$ , if $I$ is prime, then $I$ is maximal.

Proof. Fix any nontrivial prime ideal $I = \langle x \rangle$ , $x \neq 0$ . Fix any ideal $J$ such that $\langle x \rangle \subsetneq J \subseteq R$ . Since $R$ is a PID, there exists $y \in R\backslash \langle x \rangle$ such that $J = \langle y\rangle$ . We claim that $R = \langle y \rangle$ .

Since $R = \langle 1 \rangle$ , it suffices to check that $1 \in \langle y \rangle$ . Since $x \in \langle y \rangle$ , there exists $u \in R$ such that $x = uy$ . Since $uy = x \in \langle x \rangle$ and $\langle x \rangle$ is prime, either $u \in \langle x \rangle$ or $y \in \langle x \rangle$ . By hypothesis, the latter cannot hold, so we must have $u \in \langle x \rangle$ . Hence, there exists $v \in R$ such that $u = vx$ , so that $x =(vx)y = x(vy)$ . By the cancellation law, $1 = vy \in \langle y\rangle$ .

Definition 6. An element $x \in R^*$ is a unit if it has a multiplicative inverse. An element $x \in R^*$ is irreducible if $x$ is not a unit, and if $x = uv$ for some $u, v \in R$ , at least one of $u, v$ is a unit.

Theorem 5. Let $R$ be a PID and fix $x \in R^*$ . The following are equivalent:

$\langle x \rangle$ is maximal.
$\langle x \rangle$ is prime.
$x$ is irreducible.

Proof. Corollary 1 and Theorem 4 establish the equivalence of the first two points. For the third point, we first suppose $\langle x \rangle$ is prime. Suppose $x = uv$ for some $u, v \in R$ . Since $uv \in \langle x \rangle$ , we must have $u \in \langle x \rangle$ or $v \in \langle x \rangle$ . Suppose the latter without loss of generality, so that $v = wx$ for some $w \in R$ . Then

$x=uv \quad \Rightarrow \quad x = u(wx) = (uw)x.$

Since $R$ is an integral domain, by the cancellation law, $uw = 1$ . Therefore, $u$ is a unit. Similarly, if $u \in \langle x \rangle$ , then $x = vu$ implies that $v$ is a unit. In either case, we have proven that $x$ is irreducible.

Now suppose instead that $x$ is irreducible. We claim that $\langle x \rangle$ is maximal. Fix $y \in R\backslash \langle x \rangle$ such that $\langle x \rangle \subsetneq \langle y \rangle \subseteq R$ . Since $x \in \langle y \rangle$ , find $u \in R$ such that $x = uy$ . Since $x$ is irreducible, either $u$ is a unit or $y$ is a unit. If $u$ is a unit, denote its inverse by $u^{-1}$ , so that $y = u^{-1}x \in \langle x \rangle$ , a contradiction. Hence, $y$ is a unit. Denote its inverse by $y^{-1}$ , so that $1 = y^{-1}y \in \langle y \rangle$ . Hence, $R \subseteq \langle y \rangle \subseteq R$ . Since $\langle y \rangle = R$ , we conclude that $\langle x \rangle$ is maximal.

We conclude our commutative ring zoo tour by defining the unique factorisation domain, which extends our ideas of unique factorisation in $\mathbb Z$ to plausibly other kinds of rings.

Definition 7. Given $a, b \in R$ , write $a \sim b$ if there exists a unit $u \in R$ such that $a = ub$ .

We leave it as an exercise to check that $\sim$ forms an equivalence relation on $R$ . Denote elements of $R/{\sim}$ by $\bar{a} := \{b \in R : a \sim b\}$ .

Definition 8. An integral domain $R$ is a unique factorisation domain (UFD) if for any $x \in R$ , there exist unique elements $\bar{p}_1, \dots, \bar{p}_n \in R/{\sim}$ such that

$\displaystyle \bar x = \prod_{i=1}^n \bar{p}_i.$

Theorem 6. If $R$ is a PID, then $R$ is a UFD.

Proof. Stay tuned for the next post.

—Joel Kindiak, 28 Jan 26, 2310H

KindiakMath

Commutative Ring Zoo

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Commutative Ring Zoo

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