Ring Theory

Introduction

Ring theory studies various kinds of rings.

The content of ring theory covered in this article is closely related to divisibility theory in number theory. First, analogous to normal subgroups in group theory, this article introduces the kernel of a ring homomorphism, called an ideal of the ring; this is actually a generalization of the concept of numbers in number theory to general rings. Then, by extending concepts such as prime numbers, the Euclidean algorithm, and prime factorization from the ring of integers to general rings, we obtain the concepts of different types of integral domains.

Many conclusions from number theory still hold in other common rings. It can be said that part of the work in ring theory is to discuss whether these number-theoretic conclusions hold in general rings; if not, what restrictions need to be imposed on the ring for these conclusions to hold.

Notation

When no ambiguity arises, this article may omit the multiplication symbol of the ring and write the ring \((R,+,\cdot)\) as \(R\). In the ring \(R\), the additive identity is also called the zero element, denoted by \(0\); the multiplicative identity is also called the unit element, denoted by \(1\).

Rings in this article are not required to have a unit element

Note that the definition of a ring in this article does not require the presence of a unit. Some texts require rings to have a unit, in which case some statements of conclusions in this article need slight adjustments. For example, ideals in this article can be defined based on subrings, but in other texts they may need to be defined based on additive subgroups.

Ideals

Similar to the group case, we can define subrings and ring homomorphisms.

Subring

For a ring \((R,+,\cdot)\) and its subset \(S\), if \((S,+,\cdot)\) is also a ring, then \(S\) is called a subring of \(R\).

Example: the ring of integers \(\mathbf Z\)

For any integer \(n\), \(n\mathbf Z=\{nk:k\in\mathbf Z\}\) is a subring of \(\mathbf Z\).

Ring homomorphism

For rings \((R,+,\cdot)\) and \((S,\oplus,\odot)\), if \(\pi\) preserves the ring's addition and multiplication operations, i.e., for all \(r_1,r_2\in R\), \(\pi(r_1+r_2)=\pi(r_1)\oplus\pi(r_2)\) and \(\pi(r_1\cdot r_2)=\pi(r_1)\odot\pi(r_2)\) hold, then the map \(\pi:R\rightarrow S\) is called a homomorphism from ring \(R\) to ring \(S\).

When the ring definition requires a unit

If the definition of a ring requires a unit element, then the definition of a ring homomorphism often also requires mapping the unit to the unit. For homomorphisms between nonzero unital rings, this additional requirement merely ensures that the homomorphism does not map the entire unital ring to the zero element.

Example: the ring of integers \(\mathbf Z\) (continued)

The map modulo any nonzero integer \(n\), i.e., \(\pi:\mathbf Z\rightarrow\mathbf Z/n\mathbf Z\), where \(\pi(a)=\bar a\), is a ring homomorphism.

The discussion of the kernel and image of a group homomorphism can be almost verbatim transferred here. The (relative) size of the image of a homomorphism determines whether the homomorphism is surjective, while the triviality of the kernel determines whether it is injective. The kernel of a ring homomorphism is defined as follows:

Kernel of a homomorphism

The kernel of a homomorphism \(\pi:R\rightarrow S\) from ring \(R\) to ring \(S\) is \(\{r\in R:\pi(r)=0\}\), denoted \(\ker\pi\), where \(0\) is the additive identity of \(S\).

Clearly, the kernel and image of a ring homomorphism are subrings. Conversely, not all subrings can be the kernel of some ring homomorphism. Subrings that can be the kernel of a ring homomorphism are called ideals of the ring.

Ideal

For a ring \(R\) and its subring \(I\), then \(I\) is called

a left ideal of \(R\) if for all \(r\in R\), \(rI\subseteq I\), where \(rI=\{ra:a\in I\}\);
a right ideal of \(R\) if for all \(r\in R\), \(Ir\subseteq I\), where \(Ir=\{ar:a\in I\}\);
an ideal of \(R\) if \(I\) is both a left ideal and a right ideal of \(R\).

Here we require the ideal \(I\) to be closed under both left and right multiplication by elements of \(R\). This condition is natural because elements in the ideal map to zero under a ring homomorphism, and any number multiplied (left or right) by zero should equal zero, which is the required closure. Additionally, since the additive structure of a ring is an abelian group, any subgroup is a normal subgroup; and the multiplicative structure of a ring is quite primitive, imposing no extra restrictions on substructures. This shows that the condition of closure under left and right multiplication is also sufficient.

Example: the ring of integers \(\mathbf Z\) (continued)

As an example, the subring \(n\mathbf Z\) mentioned earlier is actually an ideal of \(\mathbf Z\). It is the set of all multiples of \(n\). A multiple of \(n\), multiplied by any integer, yields a multiple of \(n\). In fact, all ideals of \(\mathbf Z\) are of this form; such rings are called principal ideal domains. For general rings, some ideals are not sets of multiples of a single element; the existence of such general rings is precisely the original motivation for studying ideals (rather than simply studying multiples)[^ideal-history].

Quotient Rings

As with groups, based on ideals of a ring, we can define a quotient ring on the set of all cosets (in the additive group sense). Consider the set

\[ R/I=\{a+I:a\in R\}, \]

where the coset \(a+I=\{a+b:b\in I\}\). It can be proved that if and only if \(I\) is an ideal, the operations

\[ \begin{aligned} (a+I)+(b+I)&=(a+b)+I,\\ (a+I)(b+I)&=(ab)I \end{aligned} \]

are well-defined, i.e., the results of these operations are independent of the choice of representatives in the cosets. Under these operations, \(R/I\) forms a ring. Again consistent with the group case, we can establish the first isomorphism theorem for rings, and there exists a natural homomorphism from a ring to its quotient ring. These proofs show that ideals in rings play the same role as normal subgroups in groups in the homomorphisms of the respective structures.

First isomorphism theorem

Let \(\pi:R\rightarrow S\) be a homomorphism from ring \(R\) to ring \(S\). Then \(\ker\pi\) is an ideal of \(R\), and \(R/\ker\pi\cong\pi(R)\) is a subring of \(S\).

Natural homomorphism

For a ring \(R\) and its ideal \(I\), the map \(\pi:R\rightarrow R/I\) given by \(\pi(r)=r+I\) is a surjective homomorphism from \(R\) to \(R/I\), called the natural homomorphism from ring \(R\) to the quotient ring \(R/I\).

Example: the ring of integers \(\mathbf Z\) (continued)

As an example, the ring \(\mathbf Z/n\mathbf Z\) of congruence classes modulo \(n\) is the quotient ring of \(\mathbf Z\) modulo its ideal \(n\mathbf Z\). This also explains the meaning of the notation \(\mathbf Z/n\mathbf Z\). The modulo \(n\) map \(\pi:\mathbf Z\rightarrow\mathbf Z/n\mathbf Z\) mentioned above is the natural map here, and its kernel is precisely the ideal \(n\mathbf Z\).

In the ring case, other isomorphism theorems also hold.

Second isomorphism theorem

Let ring \(R\) have a subring \(A\) and an ideal \(B\). Then \(A+B=\{a+b:a\in A,b\in B\}\) is also a subring of \(R\), \(A\cap B\) is an ideal of \(A\), \(B\) is an ideal of \(A+B\), and \((A+B)/B\cong A/(A\cap B)\).

Third isomorphism theorem

Let ring \(R\) have ideals \(I,J\) with \(I\subseteq J\). Then \(J/I\) is also an ideal of \(R/I\), and \((R/I)/(J/I)\cong R/J\).

Correspondence theorem

Let ring \(R\) have an ideal \(I\). Then there is a bijection \(\varphi:\mathcal S\rightarrow\mathcal T\) between the set \(\mathcal S=\{S:I\subseteq S\subseteq R\}\) of all subrings of \(R\) containing \(I\) and the set \(\mathcal T=\{T:T\le R/I\}\) of all subgroups of the quotient group \(R/I\), mapping \(S\in\mathcal S\) to \(S/I\in\mathcal T\). This bijection preserves inclusion of subrings, and ideals of \(R\) always map to ideals of \(R/I\).

These theorems will play a fundamental role in discussing the structure of rings and ideals later.

Operations on Ideals

Various operations can be defined on ideals of a ring. This is similar to defining concepts like greatest common divisor and least common multiple on the divisibility structure of integers.

Operations on ideals

Let ring \(R\) have ideals \(I,J\). The following operations can be defined:

Sum of ideals: \(I+J=\{a+b:a\in I,b\in J\}\);
Product of ideals: \(IJ=\{\sum_{i=1}^na_ib_i:a_i\in I,b_i\in J\}\), i.e., the set of all finite sums of products of the form \(ab\);
Intersection of ideals: \(I\cap J\).

It is easy to verify that the results of these operations are again ideals of the ring.

Example: the ring of integers \(\mathbf Z\) (continued)

Consider the case of the ring of integers \(\mathbf Z\). For ideals \(n\mathbf Z\) and \(m\mathbf Z\), we have

\[ \begin{aligned} n\mathbf Z+m\mathbf Z&=\gcd(m,n)\mathbf Z,\\ (n\mathbf Z)(m\mathbf Z)&=(mn)\mathbf Z,\\ (n\mathbf Z)\cap(m\mathbf Z)&=\mathrm{lcm}(m,n)\mathbf Z. \end{aligned} \]

In general, for a ring \(R\) and its ideals \(I\) and \(J\), we always have

\[ IJ\subseteq I\cap J\subseteq I,J\subseteq I+J. \]

Using these definitions, the Chinese remainder theorem for integers can be generalized to general rings. But before that, we need to further generalize concepts such as prime numbers and coprimality to general rings.

Maximal Ideals

The properties of a ring can be understood through the structure of its ideals.

A nonzero ring \(R\) always has two trivial ideals: \(\{0\}\) and \(R\). If the ring \(R\) is also commutative, then a ring with only these two ideals can only be a field[^simple-ring].

Theorem

Let \(R\) be a nonzero commutative unital ring. Then \(R\) is a field if and only if \(R\) has only the trivial ideals \(\{0\}\) and \(R\).

Proof

If \(R\) is a field, then for any nonzero ideal \(I\), pick any nonzero element \(a\in I\). Then for any element \(r\in R\) of the field, we have \(r=(ra^{-1})a\in (ra^{-1})I\subseteq I\), so \(I=R\). Conversely, for any \(a\in R\) with \(a\neq 0\), we can verify that \(aR=\{ar:r\in R\}\) is an ideal, which must equal \(R\). Thus there exists \(b\in R\) such that \(ab=1\), showing that \(a\) has an inverse, so \(R\) is a field.

The condition of a commutative ring is necessary here; otherwise, we need to restrict both left and right ideals to be trivial to ensure the ring is a division ring.

This conclusion can be generalized to cases where the ring itself is not a field. However, we then need to consider quotient rings and discuss conditions under which the quotient ring of a commutative nonzero unital ring is a field. The quotient ring \(R/I\) being a field means \(R/I\) has only trivial ideals. By the correspondence theorem, this means there is no ideal of the original ring \(R\) strictly between the modded-out ideal \(I\) and \(R\). Such an ideal \(I\) is called a maximal ideal.

Maximal ideal

For a ring \(R\) and its ideal \(M\), if \(M\neq R\) and the only ideals of \(R\) containing \(M\) are \(M\) and \(R\), then \(M\) is called a maximal ideal.

Theorem

Let \(R\) be a commutative nonzero unital ring with an ideal \(M\). Then the quotient ring \(R/M\) is a field if and only if \(M\) is a maximal ideal.

Example: the ring of integers \(\mathbf Z\) (continued)

For example, the ideal \(n\mathbf Z\) in the ring of integers \(\mathbf Z\) is a maximal ideal if and only if \(n\) is a prime number. For a prime \(p\), the quotient ring \(\mathbf Z/p\mathbf Z\) is a field, also denoted \(\mathbf F_p\).

Not all rings have maximal ideals, but nonzero unital rings always have maximal ideals.

Theorem (Krull)

For an ideal \(I\neq R\) of a nonzero unital ring \(R\), there always exists a maximal ideal \(M\) of \(R\) such that \(I\subseteq M\).

Proof

The idea is to use Zorn's lemma. Consider the set \(\mathcal S\) of all proper ideals of \(R\) (i.e., ideals not equal to \(R\)) containing \(I\). Since \(I\in\mathcal S\), it is nonempty and forms a partially ordered set under inclusion. For any chain \(J_0\subseteq J_1\subseteq\cdots\subseteq J_n\subseteq\cdots\) in it, let \(J\) be their union. It is easy to verify that \(J\) is also an ideal. Moreover, \(J\neq R\); otherwise \(1\in J\), meaning there exists \(n\) such that \(1\in J_n\), contradicting that \(J_n\) is a proper ideal. Thus, by Zorn's lemma, there exists a maximal ideal \(M\supseteq I\).

Maximal ideals, analogous to irreducible elements in divisibility theory, are such because the inclusion relation of ideals corresponds to the divisibility relation of integers; having no proper superset as an ideal is equivalent to having no divisor. However, the concept of maximal ideals is broader than that of irreducible elements because not all ideals are principal ideals.

Prime Ideals

The condition for a field is stricter than that for an integral domain. Ideals that ensure the quotient ring is an integral domain are called prime ideals, analogous to the concept of prime numbers in divisibility theory.

Prime ideal

For a commutative ring \(R\) and its ideal \(P\), if \(P\neq R\) and for any elements \(a,b\in R\), whenever \(ab\in P\) we always have \(a\in P\) or \(b\in P\), then \(P\) is called a prime ideal.

This definition may seem slightly abrupt, but comparing it with the definition of prime numbers, the definition of prime ideals is also natural.

Theorem

Let \(R\) be a commutative nonzero unital ring with an ideal \(P\). Then the quotient ring \(R/P\) is an integral domain if and only if \(P\) is a prime ideal.

Proof

For a commutative nonzero unital ring \(R\), the quotient ring \(R/P\) is an integral domain if and only if \(R/P\) has no zero divisors. Denote the coset \(a+P\) as \(\bar a\). \(R/P\) having no zero divisors is equivalent to \(\bar a\bar b=\bar 0\) always implying \(\bar a=\bar 0\) or \(\bar b=\bar 0\). By the correspondence theorem, this is equivalent to \(ab\in P\) always implying \(a\in P\) or \(b\in P\).

In the ring of integers \(\mathbf Z\), \(n\mathbf Z\) is both a maximal ideal and a prime ideal if and only if \(n\) is prime. In general commutative rings, maximal ideals always imply prime ideals, but the converse is not necessarily true; this can be seen from the properties of the corresponding quotient rings.

Theorem

For a commutative nonzero unital ring \(R\), its maximal ideals are necessarily prime ideals.

Later we will see that the converse holds only in rings with good properties, sufficiently similar to the ring of integers.

Principal Ideals

Similar to the concept of subgroups generated by subsets, we often consider ideals generated by a subset in ring theory.

Ideal generated by a subset

For a nonzero unital ring \(R\) and its nonempty subset \(A\subseteq R\), if \(I\) is the smallest (under inclusion) ideal of \(R\) containing \(A\), then \(I\) is called the ideal generated by the subset \(A\), denoted \((A)\). Here \(A\) is called a generating set of \((A)\).

Principal ideal

An ideal generated by a single element \(a\in R\) is called a principal ideal, denoted \((a)\). Here \(a\) is called the generator of \((a)\).

For a set \(A\), we can give a construction of the ideal it generates. First, define

\[ \begin{aligned} RA&=\{r_1a_1+\cdots+r_na_n:r_i\in R,a_i\in A,n\in\mathbf Z\},\\ AR&=\{a_1r_1+\cdots+a_nr_n:r_i\in R,a_i\in A,n\in\mathbf Z\}. \end{aligned} \]

These are actually the left ideal and right ideal generated by \(A\), respectively. Then the ideal generated by subset \(A\) is \(RAR\). For commutative rings, these structures coincide.

All ideals \(n\mathbf Z\) in the ring of integers are principal ideals, often denoted \((n)\) below.

Integral Domains

An integral domain is a nonzero commutative unital ring with no zero divisors. This concept is a generalization of the ring of integers. However, rings obtained this way may not have sufficiently good properties to allow every conclusion from the divisibility theory of integers to be directly transferred. To generalize number-theoretic conclusions, further restrictions can be imposed on integral domains. The three most common types of integral domains are Euclidean domains, principal ideal domains, and unique factorization domains; the former concepts are strictly contained in the latter.

Divisibility Relation

First, we generalize related concepts from the divisibility theory of integers to general commutative rings.

Divisibility

Let \(R\) be a commutative ring with elements \(a,b\in R\). If there exists \(x\in R\) such that \(a=bx\), then we say \(b\) divides \(a\), denoted \(b\mid a\). Here \(b\) is called a divisor of \(a\).

Associate

Let \(R\) be a commutative ring with elements \(a,b\in R\). If they differ only by a unit, i.e., there exists a unit \(u\in R\) such that \(a=bu\), then \(a\) and \(b\) are said to be associate.

Divisibility is a partial order relation on the ring, and being associate is an equivalence relation. From the perspective of ideals, \(a\mid b\) is equivalent to \((b)\subseteq (a)\), and \(a\) and \(b\) being associate is equivalent to \((a)=(b)\). Thus, when discussing elements in a ring, we usually do not distinguish between associate elements. Similar to the integer case, the greatest common divisor of \(a\) and \(b\) in a commutative ring is defined as the infimum of \(\{a,b\}\).

Greatest common divisor

For a commutative ring \(R\) and its elements \(a,b\in R\), if there exists a nonzero element \(d\in R\) such that \(d\mid a\) and \(d\mid b\), and for any \(d'\) satisfying \(d'\mid a\) and \(d'\mid b\) we have \(d'\mid d\), then \(d\) is called a greatest common divisor of \(a\) and \(b\), denoted \(\gcd(a,b)\).

In an integral domain, the greatest common divisor is uniquely determined up to associates. The following discussion is restricted to integral domains.

We can also establish a concept analogous to prime numbers in integral domains. In integer theory, there are two equivalent definitions of prime numbers, but in general integral domains, these two definitions correspond to different concepts:

Prime element

Let \(R\) be an integral domain with a nonzero element \(p\in R\). If \((p)\) is a prime ideal, i.e., \(p\) is not a unit and \(p\mid ab\) always implies \(p\mid a\) or \(p\mid b\), then \(p\) is called a prime element.

Irreducible element

Let \(R\) be an integral domain with a nonzero element \(r\in R\). If \(r\) is not a unit and for any \(a,b\in R\) with \(r=ab\), either \(a\) or \(b\) is a unit, then \(r\) is called an irreducible element, or \(r\) is irreducible. Conversely, if \(r=ab\) and neither \(a\) nor \(b\) is a unit, then \(r\) is called reducible.

It can be shown that the principal ideal \((r)\) corresponding to an irreducible element \(r\) is necessarily maximal among all principal ideals of the ring; however, in general integral domains, not all ideals are principal, so irreducible elements and maximal ideals are not equivalent.

Similar to proving that prime ideals are maximal ideals in principal ideal domains, we can prove the following in general:

Theorem

Let \(R\) be an integral domain. If \(a\in R\) is a prime element, then \(a\) is also an irreducible element.

Proof

Let \(r\in R\) be a prime element, and let \(a,b\in R\) satisfy \(r=ab\). Since \(r\) is prime, assume without loss of generality that \(r\mid a\). Then \(a=cr=cba\). Since cancellation holds in integral domains, we have \(1=bc\), so \(b\) has an inverse \(c\). This shows \(r\) is irreducible.

The converse does not hold.

Counterexample

In the quadratic integer ring \(\mathbf Z[\sqrt{-5}]\), \(3\) is an irreducible element, but \(9=3\cdot 3=(2+\sqrt{-5})(2-\sqrt{-5})\), so it is not a prime element.

Here is a proof of this counterexample; readers unfamiliar with quadratic integer rings may first read the quadratic integer ring section. Let \(N(\cdot)\) be the norm on the quadratic integer ring. For any factorization \(3=ab\), we have \(N(a)N(b)=N(3)=9\). If neither \(a\) nor \(b\) is a unit, then \(N(a)\) and \(N(b)\) are both greater than \(1\), so necessarily \(N(a)=N(b)=3\). But there is no such element in \(\mathbf Z[\sqrt{-5}]\), i.e., \(x^2+5y^2=3\) has no integer solution. This shows \(3\) is irreducible. That \(3\) is not prime means showing \(3\) does not divide \(2\pm\sqrt{-5}\), which is obvious.

Euclidean Domains

A Euclidean domain is an integral domain that allows the Euclidean algorithm (i.e., division with remainder).

Euclidean domain

For an integral domain \(R\), if there exists a map \(N:R\setminus\{0\}\rightarrow\mathbf N\) such that for any \(a,b\in R\) with \(b\neq 0\), there exist \(q,r\in R\) with \(a=qb+r\) and either \(r=0\) or \(N(r)<N(b)\), then \(R\) is called a Euclidean domain (ED). The map \(N\) is called the norm of elements in the Euclidean domain.

Other equivalent definitions

The definition used here only defines the norm on nonzero elements. Different texts may handle the definition of Euclidean domains differently. For example, some may additionally define \(N(0)=0\); but the division with remainder does not use the value of \(N(0)\), so this is irrelevant. Also, the Wikipedia definition requires the norm \(N\) to satisfy: for any nonzero \(a,b\in R\), \(N(a)\le N(ab)\). However, it is easy to verify that if a Euclidean domain \(R\) has a norm \(N(\cdot)\) satisfying the properties required by our definition, then we can define a norm \(N'(a)=\min_{b\in R\setminus\{0\}} N(ab)\) that satisfies the additional property \(N'(a)\le N'(ab)\). Therefore, these different definitions are equivalent.

This definition is a generalization of division with remainder in integers. The existence of a norm allows measuring the relative size of remainder and divisor. Thus, in the Euclidean algorithm, the norm of the remainder keeps decreasing; since the norm takes values in natural numbers, this process must terminate when \(r=0\). Thus, we obtain the Euclidean algorithm on Euclidean domains.

Being able to perform the Euclidean algorithm means that greatest common divisors can be computed efficiently in Euclidean domains. Completely analogous to integer divisibility theory, it can be proved that the result of the Euclidean algorithm is a greatest common divisor, Bézout's theorem holds, and the coefficients can be determined by the extended Euclidean algorithm.

Theorem

For a Euclidean domain \(R\) and its elements \(a,b\in R\), the result \(d\) of applying the Euclidean algorithm to \(a\) and \(b\) is a greatest common divisor of \(a\) and \(b\), and there exist \(x,y\in R\) such that \(d=ax+by\); conversely, any element of the form \(ax+by\) is a multiple of \(d\).

Note that in the language of ring theory, all elements of the form \(ax+by\) are precisely the elements of the ideal \((a,b)\), and this theorem says that \((a,b)\) must be the principal ideal \((d)\).

In fact, ideals in Euclidean domains are always principal.

Theorem

Ideals in Euclidean domains are principal ideals.

Proof

Let \(R\) be a Euclidean domain and \(I\) an ideal. If \(I=\{0\}\), it is clearly principal. Suppose \(I\) is nonzero. By definition, there is a norm \(N(\cdot)\) on \(R\). Take a nonzero element \(d\in I\) with minimal norm. Then for any \(a\in I\), we have \(a=qd+r\) with \(r=0\) or \(N(r)< N(d)\). Since \(r=a-qd\in I\), by the choice of \(d\) we must have \(r=0\), so \(a=qd\in (d)\). This shows \(I\) is necessarily principal.

Principal Ideal Domains

An integral domain where all ideals are principal is called a principal ideal domain. This is a class of integral domains with very good properties, also very common. In these rings, the concept of ideals coincides with the concept of multiples in integers.

Principal ideal domain

For an integral domain \(R\), if every ideal is principal, then \(R\) is called a principal ideal domain (PID).

Thus, the last theorem of the previous section can be restated as:

Theorem

Euclidean domains are principal ideal domains.

In principal ideal domains, maximal ideals are equivalent to ideals generated by irreducible elements. Similar to primes and irreducible elements being equivalent in integers, in principal ideal domains these two concepts are also equivalent, so maximal ideals and prime ideals are completely equivalent.

Theorem

Let \(R\) be a principal ideal domain with a nonzero ideal \(I\). Then \(I\) is a prime ideal if and only if \(I\) is a maximal ideal.

Proof

We only need to prove that prime ideals are maximal ideals. Let \((p)\) be a nonzero prime ideal in a principal ideal domain \(R\), and suppose there is an ideal \((a)\) such that \((p)\subseteq(a)\subseteq R\). This means \(a\mid p\), so there exists \(b\in R\) with \(p=ab\). Since \((p)\) is prime, \(ab\in(p)\) implies \(a\in(p)\) or \(b\in(p)\). If \(a\in(p)\), then \((a)\subseteq (p)\), so \((a)=(p)\); if \(b\in(p)\), then \(b=cp\), so \(p=acp\). Since \(p\neq 0\), we have \(1=ac\), i.e., \(a\) has an inverse \(c\), so \((a)=R\). This shows \((p)\) is maximal.

Corollary

Let \(R\) be a principal ideal domain with a nonzero element \(r\). Then \(r\) is a prime element if and only if \(r\) is an irreducible element.

The analysis of Bézout's theorem in the previous section can be transferred to principal ideal domains.

Theorem

Let \(R\) be a principal ideal domain and \(a,b\in R\) nonzero elements. Let \(d\in R\) be a generator of the ideal \((a,b)\). Then a greatest common divisor of \(a\) and \(b\) is \(d\), unique up to associates; moreover, there exist \(x,y\in R\) such that \(ax+by=d\).

That is, Bézout's theorem still holds in principal ideal domains. The main difference between Euclidean domains and principal ideal domains regarding the existence of greatest common divisors is that in Euclidean domains, the greatest common divisor can be computed efficiently via the Euclidean algorithm, but in principal ideal domains there is generally no such efficient algorithm.

Unique Factorization Domains

A more general concept than principal ideal domains is unique factorization domains. The unique factorization theorem for integers is called the fundamental theorem of arithmetic. A similar unique factorization theorem actually holds in some integral domains that are not principal ideal domains. Such integral domains are called unique factorization domains.

Unique factorization domain

For an integral domain \(R\), if every nonzero non-unit element \(r\) can be written as \(r=p_1\cdots p_n\), where \(p_1,\cdots,p_n\) are irreducible elements (possibly repeated), and such a factorization is unique up to associates and reordering, then \(R\) is called a unique factorization domain (UFD).

The fundamental theorem of arithmetic says that the ring of integers \(\mathbf Z\) is a unique factorization domain.

Earlier we gave a counterexample where an irreducible element is not prime; the integral domain \(\mathbf Z[\sqrt{-5}]\) involved there no longer satisfies unique factorization. However, in all unique factorization domains, irreducible elements and prime elements are equivalent.

Theorem

For a unique factorization domain \(R\) and its nonzero element \(a\in R\), \(a\) is a prime element if and only if \(a\) is an irreducible element.

Proof

We only need to prove that irreducible elements are prime elements. For an irreducible element \(r\), if \(r\mid ab\), then there exists \(c\in R\) such that \(ab=rc\). Since \(R\) is a UFD, we can factor \(a,b,c\in R\) into products of irreducible elements. Comparing both sides, by uniqueness of factorization, \(r\) must be associate to some irreducible factor of \(a\) or \(b\), so \(r\) divides \(a\) or \(b\). This shows \(r\) is also prime.

All principal ideal domains are unique factorization domains.

Theorem

Principal ideal domains are unique factorization domains.

Proof

Let \(R\) be a principal ideal domain and \(r\in R\) a nonzero non-unit. To show \(r\) can be uniquely factored into a product of irreducible elements, we proceed in two steps: first prove existence of factorization, then prove uniqueness.

Existence is natural. If \(r\) is already irreducible, no further decomposition is needed; otherwise, there exist \(r_1,r_2\) such that \(r=r_1r_2\) and neither \(r_1\) nor \(r_2\) is a unit. Then if both \(r_1\) and \(r_2\) are irreducible, we stop; otherwise, further decompose those that are not irreducible. Thus, as long as not all factors are irreducible, we can continue the decomposition process. The decomposition must terminate after finitely many steps. Otherwise, the axiom of choice allows selecting an infinite chain \(\{r_{(i)}\}_{i=0}^\infty\) from \(R\) such that \(r_{(0)}=r\) and \(r_{(i+1)}\mid r_{(i)}\) for all \(i\in\mathbf N\), and these divisibility relations are strict, i.e., no elements in the chain are associate. In ideal language, this corresponds to a strictly infinite ascending chain of ideals: \(I_{0}\subset I_{1}\subset \cdots\subset I_{i}\subset\cdots\subset R\), where \(I_i=(r_{(i)})\). It is easy to verify that the union \(I=\bigcup_{i=0}^\infty I_i\) is still an ideal, hence necessarily principal. Let \(a\) be a generator of the principal ideal \(I\). Then there exists \(n\in\mathbf N\) such that \(a\in I_n\). So \(I=(a)\subseteq I_n\). This means such a strictly infinite ascending chain does not exist, so the decomposition process must terminate in finitely many steps.

Then prove uniqueness. We induct on the number of factors in the decomposition. The key inductive step is to verify that if \(r=p_1p_2\cdots p_n=q_1q_2\cdots q_m\) with \(n\le m\), then necessarily \(p_1\) is associate to some \(q_j\). Here we use the earlier result: in a PID, irreducible elements are prime. Since \(p_1\) is irreducible in \(R\), it is also prime, so by induction on the right-hand product, there must be some element \(q_j\) such that \(p_1\mid q_j\). So there exists \(c\in R\) with \(q_j=p_1c\). Since \(q_j\) is irreducible and \(p_1\) is irreducible, by definition \(c\) must be a unit, so \(p_1\) and \(q_j\) are associate. We can then cancel \(p_1\) and \(q_j\) from both sides, multiplying the difference in associate units to any remaining element. By the induction hypothesis, \(p_2\cdots p_n\) and \(q_1\cdots q_{j-1}q_{j+1}\cdots q_m\) must have the same number of irreducible factors and are the same up to associates. The theorem is proved.

Finally, existence of greatest common divisors still holds in unique factorization domains.

Theorem

Let \(R\) be a unique factorization domain with nonzero elements \(a,b\in R\). Suppose they can be factored as \(a=up_1^{r_1}\cdots p_n^{r_n}\) and \(b=vp_1^{s_1}\cdots p_n^{s_n}\), where \(u,v\) are units, \(p_1,\cdots,p_n\) are distinct irreducible elements, and \(r_i,s_i\) are natural numbers. Then a greatest common divisor of \(a\) and \(b\) is \(d=p_1^{\min\{r_1,s_1\}}\cdots p_n^{\min\{r_n,s_n\}}\).

This actually shows that the property of existence of greatest common divisors is weaker than unique factorization[^gcd-domain].

Ring Theory

Introduction

Ideals

Quotient Rings

Operations on Ideals

Maximal Ideals

Prime Ideals

Principal Ideals

Integral Domains

Divisibility Relation

Euclidean Domains

Principal Ideal Domains

Unique Factorization Domains

Example: Quadratic Integer Rings