Integral Domains

This post was translated from Korean by LLM (Kimi). The translation may contain errors or awkward sentences. The Korean original is the source of truth.

In the posts in this category, we examine properties of rings in a bit more detail. The first topic we cover is integral domains. ([Algebraic Structures] §Field of Fractions, ⁋Definition 5)

Euclidean Domains

Definition 1 Fix an integral domain \(A\). If a function \(N : A \to \mathbb{Z}^{\geq0}\) satisfies \(N(0) = 0\), we call it a norm on \(R\). If \(N(a) > 0\) holds for all \(a \neq 0\), we call this norm a positive norm.

The condition for a norm is very weak; for example, given the same integral domain, there are many ways to define a norm on it.

Definition 2 An integral domain \(A\) is called a Euclidean domain if there exists a norm \(N\) on \(A\) such that for any \(a, b \in A\) (with \(b \neq 0\)), there exist elements \(q, r \in A\) satisfying

\[a = qb + r \qquad\text{with $r = 0$ or $N(r) < N(b)$}\]

Here \(q\) is called the quotient and \(r\) the remainder.

If we replace \(A\) with the integers \(\mathbb{Z}\) and view \(N\) as the absolute value function \(\lvert-\rvert:\mathbb{Z} \rightarrow \mathbb{Z}^{\geq0}\), this is exactly the division algorithm for integers that we know well. As another example, if we define \(N\) on the polynomial ring \(\mathbb{K}[\x]\) to be the degree function, we obtain the polynomial division algorithm. Finally, any field \(\mathbb{K}\) is a Euclidean domain, obtained via the function \(N\) sending every \(x\in\mathbb{K}\) to \(0\). The reason this satisfies the above condition is that every (nonzero) element of a field always divides any other element.

Proposition 3 Every ideal of a Euclidean domain is principal. More precisely, for any nonzero ideal \(\mathfrak{a}\) of a Euclidean domain \(A\), if we let \(a\) be an element of \(\mathfrak{a}\) with minimal norm, then \(\mathfrak{a}=a\).

Proof

If \(\mathfrak{a}\) is the zero ideal there is nothing to prove, so assume \(\mathfrak{a}\neq 0\).

The image of \(\mathfrak{a}\setminus \{0\}\) under \(N\) is a subset of \(\mathbb{Z}^{\geq 0}\), which is a well-ordered set, so we can choose a nonzero element \(a\) of \(\mathfrak{a}\) with minimal norm.

Since \((a)\subseteq \mathfrak{a}\) is obvious, to show \((a)=\mathfrak{a}\) it suffices to show the reverse inclusion. Applying the division algorithm to an arbitrary element \(x\) of \(\mathfrak{a}\), we have

\[x=qa+r,\qquad\text{with $r = 0$ or $N(r) < N(a)$}\]

Then \(r = x - qa\), and since both \(x\) and \(qa\) lie in \(\mathfrak{a}\), so does \(r\). Now by the minimality assumption on the norm of \(a\), \(N(r) < N(a)\) is impossible, so \(r\) must be \(0\). Thus \(x=qa\), and from this we know \(\mathfrak{a} = (a)\).

Now let us define the following.

Definition 4 Let \(A\) be a commutative ring and let \(a, b \in A\) (\(b \neq 0\)). We define the following.

1. We say \(a\) is a multiple of \(b\) if there exists \(x \in A\) such that \(a = bx\). In this case we say \(b\) divides \(a\), and write \(b \mid a\).
2. A greatest common divisor of \(a\) and \(b\) is a nonzero element \(d\) satisfying the following conditions.
   • \(d \mid a\) and \(d \mid b\), and
   • if \(d'\) satisfies \(a \mid d'\) and \(b \mid d'\), then always \(d \mid d'\)

The greatest common divisor of \(a\) and \(b\) is denoted \(\gcd(a, b)\) or simply \((a, b)\).

By definition, \(b \mid a\) holding in the ring \(A\) is equivalent to \((a) \subset (b)\). In particular, if \(d\) is a common divisor of \(a\) and \(b\), then \((d)\) must contain \((a, b)\). Thus the above two conditions can be translated into the language of ideals as follows:

• \[\mathfrak{a} \subseteq (d)\]
• \((d) \subseteq (d')\) for any principal ideal \((d')\) such that \(a, b \in (d')\)

That is, the greatest common divisor of \(a\) and \(b\) (if it exists) is the element generating the smallest principal ideal containing \(a\) and \(b\). An integral domain in which this is always possible is called a GCD domain, but this definition will not appear separately in our discussion.

From the above discussion we obtain the following.

Proposition 5 Let \(a, b \in A\) in a commutative ring \(A\) be nonzero. If the ideal \((a, b)\) generated by \(a\) and \(b\) is a principal ideal \((d)\) generated by some element \(d \in A\), then \(d\) is the greatest common divisor of \(a\) and \(b\).

Then the greatest common divisor is uniquely determined. We must be somewhat careful when speaking of this uniqueness, because for example in the integers \((2)\) and \((-2)\) are the same ideal.

Proposition 6 Let \(A\) be an integral domain. Suppose two elements \(d, d' \in A\) generate the same principal ideal, i.e., \((d) = (d')\). Then there exists a unit \(u \in A\) such that \(d' = ud\).

Proof

If \(d = 0\) or \(d' = 0\) the claim is obvious, so assume \(d, d'\) are both nonzero. That is, since \((d) = (d')\), there exist \(x, y \in A\) such that

\[d = xd',\qquad d' = yd\]

are satisfied. Then from \(d = xyd\) we obtain \((1-xy)d=0\). Now from the assumption that \(A\) is an integral domain and \(d \neq 0\), we know \(xy = 1\), and therefore \(x\) and \(y\) are units and inverses of each other.

Then, just as Bézout’s lemma for the integers, the following holds for Euclidean domains.

Theorem 7 Let \(A\) be a Euclidean domain. Let \(a, b \in A\) be nonzero elements, and let \(r_n\) be the last nonzero remainder when the process of Definition 2 is repeated until it can no longer continue. Then the following hold:

1. \(r_n\) is the greatest common divisor of \(a\) and \(b\).
2. \((r_n)\) is the same ideal as \((a, b)\). In particular, \(r_n\) can be written as an \(A\)-linear combination of \(a\) and \(b\); that is, there exist \(x, y \in A\) satisfying \(r_n=ax+by\).

Proof

Let us show that \(r_n\) divides \(a\) and \(b\). The last step of the Euclidean algorithm is given by the equation

\[r_{n-1} = q_n r_n\]

and in general takes the form

\[r_{k-2}=q_{k}r_{k-1}+r_{k}\]

where we set \(r_{-2}=a\), \(r_{-1}=b\). Our claim is that all remainders \(r_k\) are divisible by \(r_n\). To show this, we use induction following the equations backwards.

First, \(r_n\mid r_{n-1}\) is obvious. Thus, assuming \(r_n \mid r_k\) and \(r_n \mid r_{k-1}\), we see from the above equation that \(r_n\) also divides \(r_{k-2}\), and hence by induction we know that \(r_n\) divides each of \(a\) and \(b\). That is, since \((a), (b)\subset (r_n)\), we know \((a,b)\subset (r_n)\). On the other hand, for \(r_n\) to be the greatest common divisor of \(a\) and \(b\), we must have \((a,b)=(r_n)\), and this is obvious because the above equations allow us to express \(r_n\) as an \(A\)-linear combination of \(a\) and \(b\).

Principal Ideal Domains

Now we define the following.

Definition 8 A principal ideal domain is an integral domain in which every ideal is principal.

Then from Proposition 3 we know that every Euclidean domain is always a PID. However, the converse does not hold. The division algorithm gives a concrete method for obtaining the greatest common divisor of two elements \(a, b\), but by Proposition 5 the following holds.

Corollary 9 Let \(A\) be a principal ideal domain and let \(a, b \in A\) be nonzero elements. Let \(d\) be a generator of the principal ideal \((a, b)\) generated by \(a, b\). Then the following hold:

1. \(d\) is the greatest common divisor of \(a\) and \(b\).

2. \(d\) can be written as an \(A\)-linear combination of \(a\) and \(b\). That is, there exist \(x, y \in A\) such that

   \[d = ax + by\]

   are satisfied.

3. \(d\) is unique in the sense of Proposition 6.

Proof

This follows from Proposition 5 and Proposition 6; the second result is obvious from the assumption \((a,b)=(d)\).

One of the useful properties of a principal ideal domain is that every prime ideal is always maximal.

Proposition 10 Every nonzero prime ideal of a principal ideal domain \(A\) is a maximal ideal.

Proof

Suppose, for contradiction, that for a nonzero prime ideal \(\mathfrak{p} = (p)\) of \(A\) there exists an ideal \(\mathfrak{m}\) with \(\mathfrak{p} \subsetneq \mathfrak{m} = (m)\). Then first, since \(p \in \mathfrak{m} = (m)\), we have \(p = rm\) for some \(r\in A\), and from the assumption that \(\mathfrak{p}\) is a prime ideal we have \(r\in \mathfrak{p}\) or \(m\in\mathfrak{p}\); but by the assumption \(\mathfrak{p} \subsetneq \mathfrak{m}\) we must have \(m\not\in \mathfrak{p}\). However, if \(r \in \mathfrak{p} = (p)\), then \(r = ps\) for some \(s\in A\), and thus \(p = rm = psm\), so \(1 = sm\). That is, \(m\) is a unit, and this contradicts the assumption that \(\mathfrak{m}\) is a maximal ideal. (By definition, \(A\) itself is not a maximal ideal.)

Unique Factorization Domains

Now we finally define unique factorization domains. To do so, let us first organize the terminology.

Definition 11 Let \(A\) be an integral domain.

1. Fix a nonzero, non-unit \(r\in A\). If whenever \(r = ab\) for some \(a, b\), at least one of \(a\) or \(b\) is necessarily a unit, we call \(r\) irreducible. Otherwise \(r\) is called reducible.
2. Fix a nonzero, non-unit \(p \in A\). If whenever \(p \mid ab\) for some \(a, b\), we always have \(p \mid a\) or \(p \mid b\), we call \(p\) prime.
3. For two elements \(a, b\), if there exists a unit \(u\) in \(A\) such that \(a=ub\), we say they are associate in \(R\).

For example, consider \(\mathbb{Z}\). We would like to claim that any element of \(\mathbb{Z}\), for example \(10\), can be uniquely factorized into the form \(2\times 5\). However, because of the unit of \(\mathbb{Z}\) other than \(1\), namely \(-1\), the equations

\[10=2\times 5=(-2)\times (-5)=(-1)^2\times 2\times 5=\cdots\]

allow us to express \(10\) as products of this form, and we would like to regard these expressions as not genuinely different. This is why in the above definition we regard two elements differing by a unit as (essentially) the same.

Meanwhile, the difference between irreducible and prime is somewhat subtle, and to properly define unique factorization domains we must clearly distinguish them. First, the following proposition is obvious.

Proposition 12 In an integral domain \(A\), a prime element is always irreducible.

Proof

Assume \(p \in A\) is a prime element, and let \(p = ab\) for some \(a, b\). Then \(p\mid a\) or \(p\mid b\); without loss of generality assume \(p\mid a\). That is, there exists \(r\in A\) such that \(a=pr\). Now from \(p=ab=prb\), we have \(p(1-rb)=0\); since \(A\) is an integral domain and \(p\) is nonzero, \(1-rb=0\). Thus \(b\) is a unit.

However, the converse does not always hold.

Example 13 First, define the following norm on \(A=\mathbb{Z}[\sqrt{-5}]\):

\[N(a + b\sqrt{-5}) := a^2 + 5b^2\]

then we know that the equation

\[N(xy) = N(x)N(y)\]

holds for all \(x,y\in A\). From this, if \(x\) is a unit of \(A\) then necessarily \(N(x)=1\), and the converse also holds.

Now consider the element \(3\) of \(A\). Then for \(xy=3\) with \(x\) and \(y\) simultaneously non-units, we must have

\[N(x) N(y)=N(3) = 9\]

\(N(x)=N(y)=3\). However, by the definition of \(N\) there is no element satisfying this, so \(3\) is irreducible.

But \(3\) is not prime. This is because

\[3 \mid (2 + \sqrt{-5})(2 - \sqrt{-5}) = 4 + 5 = 9\]

but \(3\) divides neither \(2 + \sqrt{-5}\) nor \(2 - \sqrt{-5}\).

However, in a PID this always holds.

Proposition 14 In a PID \(A\), being irreducible and being prime are equivalent for a nonzero element.

Proof

Fix an irreducible element \(p \in A\). What we must show is that \(p\) is prime, i.e., that \((p)\) is a prime ideal. Since \(A\) is a P.I.D., any ideal can be written in the form \((m)\). Now if \((p) \subseteq (m)\), then \(p = rm\) for some \(r \in A\). But \(p\) is irreducible, so one of \(r\) or \(m\) must be a unit.

• If \(r\) is a unit, then \(p\) and \(m\) are associate, and \((p) = (m)\).
• If \(m\) is a unit, then \((m) = A\).

Therefore the ideals containing \((p)\) are only \((p)\) itself or \(A\), so \((p)\) is maximal; and in a P.I.D. maximal ideals are always prime, so \((p)\) is a prime ideal. Hence \(p\) is a prime element.

Therefore \(A=\mathbb{Z}[\sqrt{-5}]\) of Example 13 is not a PID.

Example 15 Let us show that \(A=\mathbb{Z}[\sqrt{-5}]\) is actually not a PID using the definition. Our claim is that the ideal

\[\mathfrak{a}=(3, 1+\sqrt{-5})\]

is non-principal. Let us continue using the norm defined in Example 13 above. Then

\[N(3)=9,\qquad N(1+\sqrt{-5})=6\]

so if there exists \(x\in A\) such that \(\mathfrak{a}=(a)\), then \(N(x)\) must be a divisor of \(3\). However, as seen in Example 13, there is no \(x\in A\) satisfying \(N(x)=3\), so the only possibility is \(N(x)=1\), and therefore \((3, 1+\sqrt{-5})\) would be the unit ideal.

However, \(2\not\in \mathfrak{a}\). To show this, suppose for some \(x, y \in A\) that

\[2=3x+(1+\sqrt{-5})y\]

holds. Writing \(x = a_1 + a_2\sqrt{-5}\), \(y = b_1 + b_2\sqrt{-5}\) and expanding the right-hand side,

\[3x + (1+\sqrt{-5})y = (3a_1 + b_1 - 5b_2) + (3a_2 + b_1 + b_2)\sqrt{-5}\]

for this to equal \(2\) the following system of two equations

\[\begin{cases}3a_1 + b_1 - 5b_2 = 2 \\3a_2 + b_1 + b_2 = 0\end{cases}\]

must hold. But substituting \(b_1 = -3a_2 - b_2\) from the second equation into the first,

\[3a_1 - 3a_2 - 6b_2 = 2\]

we obtain an equation whose left-hand side is a multiple of \(3\) while the right-hand side is not, which is a contradiction. Thus \((3, 1+\sqrt{-5}) \neq (1)\).

Now let us define the following.

Definition 16 An integral domain \(A\) is called a unique factorization domain if the following two conditions hold for every non-zero, non-unit \(a\in A\).

1. \(a\) can be expressed as a finite product of irreducible elements of \(A\). That is, there exist irreducibles \(p_1, \dots, p_n \in A\) such that

   \[a = p_1 p_2 \cdots p_n\]

   can be written.

2. The above expression is unique up to associate. That is, if \(a = q_1 q_2 \cdots q_m\) is another such expression, then \(m = n\) and after a suitable reordering each \(p_i\) and \(q_i\) are associate.

Then the following holds.

Proposition 17 In a UFD \(A\), being irreducible and being prime are equivalent for a nonzero element.

Proof

It suffices to show that an irreducible element is always prime.

Choose an irreducible element \(p \in A\), and suppose \(p \mid ab\) for some \(a, b \in A\). We must show that \(p \mid a\) or \(p \mid b\). First, since \(A\) is a UFD, we can write \(a\), \(b\), and \(ab\) as products of irreducibles. Using this, write

\[a = p_1 \cdots p_k,\quad b = q_1 \cdots q_l\]

then \(ab = p_1 \cdots p_k q_1 \cdots q_l\). Here \(p \mid ab\) means there exists \(c \in A\) such that \(ab = pc\). Now \(p\) is also irreducible and \(ab\) is expressed as a product of irreducibles, so by the definition of a UFD, \(p\) is associate to one of the factors \(p_i\) or \(q_j\) of \(ab\), and accordingly \(p\) divides \(a\) or \(b\).

A UFD is, by its definition, an integral domain in which any element can be factored. The nice thing about prime factorizations of two integers is that their greatest common divisor comes out immediately from them.

Proposition 18 Let \(A\) be a unique factorization domain. Suppose \(a, b \in A\) are nonzero elements with the following prime factorizations:

\[a = u \cdot p_1^{e_1} p_2^{e_2} \cdots p_n^{e_n}, \qquad b = v \cdot p_1^{f_1} p_2^{f_2} \cdots p_n^{f_n}\]

where \(u, v \in A^\times\) are units, \(p_1, \dots, p_n\) are distinct irreducibles (or primes), and \(e_i, f_i \geq 0\). Then the element defined by

\[d = p_1^{\min(e_1, f_1)} p_2^{\min(e_2, f_2)} \cdots p_n^{\min(e_n, f_n)}\]

is the greatest common divisor of \(a\) and \(b\). (If all exponents are \(0\), then \(d = 1\).)

Proof

First, since the exponent \(\min(e_i, f_i)\) of each prime factor \(p_i\) of \(d\) is the minimum number of times \(p_i\) appears in the factorizations of both \(a\) and \(b\), it is obvious that \(d \mid a\) and \(d \mid b\). That is, \(d\) is a common divisor.

Now let \(c\) be an arbitrary common divisor of \(a\) and \(b\). Then the prime factorization of \(c\) is as follows:

\[c = q_1^{g_1} \cdots q_m^{g_m}\]

Since \(c \mid a, b\), each \(q_j\) must be associate to some prime factor of \(a\) or \(b\). That is, \(\{q_1, \dots, q_m\} \subseteq \{p_1, \dots, p_n\}\).

Moreover, each exponent \(g_j\) cannot exceed the exponent of \(q_j\) in \(a\) and \(b\), so \(g_j \leq \min(e_j, f_j)\). Therefore \(c \mid d\), and \(d\) is the greatest common divisor of \(a\) and \(b\).

Then the following theorem ties together the three definitions we have covered so far.

Theorem 19 Every Euclidean domain is a principal ideal domain, and every principal ideal domain is a unique factorization domain. In particular, every Euclidean domain is a unique factorization domain.

Proof

What we must show is only that every PID is a UFD. Fix a PID \(A\) and a non-zero, non-unit \(r\in A\). First, we will inductively construct a chain of prime ideals

\[(r)=(r_0) \subsetneq (r_1) \subsetneq (r_2) \subsetneq \cdots\]

If \(r\) is irreducible we end the process here; otherwise there exist non-units \(r_1, r_2\) such that \(r = r_1 r_2\). If both of these are again irreducible we terminate the process here; otherwise we can obtain the above chain of ideals by repeating this process.

Now set \(\mathfrak{a}=\bigcup_{i=0}^\infty (r_i)\). That \(\mathfrak{a}\) is an ideal is obvious, and by the assumption that \(A\) is a PID there exists \(a\in A\) such that \(\mathfrak{a}=(a)\). Then for some \(n\) we must have \(a\in (r_n)\), and then from this \(n\) onwards \((r_n)\) always contains \(a\), contradicting \((r_n)\subsetneq (r_{n+1})\).

For uniqueness, in a manner similar to Proposition 18, assume two expressions

\[r = p_1 \cdots p_m = q_1 \cdots q_n\]

are given, and find the \(q_j\)’s associate to the \(p_1\)’s in order.