§ 1. Principal ideal domains

1. Definition of a principal ideal domain

Recall (I, p. 104) that an ideal of a commutative ring $A$ is said to be principal if it has the form $(a) = Aa$ for some $a \in A$ .

Definition 1

A principal ideal domain is an integral domain (I, p. 116) in which every ideal is principal.

Example

The ring $\mathbf{Z}$ of rational integers is a principal ideal domain (I, p. 111). If $K$ is a commutative field, then the polynomial ring $K[X]$ in one indeterminate over $K$ is a principal ideal domain (IV, p. 11, Prop. 11); the same is true of the ring of formal power series $K[[X]]$ , for every ideal of this ring has the form $(X^n)$ (IV, p. 38, Prop. 12). \* The ring of integers in a p-adic field is a principal ideal domain. \

If $\mathbf{Q}(i)$ denotes the field obtained from the field $\mathbf{Q}$ of rational numbers by adjoining a root $i$ of the irreducible polynomial $X^2 + 1$ , then the elements $a + bi$ of $\mathbf{Q}(i)$ , where $a$ and $b$ are rational integers, form a subring $\mathbf{A}$ of $\mathbf{Q}(i)$ , called « the Gaussian integers », which is a principal ideal domain (VII, p. 50, Ex. 7). By contrast, in the field $\mathbf{Q}(\rho)$ , where $\rho$ is a root of $X^2 + 5$ , the subring $B$ consisting of elements $a + bp$ (a and b rational integers) is not a principal ideal domain (VII, p. 51, Ex. 12).

The polynomial ring $K[X, Y]$ in two indeterminates over a field $K$ is not a principal ideal domain. Indeed the nonzero constants are the only elements which divide both $X$ and $Y$ , and none of them generates the ideal generated by $X$ and $Y$ .

2. Divisibility in principal ideal domains

Let $A$ be a principal ideal domain and let $K$ be its field of fractions (I, p. 116); we will see that the ordered group $\mathscr{P}^*$ of principal fractional ideals (VI, p. 6) of $K$ is lattice ordered; more precisely:

Proposition 1

Let $K$ be the field of fractions of a principal ideal domain $A$ , and let $(x_i)_{i \in I}$ be a family of elements of $K$ having a common denominator $b \in K^*$ (in other words $bx_i \in A$ for all $i$ ). Then:

a) The family $(x_i)$ has a gcd in $K$ . b) Every gcd of $(x_i)$ can be expressed in the form $d = \sum a_i x_i$ , where the $a_i$ are elements of $A$ , all but finitely many of which are zero.

Indeed the ideal $\sum Abx$ , of $A$ is principal, and so of the form $Ad'$ . Put $d' = bd \ (d \in K)$ . From the relation $d' = \sum a_i b x_i$ , we deduce $d = \sum a_i x_i$ , where $a_i \in A$ . Hence every common divisor of the $x_i$ divides $d$ . On the other hand, since $bd$ is a common divisor of the $bx$ , by construction, it follows that $d$ is a common divisor of the $x_i$ .

Remark

Prop. 1 applies with no restrictions to an arbitrary family $(x_i)$ of elements of $A$ (take $b = 1$ ), and also to any finite family $(x_i)$ of elements of $K$ (if $x_i = c_i b_i$ with $c_i \in A$ and $b_i \in A$ , then take $b$ to be the product of the $b_i$ ).

Corollary

— Let $(x_i)$ be an arbitrary family of elements of a principal ideal domain $A$ contained as a subring in an integral domain $B$ , and let $d$ be a gcd of the family $(x_i)$ in $A$ . Then the family $(x_i)$ has gcd's in $B$ , and $d$ is one of them.

Indeed $d$ is a common divisor of the $x_i$ in $B$ . On the other hand the relation $d = \sum a_i x_i$ shows that every common divisor of the $x_i$ in $B$ divides $d$ .

An important application of this corollary is when $A = K[X]$ and $B = E[X]$ , where $K$ is a field and $E$ is an extension of $K$ (IV, p. 12, Cor. 1).

The first assertion of Prop. 1 shows that the ordered group $\mathscr{P}^*$ is lattice ordered ( $VI,$ p. 10). In particular every finite family of elements of $K$ admits an lcm. We can thus apply the results denoted (DIV) in $VI$ , pp. 10 to 17 to principal ideal domains.

The following result is a consequence of the second assertion of Prop. I :

Theorem 1 (« Bezout's identity »)

— Elements $x_i \ (i \in I)$ of a principal ideal domain $A$ are setwise coprime if and only if there exist elements $a_i \ (i \in I)$ of $A$ , all but finitely many of them zero, such that $\sum a_i x_i = 1$ .

The necessity of the condition is just Prop. 1. Conversely, if $\sum a_i x_i = 1$ then every common divisor of the $x_i$ divides 1, and so 1 is a gcd of the $x_i$ .

Proposition 2

— Let $a, b, d, m$ and $p$ be elements of the field of fractions $K$ of a principal ideal domain $A$ . a) « $d$ is a gcd of $a$ and $b$ » is equivalent to « $(d) = (a) + (b)$ ». b) « $m$ is an lcm of $a$ and $b$ » is equivalent to « $m = (a) \cap (b)$ ». c) « $p$ is an irreducible element of $A$ » is equivalent to « $(p)$ is a nonzero maximal ideal of $A$ » and to « $(p)$ is a nonzero prime ideal of $A$ ».

We have already proved a) (Prop. 1). Since the common multiples of $a$ and $b$ are the elements of $(a) \cap (b)$ , and since, by hypothesis, the ideal $(a) \cap (b)$ is principal, say $(a) \cap (b) = (m)$ , it follows that $m$ is an Icm of $a$ and $b$ , which proves b). Finally, to say that $p \neq 0$ is an irreducible element of $A$ means by definition (VI, p. 17) that $(p)$ is a maximal element of the family of principal ideals $\neq A$ of $A$ , ordered by inclusion; since $A$ has no ideals other than the principal ideals, this means that $(p)$ is a maximal ideal of $A$ , whence c), by the remark in VI, p. 17.

In a principal ideal domain $A$ , the sum (resp. the intersection) of a finite number of ideals is also called the gcd (resp. lcm) of these ideals.

Proposition 3

Let $a, b, c$ be elements of the field of fractions of a principal ideal domain $A$ , and let $d$ be a gcd of $a$ and $c$ ; then the congruence $ax \equiv b \pmod{c}$ has a solution $x_0 \in A$ if and only if $d$ divides $b$ ; in this case the elements $x$ of $A$ which satisfy $ax \equiv b \pmod{c}$ are precisely those which satisfy $x \equiv x_0 \pmod{cd}$ *.

If $ax \equiv b \pmod{c}$ , with $x \in A$ , then there exists $y \in A$ such that $b = ax + cy$ , so $d$ divides $b$ . Conversely, if $d$ divides $b$ then $b = ax_0 + cy_0$ with $x_0, y_0 \in A$ (Prop. 1), so $ax_0 \equiv b \pmod{c}$ ; moreover, the relation $ax \equiv b \pmod{c}$ is then equivalent to $a(x - x_0) \equiv 0 \pmod{c}$ ; putting $a = da'$ and $c = dc'$ , the latter becomes $a'(x - x_0) \equiv 0 \pmod{c'}$ . But this is equivalent (for $x \in A$ ) to $x - x_0 \equiv 0 \pmod{c'}$ , since $a'$ and $c'$ are coprime (VI, p. 14, Prop. 10 (DIV) and VI, p. 15, Cor. 2 to Prop. 11 (DIV)).

Proposition 4

Let $(a_i)$ , $i \leq n$ be a finite family of pairwise coprime elements of the principal ideal domain $A$ . Then the canonical homomorphism (I, p. 110) from $A / \left( \prod_{r=1}^n a_r \right)$ into the product $\prod_{r=1}^n A/(a_r)$ is an isomorphism of $A$ -algebras.

This follows from I, p. 110, Prop. 9 and from Prop. 2 a).

The conclusion of Prop. 4 is not valid if it is not assumed that the $a_i$ are pairwise coprime (cf. VII, p. 24, Prop. 9).

3. Decomposition into irreducible factors in principal ideal domains

We will now apply the results of VI, p. 18, relating to decomposition into irreducible elements, to principal ideal domains. By Prop. 2, an element $p \neq 0$ of $A$ is irreducible if and only if the ring $A/(p)$ is a field (I, p. 115, Cor. 1), that is if and only if the congruence $ax \equiv b \pmod{p}$ admits a solution in $A$ for each $b \in A$ and for each $a \in A$ which is not a multiple of $p$ .

Definition 2

Let $A$ be an integral domain. A system of representatives of irreducible elements of $A$ is a family $(p_\alpha)$ of irreducible elements of $A$ such that every irreducible element of $A$ is an associate of precisely one $p_\alpha$ .

Theorem 2

— Let $\mathbf{A}$ be a principal ideal domain and let $(p_\alpha)$ be a system of representatives of irreducible elements of $\mathbf{A}$ . Then every nonzero element $x$ of the field of fractions of $A$ can be uniquely expressed in the form

x = u \prod_{\alpha} p_{\alpha}^{n_{\alpha}},

where $u$ is an invertible element of $\mathbf{A}$ , and where the $n_{\alpha}$ are integers, all but finitely many of which are zero. For $x$ to belong to $\mathbf{A}$ it is necessary and sufficient that all the $n_{\alpha}$ be positive.

We will use the theorem about decomposition as a sum of irreducible elements (VI, p. 18, Th. 2), of which the above statement is only a translation. Since $\mathscr{P}^*$ is a lattice ordered group it will be sufficient for us to show that every nonempty set of principal ideals of $\mathbf{A}$ contains a maximal element, in order to check that the hypotheses of this theorem are indeed satisfied; now this follows from the next Lemma:

Lemma 1

Let $A$ be a ring such that every left ideal of $A$ is finitely generated. Then every nonempty set $\Phi$ of left ideals of $\mathbf{A}$ , ordered by inclusion, has a maximal element.

By Zorn's Lemma (Set Theory, III, p. 154, Th. 2) it is enough to prove that $\Phi$ is inductive. Now if $(a_\lambda)$ is a totally ordered family of elements of $\Phi$ then the union $a$ of the ideals $a_i$ is a left ideal of $\mathbf{A}$ , and so admits a finite system of generators $(a_i)_{1 \leq i \leq n}$ . Since each $a_i$ belongs to an ideal $a_{\lambda_i}$ , and since the family $(a_\lambda)$ is totally ordered, the $a_i$ all belong to the largest of the ideals $a_{\lambda_i}$ , say $a_1$ . Then $a = a_1$ belongs to $\Phi$ , which is thus indeed an inductive set.

Later we will study those rings $B$ , called noetherian rings, such that every nonempty set of ideals of $B$ contains a maximal element.

Remark

The family $(u, (n_\alpha))$ is called the decomposition of $x$ into irreducible factors; by abuse of language, we also say that the formula (1) is the decomposition of $x$ into irreducible factors. If $x = u \prod_{\alpha} p_{\alpha}^{n_{\alpha}}$ and $y = v \prod_{\alpha} p_{\alpha}^{m_{\alpha}}$ are the decompositions of $x$ and $y$ into irreducible factors, then a necessary and sufficient condition for $x$ to divide $y$ is that $n_{\alpha} \leq m_{\alpha}$ , for all $\alpha$ ; from this we deduce the formulae

\gcd(x, y) = \prod_{\alpha} p_{\alpha}^{\inf(n_{\alpha}, m_{\alpha})}

\operatorname{lcm}(x, y) = \prod_{\alpha} p_{\alpha}^{\sup(n_{\alpha}, m_{\alpha})}

The property expressed by Th. 2 is true for a more general class of rings than principal ideal domains; we will study them later as unique factorisation domains; and we will see that polynomial rings and formal power series rings in arbitrarily many indeterminates are unique factorisation domains (Comm. Alg., VII, § 3).

4. Divisibility of rational integers

As was pointed out in section 1, the ring $\mathbf{Z}$ of rational integers is a principal ideal domain; its field of fractions is $\mathbf{Q}$ . The multiplicative group $U$ of invertible elements of $\mathbf{Z}$ has two elements 1 and $-1$ . The group $Q_+^*$ of rational numbers $> 0$ contains precisely one element from each class of associate elements of $\mathbf{Q}$ ; it is thus isomorphic to the multiplicative group $\mathscr{P}^* = Q^*/U$ of principal fractional ideals of $\mathbf{Q}$ , with which it is usually identified. In particular, whenever gcd or lcm is used in the field $\mathbf{Q}$ (with respect to the ring $\mathbf{Z}$ ), it is understood that these are elements $\geqslant 0$ ; this convention allows us to speak of the gcd and the lcm of a family of rational numbers.

The irreducible integers $> 0$ in $\mathbf{Z}$ are precisely those we have called prime numbers (I, p. 50) (they are sometimes called rational prime numbers); every irreducible element of $\mathbf{Z}$ is thus of the form $p$ or $-p$ , where $p$ is a prime number, and the set $P$ of prime numbers is a system of representatives of irreducible elements of $\mathbf{Z}$ .

Proposition 5

— The set of prime numbers is infinite.

Indeed, given an arbitrary finite family $(p_i)$ ( $1 \leq i \leq n$ ) of distinct prime numbers, any prime divisor $q$ of the number $\left( \prod_{i=1}^n p_i \right) + 1$ (which is $> 1$ ) is distinct from all the $p_i$ , for otherwise it would divide 1.

5. Divisibility of polynomials in one indeterminate over a field

The polynomial ring $K[X]$ in one indeterminate over a commutative field $K$ is a principal ideal domain (IV, p. 11, Prop. 11). Its field of fractions is the field $K(X)$ of rational functions in $X$ with coefficients from $K$ . The ring $K[X]$ contains the subring of polynomials of degree 0, that is the field of constants, which is identified with $K$ ; the elements of $K^*$ are invertible in $K$ , and hence in $K[X]$ ; conversely the formula $\deg(uv) = \deg(u) + \deg(v)$ shows that every invertible polynomial has degree 0; the group $U$ of invertible elements of $K[X]$ is thus precisely $K^*$ . Thus two associate polynomials differ only by a nonzero constant factor; in particular every class of associate polynomials contains a unique monic polynomial. The subgroup of the multiplicative group $K(X)^*$ generated by the monic polynomials thus contains a unique element from each class of associate rational functions, and consequently is isomorphic to the group

\mathscr{P}^* = K(X)^*/U

of principal fractional ideals of $K(X)$ . In particular, whenever gcd or lcm in the field $K(X)$ (with respect to the ring $K[X]$ ) is mentioned, it will usually be understood that these are quotients of monic polynomials (or 0); this convention allows us to speak of the gcd or the lcm of a family of rational functions.

The irreducible elements of $K[X]$ are precisely the irreducible polynomials in the usual sense (IV, p. 13, Def. 2), and the set of monic irreducible polynomials is a system of representatives of irreducible elements of $K[X]$ .

A first degree polynomial is always irreducible. If $K$ is an algebraically closed field then the converse is true (V, p. 19, Prop. 1); thus in this case every polynomial $p(X)$ of degree $n$ in $K[X]$ can be written uniquely (up to the order of the factors) in the form

p(X) = c(X - a_1)(X - a_2) \ldots (X - a_n)

where $c$ and the $a_i$ are elements of $K$ .

Proposition 6

For any field $K$ , the set of monic irreducible polynomials in $K[X]$ is infinite.

Indeed, given an arbitrary nonempty finite family $(p_i)$ ( $1 \leq i \leq n$ ) of distinct monic irreducible polynomials, the polynomial $\left( \prod_{i=1}^n p_i \right) + 1$ is not invertible, and any monic irreducible factor $q$ of this polynomial is necessarily distinct from all the $p_i$ , otherwise it would divide 1.

Exercises

See the exercises for § 1.