Principal ideal domain

Examples include: • K: any field, • \mathbb{Z}: the ring of integers, • K[x]: rings of polynomials in one variable with coefficients in a field. (The converse is also true, i.e. if A[x] is a PID then A is a field.) Furthermore, a ring of formal power series in one variable over a field is a PID since every ideal is of the form (x^k), • \mathbb{Z}[i]: the ring of Gaussian integers, • \mathbb{Z}[\omega] (where \omega is a primitive cube root of 1): the Eisenstein integers, • Any discrete valuation ring, for instance the ring of -adic integers \mathbb{Z}_p. Non-examples Examples of integral domains that are not PIDs: • \mathbb{Z}[\sqrt{-3}] is an example of a ring that is not a unique factorization domain, since 4 = 2\cdot 2 = (1+\sqrt{-3})(1-\sqrt{-3}). Hence it is not a principal ideal domain because principal ideal domains are unique factorization domains. Also, \langle 2, 1+\sqrt{-3} \rangle is an ideal that cannot be generated by a single element. • \mathbb{Z}[x]: the ring of all polynomials with integer coefficients. It is not principal because \langle 2, x \rangle is an ideal that cannot be generated by a single polynomial. • K[x, y, \ldots], the ring of polynomials in at least two variables over a ring is not principal, since the ideal \langle x, y \rangle is not principal. • Most rings of algebraic integers are not principal ideal domains. This is one of the main motivations behind Dedekind's definition of Dedekind domains, which allows replacing unique factorization of elements with unique factorization of ideals. In particular, many \mathbb{Z}[\zeta_p], for the primitive p-th root of unity \zeta_p, are not principal ideal domains. The class number of a ring of algebraic integers gives a measure of "how far away" the ring is from being a principal ideal domain. ==Modules==

The key result is the structure theorem: If R is a principal ideal domain, and M is a finitely generated R-module, then M is a direct sum of cyclic modules, i.e., modules with one generator. The cyclic modules are isomorphic to R/xR for some x\in R (notice that x may be equal to 0, in which case R/xR is R). If M is a free module over a principal ideal domain R, then every submodule of M is again free. This does not hold for modules over arbitrary rings, as the example (2,X) \subseteq \mathbb{Z}[X] of modules over \mathbb{Z}[X] shows. ==Properties==

In a principal ideal domain, any two elements have a greatest common divisor, which may be obtained as a generator of the ideal . All Euclidean domains are principal ideal domains, but the converse is not true. An example of a principal ideal domain that is not a Euclidean domain is the ring \mathbb{Z}\bigl[\tfrac12\bigl(1+\sqrt{-19}~\!\bigr)\bigr], this was proved by Theodore Motzkin and was the first case known. In this domain no and exist, with , so that \bigl(1+\sqrt{-19~\!}\bigr)=(4)q+r, despite 1+\sqrt{-19} and 4 having a greatest common divisor of . Every principal ideal domain is a unique factorization domain (UFD). The converse does not hold since for any UFD , the ring of polynomials in 2 variables is a UFD but is not a PID. (To prove this look at the ideal generated by \left\langle X,Y \right\rangle. It is not the whole ring since it contains no polynomials of degree 0, but it cannot be generated by any one single element.) • Every principal ideal domain is Noetherian. • In all unital rings, maximal ideals are prime. In principal ideal domains a near converse holds: every nonzero prime ideal is maximal. • All principal ideal domains are integrally closed. The previous three statements give the definition of a Dedekind domain, and hence every principal ideal domain is a Dedekind domain. Let A be an integral domain, the following are equivalent. • A is a PID. • Every prime ideal of A is principal. • A is a Dedekind domain that is a UFD. • Every finitely generated ideal of A is principal (i.e., A is a Bézout domain) and A satisfies the ascending chain condition on principal ideals. • A admits a Dedekind–Hasse norm. Any Euclidean norm is a Dedekind-Hasse norm; thus, (5) shows that a Euclidean domain is a PID. (4) compares to: • An integral domain is a UFD if and only if it is a GCD domain (i.e., a domain where every two elements have a greatest common divisor) satisfying the ascending chain condition on principal ideals. An integral domain is a Bézout domain if and only if any two elements in it have a gcd that is a linear combination of the two. A Bézout domain is thus a GCD domain, and (4) gives yet another proof that a PID is a UFD. == See also ==