Principal ring

In Mathematical, a principal ring is a just Anneau whose each Idéal is principal, i.e., is generated by a single element.

Examples

The following rings are principal:

the ring $\ mathbb {Z}$ of the relative entireties.
the ring of the Polynomial S in $X$ on a body $\ mathbb K$ : $\mathbb K$ .

These two examples are Euclidean rings. It is a general result which any Euclidean ring is principal. The reciprocal one is false in general. On the other hand, it is true that if a ring of polynomial A with coefficients in a commutative ring is a principal ring, then the ring has is a body (and thus the ring of polynomial is Euclidean).

The pleasant properties of the Euclidean rings make that are sought, in Arithmétique, of the results on the number of Corps of numbers, whose ring of the entireties is principal. An important invariant, the Group of the classes, makes it possible to test this property, and the problem is in particular solved for the imaginary quadratic bodies, like a particular case of the Problème of the number of classes for the imaginary quadratic bodies.

First properties

The principal rings check the Identité of Bézout:

* If has and B is two elements of has not having other common dividers that the elements of the group of the units of the ring, then there exists U and v elements of the group such as has . U + B . v = 1.

This property results owing to the fact that the ideal generated by has and B is principal, and any generator of this ideal is dividing commun run with has and B , therefore is invertible and generates the entire ring. In particular, the element 1 belongs to this ideal, which involves the relation.

Once established the definitions of pgcd and ppcm, the identity of Bézout takes a a little different form: the equation has . X + B . there = C admits solutions if and only if C is a multiple of the pgcd of has and of B .

* If has is first, then has /a. has is a body.

Either B an element whose class in the Anneau quotient is nonnull, then B is not element of has . has . As has is first, there do not exist other dividers the common only elements of the group of the units. The identity of Bézout, by passage to the classes shows whereas B is invertible.
The Lemme of Euclide also is checked:

* is has , B and C three elements of has such as has divides B . C and such as there do not exist other dividers commun run with has and with B only elements of the group of the units. Then has is a divider of C .

Indeed, the identity of Bézout ensures the existence of two elements of has , U and v such as has . U + B . v = 1. The multiplication by C of the two members of this equality makes it possible to write (I) has . U . C + B . C . v = C . Moreover, has is a divider of B . C , which results in the existence of an element D of has such as: (II) B . C = has . D . The equalities (I) and (II) show the following equality: has . ( U . C + v . D ) = C . This shows that has divides C and the lemma of Euclide is well checked.

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