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'image ideal of divisor'
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| Title of object: |
image ideal of divisor |
| Canonical Name: |
ImageIdealOfDivisor |
| Type: |
Topic |
| Created on: |
2008-05-06 14:59:29 |
| Modified on: |
2008-05-06 18:39:44 |
| Classification: |
msc:11A51, msc:13A05, msc:13A15 |
| Defines: |
image ideal, ideal determined by the divisor |
Revision comment (for changes between this and next version):
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Content:
\textbf{Theorem.}\, If an integral domain $\mathcal{O}$ has a divisor theory \,$\mathcal{O}^* \to \mathfrak{D}$,\, then the subset $[\mathfrak{a}]$ of $\mathcal{O}$, consisting of 0 and all elements divisible by a divisor $\mathfrak{a}$, is an ideal of $\mathcal{O}$.\, The mapping
$$\mathfrak{a} \mapsto [\mathfrak{a}]$$
from the set $\mathfrak{D}$ of divisors into the set of ideals of $\mathcal{O}$ is injective and maps any principal divisor $(\alpha)$ to the principal ideal $(\alpha)$.\\
The ideal $[\mathfrak{a}]$ may be called the {\em image ideal} of $\mathfrak{a}$ or the {\em ideal determined by the divisor} $\mathfrak{a}$.\\
\textbf{Remark.}\, There are integral domains $\mathcal{O}$ having a divisor theory but also having ideals which are not of the form $[\mathfrak{a}]$ (for example a polynomial ring in two indeterminates and its ideal formed of the polynomials without constant term).\, Such rings have ``too much ideals''.\; On the other hand, in some integral domains the monoid of principal ideals cannot be embedded into a free monoid; thus those rings cannot have a divisor theory. |
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