# proof that a domain is Dedekind if its ideals are products of maximals

Let $R$ be an integral domain^{}. We show that it is a Dedekind domain if and only if every nonzero proper ideal^{} can be expressed as a product^{} of maximal ideals^{}.
To do this, we make use of the characterization of Dedekind domains as integral domains in which every nonzero integral ideal is invertible (proof that a domain is Dedekind if its ideals are invertible).

First, let us suppose that every nonzero proper ideal in $R$ is a product of maximal ideals.
Let $\U0001d52a$ be a maximal ideal and choose a nonzero $x\in \U0001d52a$. Then, by assumption^{},

$$(x)={\U0001d52a}_{1}\mathrm{\cdots}{\U0001d52a}_{\U0001d52b}$$ |

for some $n\ge 0$ and maximal ideals ${\U0001d52a}_{k}$. As $(x)$ is a principal ideal^{}, each of the factors ${\U0001d52a}_{k}$ is invertible.
Also,

$${\U0001d52a}_{1}\mathrm{\cdots}{\U0001d52a}_{\U0001d52b}\subseteq \U0001d52a.$$ |

As $\U0001d52a$ is prime, this gives ${\U0001d52a}_{k}\subseteq \U0001d52a$ for some $k$. However, ${\U0001d52a}_{k}$ is maximal so must equal $\U0001d52a$, showing that $\U0001d52a$ is indeed invertible. Then, every nonzero proper ideal is a product of maximal, and hence invertible, ideals and so is invertible, and it follows that $R$ is Dedekind.

We now show the reverse direction, so suppose that $R$ is Dedekind.
Proof by contradiction^{} will be used to show that every nonzero ideal is a product of maximals, so suppose that this is not the case.
Then, as $R$ is defined to be Noetherian^{} (http://planetmath.org/Noetherian), there is an ideal $\U0001d51e$ maximal (http://planetmath.org/MaximalElement) (w.r.t. the partial order^{} of set inclusion) among those proper ideals which are not a product of maximal ideals.
Then $\U0001d51e$ cannot be a maximal ideal itself, so is strictly contained in a maximal ideal $\U0001d52a$ and, as $\U0001d52a$ is invertible, we can write $\U0001d51e=\U0001d52a\U0001d51f$ for an ideal $\U0001d51f$.

Therefore $\U0001d51e\subseteq \U0001d51f$ and we cannot have equality, otherwise cancelling $\U0001d51e$ from $\U0001d51e=\U0001d52a\U0001d51e$ would give $\U0001d52a=R$. So, $\U0001d51f$ is strictly larger than $\U0001d51e$ and, by the choice of $\U0001d51e$, is therefore a product of maximal ideals. Finally, $\U0001d51e=\U0001d52a\U0001d51f$ is then also a product of maximal ideals.

Title | proof that a domain is Dedekind if its ideals are products of maximals |
---|---|

Canonical name | ProofThatADomainIsDedekindIfItsIdealsAreProductsOfMaximals |

Date of creation | 2013-03-22 18:35:04 |

Last modified on | 2013-03-22 18:35:04 |

Owner | gel (22282) |

Last modified by | gel (22282) |

Numerical id | 5 |

Author | gel (22282) |

Entry type | Proof |

Classification | msc 13F05 |

Classification | msc 13A15 |

Related topic | DedekindDomain |

Related topic | MaximalIdeal |

Related topic | FractionalIdeal |