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

We show that for an integral domain $R$, the following are equivalent.

1. 1.

$R$ is a Dedekind domain.

2. 2.

every nonzero proper ideal is a product of maximal ideals.

3. 3.

every nonzero proper ideal is a product of prime ideals.

For the equivalence of 1 and 2 see proof that a domain is Dedekind if its ideals are products of maximals. Also, as every maximal ideal is prime, it is immediate that 2 implies 3. So, we just need to consider the case where 3 is satisfied and show that 2 follows, for which it enough to show that every nonzero prime ideal is maximal.

We first suppose that $\mathfrak{p}$ is an invertible (http://planetmath.org/FractionalIdeal) prime ideal and show that it is maximal. To do this it is enough to show that any $a\in R\setminus\mathfrak{p}$ gives $\mathfrak{p}+(a)=R$. First, we have the following inclusions,

 $\mathfrak{p}\subsetneq\mathfrak{p}+(a^{2})\subseteq\mathfrak{p}+(a).$

Then, consider the prime factorizations

 $\displaystyle\mathfrak{p}+(a)=\mathfrak{p}_{1}\cdots\mathfrak{p}_{m},$ (1) $\displaystyle\mathfrak{p}+(a^{2})=\mathfrak{q}_{1}\cdots\mathfrak{q}_{n}.$ (2)

We write $\bar{a}$ for the image of $a$ under the natural homorphism (http://planetmath.org/NaturalHomomorphism) $R\rightarrow R/\mathfrak{p}$ and $\mathfrak{\bar{a}}$ for the image of any ideal $\mathfrak{a}$. Equations (1) and (2) give

 $\mathfrak{\bar{q}}_{1}\cdots\mathfrak{\bar{q}}_{n}=(\bar{a})^{2}=(\mathfrak{% \bar{p}}_{1}\cdots\mathfrak{\bar{p}}_{m})^{2}.$ (3)

As $\mathfrak{p}$ is strictly contained in $\mathfrak{p}+(a)$ and $\mathfrak{p}+(a^{2})$, it must also be strictly contained in $\mathfrak{p}_{k}$ and $\mathfrak{q}_{k}$. So, $\mathfrak{\bar{p}}_{k},\mathfrak{\bar{q}}_{k}$ are nonzero prime ideals, and by uniqueness of prime factorization (see, prime ideal factorization is unique) Equation (3) gives $n=2m$ and $\mathfrak{\bar{p}}_{k}=\mathfrak{\bar{q}}_{k}=\mathfrak{\bar{q}}_{k+m}$, after reordering of the factors. So $\mathfrak{p}_{k}=\mathfrak{q}_{k}=\mathfrak{q}_{k+m}$ and,

 $\mathfrak{p}+(a^{2})=\mathfrak{q}_{1}\cdots\mathfrak{q}_{n}=\left(\mathfrak{p}% _{1}\cdots\mathfrak{p}_{m}\right)^{2}=\left(\mathfrak{p}+(a)\right)^{2}=% \mathfrak{p}^{2}+a\mathfrak{p}+(a^{2}).$ (4)

Then, $a\not\in\mathfrak{p}$ gives $\mathfrak{p}\cap(a^{2})\subseteq a\mathfrak{p}$ and taking the intersection of both sides of (4) with $\mathfrak{p}$,

 $\mathfrak{p}=\mathfrak{p}^{2}+a\mathfrak{p}+\mathfrak{p}\cap(a^{2})\subseteq% \mathfrak{p}^{2}+a\mathfrak{p}=\left(\mathfrak{p}+(a)\right)\mathfrak{p}.$

But $\mathfrak{p}$ was assumed to be invertible, and can be cancelled giving $R\subseteq\mathfrak{p}+(a)$, showing that $\mathfrak{p}$ is maximal.

Now let $\mathfrak{p}$ be any prime ideal and $a\in\mathfrak{p}\setminus\{0\}$. Factoring into a product of primes

 $\mathfrak{p}_{1}\cdots\mathfrak{p}_{n}=(a)\subseteq\mathfrak{p},$ (5)

each of the $\mathfrak{p}_{k}$ is invertible and, by the above argument, must be maximal. Finally, as $\mathfrak{p}$ is prime, (5) gives $\mathfrak{p}_{k}\subseteq\mathfrak{p}$ for some $k$, so $\mathfrak{p}=\mathfrak{p}_{k}$ is maximal.

Title proof that a domain is Dedekind if its ideals are products of primes ProofThatADomainIsDedekindIfItsIdealsAreProductsOfPrimes 2013-03-22 18:35:07 2013-03-22 18:35:07 gel (22282) gel (22282) 5 gel (22282) Proof msc 13A15 msc 13F05 DedekindDomain PrimeIdeal