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.
$R$ is a Dedekind domain.
2.
every nonzero proper ideal is a product of maximal ideals.
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 idealsare 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, $a\ot\\in\\mathfrak{p}$ 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.