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Is it true that, in a Dedekind domain, all maximal ideals are prime?

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3 Answers 3

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Maximal ideals are prime in every ring. Probably you mean the converse, which is true since "divides = contains" in Dedekind domains. So an ideal properly containing a prime ideal would be one properly dividing it, so it can only be the unit ideal $\langle 1 \rangle$.

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Thanks, I realize this was a silly question now, since M maximal in R implies R/M a field then R/M an integral domain iff M prime...a question on a test I took in undergrad. But thanks anyhow. –  Jason Smith Feb 24 '11 at 22:01

In any commutative ring with identity, all maximal ideals are prime. An ideal is maximal if and only if the quotient ring is a field; an ideal is prime if and only if the quotient ring is an integral domain. Fields are always integral domains.

(In fact, in any ring with identity, commutative or not, maximal ideals are prime ideals: if $\mathfrak{A}\mathfrak{B}\subseteq \mathfrak{M}$, with $\mathfrak{M}$ maximal, and $\mathfrak{A}\not\subset\mathfrak{M}$, then $\mathfrak{M}+\mathfrak{A}=R$; hence $$\mathfrak{B} = R\mathfrak{B} = (\mathfrak{A}+\mathfrak{M})\mathfrak{B} = \mathfrak{A}\mathfrak{B} + \mathfrak{MB}\subseteq \mathfrak{M},$$ so $\mathfrak{B}\subseteq\mathfrak{M}$, proving that $\mathfrak{M}$ is prime.)

Did you perhaps mean it the other way?

Is it true that in a Dedekind Domain, all nonzero prime ideals are maximal?

The answer there is "yes" as well, but the reason will depend on which definition of Dedekind domain you have. One definition is that they are domains in which every ideal is finitely generated, every nonzero prime ideal is maximal, and the domain is integrally closed in its field of fractions, which would make the answer true "by definition", but there are other, equivalent, definitions. So if this is what you meant to ask, you'll have to specify what your definition is if you want elucidation.

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The third time you have "maximal", you mean "prime". –  Chris Eagle Feb 24 '11 at 20:45
    
@Chris: So I did; thanks. –  Arturo Magidin Feb 24 '11 at 20:45

It is true in every ring that maximal ideals are prime (because the quotient is a field, hence also an integral domain).

What is true in Dedekind domains however (by definition) is that every nonzero prime ideal is maximal. See also the Wikipedia article. http://en.wikipedia.org/wiki/Dedekind_domain

Edit: To add a little more substance to this answer, here are two examples of Dedekind domains. First, the obvious one: $\mathbb{Z}$. Indeed, for every prime ideal $(p)$, we have $\mathbb{Z}/(p)$ is a finite integral domain, hence a field. So every non-zero prime ideal is maximal. It is easy to see that $\mathbb{Z}$ is Noetherian (in fact, of dimension 1), and integrally closed (as is any PID).

Second example: $\mathbb{Z}[i]\cong \mathbb{Z}[x]/(x^2+1)$. By Hilbert's basis theorem, this is Noetherian (overkill-argument!), it is integrally closed (since it is the number field of $\mathbb{Q}(i)$.

Note that the algebraic closure of $\mathbb{Z}$, namely $\bar{\mathbb{Z}}$ is not a Dedekind domain. Unique factorization fails miserably: $a=\sqrt{a} \sqrt{a}=\sqrt{\sqrt{a}}\sqrt{\sqrt{a}} \sqrt{\sqrt{a}}\sqrt{\sqrt{a}}$ and so on. Another way to see it is to consider the ideals $(a)^{\frac{1}{i}}$ for $i=1,2,3,\ldots$. This is a non-terminating increasing chain of ideals, hence $\bar{\mathbb{Z}}$ is not Noetherian.

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You mean "that every nonzero prime ideal is maximal" in the second paragraph. –  Arturo Magidin Feb 24 '11 at 20:11
    
Of course I do. Thanks for the correction. –  Fredrik Meyer Feb 24 '11 at 20:24
    
since you said you meant "nonzero prime ideal" -- and since that is needed to make your statement correct -- I edited it in. –  Pete L. Clark Feb 25 '11 at 4:37
    
@Frederik: $\overline{\mathbb{Q}}$ is a field so is Noetherian. (It may also be Dedekind: this depends upon your convention as to whether fields count as Dedekind domains. I would say that they do, albeit as a trivial case.) Probably you mean $\overline{\mathbb{Z}}$ instead: this is indeed a non-Noetherian domain, as your argument shows with, say, $a = 2$. –  Pete L. Clark Feb 25 '11 at 5:29
    
I am silly of course. Corrected. –  Fredrik Meyer Feb 25 '11 at 5:52

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