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If $R$ is a commutative integral domain and it also satisfies descending chain condition on its ideals then how will we show that such ring $R$ will be a field?

Mat He Mat Cian
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1 Answers1

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If we assume that $R$ has a unit $1_R$ which acts as a multiplicative identity element, i.e. $1_Rs = s1_R = s$ for all $s \in R$, which seems to be the typical assumption according to this widipedia article, then we can argue as follows:

For any $0 \ne a \in R$, consider the principal ideals $a^iR = \langle a^i \rangle$. Since $a^{n + 1}r = a^n(ar)$ for any $r \in R$, we have $a^{n + 1}R \subseteq a^nR$; thus we have a descending chain of ideals

$aR \supseteq a^2R \supseteq . . . \supseteq a^nR \supseteq a^{n + 1}R \supseteq . . . , \tag{1}$

which must stabilize since $R$ is Artinian. Thus we can assume

$a^{m + 1}R = a^mR \tag{2}$

for some positive integer $m$. Then since $a^m = a^m1_R \in a^mR$, there must exist $s \in R$ with

$a^{m + 1}s = a^m, \tag{3}$

or

$a^m(1_R - as) = 0. \tag{4}$

Using the ID properties of $R$, this yields

$as = 1_R, \tag{5}$

since $a \ne 0 \Rightarrow a^m \ne 0$, again since $R$ is an integral domain. We have shown that every $0 \ne a \in R$ has a multiplicative inverse; hence, $R$ is a field. QED

Hope this helps. Cheerio,

and as always,

Fiat Lux!!!

Bowei Tang
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Robert Lewis
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    Thank you :) you are amazing Robert. Let me grasp it. – Mat He Mat Cian Nov 20 '13 at 05:31
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    @Mat He Mat Cian: Aw, shucks! But glad to be of service, and a sincere thanks for the kind words! – Robert Lewis Nov 20 '13 at 05:34
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    Alternatively, we could also start by observing $R$ has a minimal right ideal $xR\neq{0}$ (necessarily cyclic by minimality.) Since $(xR)^2=xR$, there exists $r$ such that $x=x^2r$, and then the idea about $x(1-xr)$ applies again. – rschwieb Nov 20 '13 at 14:13
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    @RobertLewis:What is the example of such Integral domain?? – Styles Dec 30 '17 at 17:06
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    @PKStyles: well, the question is sort of ironic in that if you had such an integral domain then it would be a field $\Bbb F$, so the only ideals would be ${0}$ and $\Bbb F$ itself, and the whole proof sort of collapses. So I guess I would say an example of such a domain would be any field $\Bbb F$. Clearly $\Bbb F$ is trivially Artinian. – Robert Lewis Dec 30 '17 at 17:15
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    @RobertLewis:I saw this post last night,from that moment i'm thinking of an example... every time i end up with a field.Don't you think that such problems are total wastage of time,as the only ideals of Fields are trivial ones?? – Styles Dec 30 '17 at 17:22
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    @PKStyles: no, not a waste of time. The point is that, a priori, we don't know if such integral domains are fields. So with this proof we deduce the structure of such domains, and then if we encounter such conditions as in the hypothesis of the problem, we know we are dealing with a field. It is sort of a standard classification, I think. – Robert Lewis Dec 30 '17 at 17:27
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    @RobertLewis:you're a good teacher!!!Fiat Lux!!! – Styles Dec 30 '17 at 17:30
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    @PKStyles. Thank you, sir. Fiat Lux!!! – Robert Lewis Dec 30 '17 at 17:38
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    @RobertLewis Why doesn't this same proof work to show that an integral domain that is Noetherian is a field? – Junglemath Jun 25 '20 at 00:02
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    @Junglemath: The first thing which comes to mind is that the ideals $a^nR$ form a descending chain but not an ascending one. Right? From there I can't see how to force $a^{n + 1}R = a^nR$, which is what makes this proof fly. OK? I mean, the $a^nR$ can be viewed as an ascending chain, but the finiteness condition doesn't show it stabilizes before we hit $aR$, so the argument breaks down. Seen? – Robert Lewis Jun 25 '20 at 00:12