Commutative integral domain with d.c.c. is a field 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?
 A: 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!!!
