Commutative integral domain with d.c.c. is a field [duplicate]

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?

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!!!