Question Prove that: in a commutative Artinian ring, prime ideals are maximal.
I need some help understanding the proof. The setting is a commutative Artinian ring, so this ring satisfies the D.C.C on ideals. Here's the proof I'm trying to understand:
Suppose P is a prime ideal and let x $\notin$ P.
Consider (x$^{m}$), the family of power-ideals of R. Clearly these decrease, so for some n, we will have that (x$^{n}$) = (x$^{n+1}$). (since R is Artinian)
Then for some ring element r, we have that
x$^{n}$ = rx$^{n+1}$
thus
x$^{n}$ – rx$^{n+1}$ = 0
thus
x$^{n}$(1 – rx) = 0
But x$^{n}$ $\notin$ P, so 1 – rx $\in$ P, thus R = P + Rx, thus P is maximal.
My main problem is why we are applying the def'n of prime to the difference x$^{n}$ – rx$^{n+1}$. How do we know it is in P? Also, how do we get that R = P + Rx from 1 – rx $\in$ P?
Thank you!
Best Answer
$x^n-rx^{n+1}=0\in P$ because every ideal contains $0$.
Also, from $1-rx\in P$ you get first $1\in P+Rx$ and then $ R=P+Rx$ just multiplying both hands by the generic element in $R$.