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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!

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1 Answer 1

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$x^n-rx^{n+1}=0\in P$ 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 $R$.

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  • $\begingroup$ That makes sense, thank you! $\endgroup$ Mar 15, 2016 at 1:14
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    $\begingroup$ I realized I'm still not 100% clear about why R = P + Rx implies that P is maximal. I'm sorry if it's something really simple that I'm missing. $\endgroup$ Mar 17, 2016 at 16:49
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    $\begingroup$ @NinosławCiszewski: because that shows that $x$ is invertible modulo $P$, i.e. that every non zero element in the quotient $R/P$ is invertible. This implies that $R/P$ is a field and so $P$ is maximal. $\endgroup$ Mar 18, 2016 at 23:54

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