prove prime ideal is maximal

**ziggychick** · January 27th 2009, 01:24 PM

Hi,

I have to show that for a commutative ring with unity, every prime ideal is a maximal ideal.

I'm having trouble. My idea was to let M be a prime ideal. Assume M is not maximal, that is, assume there exists a proper ideal N in R so N contains M, and then derive a contradiction.

I'm getting nowhere with this idea though... Am I on the wrong track?

ANY help would be greatly apprectiated. Thanks. :-)

**GaloisTheory1** · January 27th 2009, 02:06 PM

Originally Posted by ziggychick

Hi,

I have to show that for a commutative ring with unity, every prime ideal is a maximal ideal.

I'm having trouble. My idea was to let M be a prime ideal. Assume M is not maximal, that is, assume there exists a proper ideal N in R so N contains M, and then derive a contradiction.

I'm getting nowhere with this idea though... Am I on the wrong track?

ANY help would be greatly apprectiated. Thanks. :-)

i thought it was true that maximals ideals are always prime. and i thought prime ideals are only maximal in a PID. In this case, with have a commutative ring with unity, so I don't think this is true. But, here is how to prove that maximal ideals are prime:

let $A$ be a ring and m a maximal ideal. suppose $x \not \in m$ and $y \not \in m$ . then $m \subset (x) +m$ and $m \subset (y) +m.$ since $m$ is maximal, $(x) +m=(y) +m=A.$ This $A=A^2=((x) +m)((y) +m)=(x)(y)+(x)m+y(m)+m^2 \subseteq (xy)+m \subseteq A.$ Thus $A=(xy)+m \supset m$ . so $xy \not \in m$ . so m is prime. QED.

**ziggychick** · January 27th 2009, 03:12 PM

sorry, in my first post i forgot to state the ring must be finite.

Here's the actual question. "Let R be a finite commutative ring with unity. Show every prime ideal of R is a maximal ideal."

**NonCommAlg** · January 27th 2009, 03:55 PM

Originally Posted by ziggychick

sorry, in my first post i forgot to state the ring must be finite.

Here's the actual question. "Let R be a finite commutative ring with unity. Show every prime ideal of R is a maximal ideal."

well, next time make sure that you post your question correctly, so you won't waste your and other member's time!

here's how to solve the problem: suppose $P$ is a prime ideal of $R$ and

$x \in R - P.$ consider the descending chain of ideals $<x> \supseteq <x^2> \supseteq <x^3> \supseteq \cdots.$ since $R$ is finite, there is $n \in \mathbb{N}$ such that $<x^n>=<x^{n+1}> = \cdots.$ so $x^n=rx^{n+1},$ for some $r \in R.$

thus $x^n(rx - 1) = 0 \in P,$ and so either $x^n \in P$ or $rx - 1 \in P.$ but $x^n \in P$ implies that $x \in P,$ which contradicts $x \in R-P.$ thus $rx - 1 \in P$ and therefore $<x> + \ P=R. \ \Box$

**ziggychick** · January 27th 2009, 04:17 PM

Thanks for you help but I figured out an easier way to do it.... If you're interested:

Let P be a prime ideal of a finite ring R.
Then R/P is an integral domain.
Since R is finite, R/P is finite.
So R/P is a field (finite integral domains are fields).

Thanks anyway.

**NonCommAlg** · January 27th 2009, 04:50 PM

Originally Posted by ziggychick

Thanks for you help but I figured out an easier way to do it.... If you're interested:

Let P be a prime ideal of a finite ring R.
Then R/P is an integral domain.
Since R is finite, R/P is finite.
So R/P is a field (finite integral domains are fields).

Thanks anyway.

of course ... but i had no idea that you knew this result, so i gave you an elementary proof. it's also important to note that my solution (but not yours!) can be applied to a much larger class of

commutative rings, i.e. Artinian rings, to get the same result. you'll probably see this later in your course.

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