# Thread: prove prime ideal is maximal

1. ## prove prime ideal is maximal

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. :-)

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

3. 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."

4. 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 $\displaystyle P$ is a prime ideal of $\displaystyle R$ and

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

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

5. 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.

6. 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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### If R is a finite commutative ring with unity, prove that every prime ideal of R is a maximal ideal of R.

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