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Math Help - prove prime ideal is maximal

  1. #1
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    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. :-)
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  2. #2
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    Quote Originally Posted by ziggychick View Post
    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.
    Last edited by GaloisTheory1; January 27th 2009 at 02:41 PM.
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  3. #3
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    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."
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  4. #4
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    Quote Originally Posted by ziggychick View Post
    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
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  5. #5
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    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.
    Last edited by ziggychick; January 27th 2009 at 04:25 PM. Reason: found out answer
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  6. #6
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    Quote Originally Posted by ziggychick View Post
    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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