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Math Help - Prime ideals

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    Prime ideals

    Let R be an integral domain where ideals are only of form aR={ar|r is an element of R} for some a in R. Prove that every nonzero prime ideal in R is a maximal ideal.

    Besides knowing that every maximal ideal in a commutative ring with unity is a prime ideal, i do not know where to go.

    How about this: assume that not every prime ideal in R is a maximal ideal. Then for some prime ideal I in R there is another prime ideal N such that I is contained in N and N is contained in R.
    For some a,b in I then ab is in both I and N. This violates assumption that they are prime, so every prime must be maximal?

    I have no idea if this is good and advice/help would be really great.
    Last edited by mr fantastic; May 22nd 2009 at 03:51 PM. Reason: Restored original question deleted by OP
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    Every non zero prime ideal in a PID is a maximal Ideal

    Quote Originally Posted by curiousmuch View Post
    Let R be an integral domain where ideals are only of form aR={ar|r is an element of R} for some a in R. Prove that every nonzero prime ideal in R is a maximal ideal.

    Besides knowing that every maximal ideal in a commutative ring with unity is a prime ideal, i do not know where to go.

    How about this: assume that not every prime ideal in R is a maximal ideal. Then for some prime ideal I in R there is another prime ideal N such that I is contained in N and N is contained in R.
    For some a,b in I then ab is in both I and N. This violates assumption that they are prime, so every prime must be maximal?

    I have no idea if this is good and advice/help would be really great.

    Let (P) be a nontrivial prime ideal, and let I be an Ideal containing P

    So we have P \subset I \subset R Now for P to be maximal either P = I or I = R

    Since I is generated by some element a of the Ring (a) = I

    Now p \in I so p=ar for some r \in R Now since P is a prime Ideal and am \in (p) either a \in p or r \in p.

    So we have two cases.

    Case I a \in p then (I)=(P).

    Case II  r \in p then r=ps

    now we know that p=ar as well, so we sub this into above to get

    r=(ar)s \iff r-ars=0 \iff r(1-as)=0 and Since we are in an integral domain and r\ne 0 we know that

    1-as=0 \iff 1=as so a,s are both units

    so this implies that I =R and we are done
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    Hey thanks for the great help, but I was a bit confused about one variable: where did the r=ps come from in analyzing the second case. Thanks.
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    Quote Originally Posted by curiousmuch View Post
    Hey thanks for the great help, but I was a bit confused about one variable: where did the r=ps come from in analyzing the second case. Thanks.
    Since r is in the prime ideal genterated by p it must be a multiple of p.

    Remember every ideal is principle (generated by all multiples of an elemtent)


    i.e there exists an s \in R such that  sp=r
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