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Math Help - Algebra, Problems For Fun (18)

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    Algebra, Problems For Fun (18)

    Let R be a commutative ring with identity and a \in R. Suppose that every prime ideal of R is maximal. Prove that there exist b \in R and n \in \mathbb{N} such that a^n(1+ab)=0.
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    MHF Contributor Bruno J.'s Avatar
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    The problem is trivial if (0) is prime, for then it is maximal and R is a field, so that b=-a^{-1} always works for nonzero a.
    The problem is only hard when R has zero divisors. We can suppose a is neither a unit nor 0, those cases being trivial.

    If a,a^2,... are not all distinct then for some 0<s<t we have a^s=a^{s+t} \Rightarrow a^s(1-a^t)=0 then b=-a^{t-1} works.
    If they are all distinct then I don't know. I suspect this is impossible but I can't see why that would be. I'd love the solution!
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    Quote Originally Posted by Bruno J. View Post
    The problem is trivial if (0) is prime, for then it is maximal and R is a field, so that b=-a^{-1} always works for nonzero a.
    The problem is only hard when R has zero divisors. We can suppose a is neither a unit nor 0, those cases being trivial.

    If a,a^2,... are not all distinct then for some 0<s<t we have a^s=a^{s+t} \Rightarrow a^s(1-a^t)=0 then b=-a^{t-1} works.
    If they are all distinct then I don't know. I suspect this is impossible but I can't see why that would be. I'd love the solution!
    consider S=\{a^m(1+ax): \ m \in \mathbb{N}, \ x \in R \}. if 0 \in S, then we are done. hence we may assume that 0 \notin S. consider C=\{I \lhd R: \ I \cap S= \emptyset \}, partially ordered by \subseteq, and

    I \lhd R means " I is an ideal of R." now C \neq \emptyset because (0) \in C. see that if \{I_{\alpha} \} is a totally ordered subset of C, then \bigcup I_{\alpha} \in C, i.e. every chain in C has a maximal element.

    therefore by Zorn's lemma C has a maximal element, say P. the claim is that P is prime: so suppose P is not prime. then there exist some u,v \in R-P such that uv \in P.

    let I=P+Ru, \ J=P+Rv. since P is a maximal element of C and P \subset I, \ P \subset J, we have I \notin C, \ J \notin C. so I \cap S \neq \emptyset and J \cap S \neq \emptyset. let s \in I \cap S, \ t \in J \cap S.

    since S is multiplicatively closed, we have st \in IJ \cap S. but, since uv \in P, we have IJ \subseteq P and thus st \in P \cap S = \emptyset. contradiction! so P is prime and therefore maximal. now,

    since a \in S, we must have a \notin P. thus P+Ra=R, because P is a maximal ideal. hence p+ra=1, for some p \in P, \ r \in R. but then we will get a contradiction:

    pa=a(1-ra) \in P \cap S = \emptyset. this contradiction means that our initial assumption that 0 \notin S is false and the proof is complete.
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