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Thread: Algebra, Problems For Fun (18)

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    Algebra, Problems For Fun (18)

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

    If $\displaystyle a,a^2,...$ are not all distinct then for some $\displaystyle 0<s<t$ we have $\displaystyle a^s=a^{s+t} \Rightarrow a^s(1-a^t)=0$ then $\displaystyle b=-a^{t-1}$ 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 $\displaystyle (0)$ is prime, for then it is maximal and $\displaystyle R$ is a field, so that $\displaystyle b=-a^{-1}$ always works for nonzero $\displaystyle a$.
    The problem is only hard when $\displaystyle R$ has zero divisors. We can suppose $\displaystyle a$ is neither a unit nor 0, those cases being trivial.

    If $\displaystyle a,a^2,...$ are not all distinct then for some $\displaystyle 0<s<t$ we have $\displaystyle a^s=a^{s+t} \Rightarrow a^s(1-a^t)=0$ then $\displaystyle b=-a^{t-1}$ 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 $\displaystyle S=\{a^m(1+ax): \ m \in \mathbb{N}, \ x \in R \}.$ if $\displaystyle 0 \in S,$ then we are done. hence we may assume that $\displaystyle 0 \notin S.$ consider $\displaystyle C=\{I \lhd R: \ I \cap S= \emptyset \},$ partially ordered by $\displaystyle \subseteq$, and

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

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

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

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

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

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