# Thread: Algebra, Problems For Fun (18)

1. ## 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 $a^n(1+ab)=0.$

2. 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 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!

3. Originally Posted by Bruno J.
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 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.