exercise in commutative algebra

Question

exercise: let $A$ be an integral domain in which all prime ideals are maximal (but may not be Dedekind in general). let $P_1, P_2$ be 2 distinct prime ideals. prove: for any positive integer $m, n$, $P_1^m+P_2^n=A$.

so the hint is to prove this first for arbitrary $m$ when $n=1$. I finished this step by observing that if $P_1^m$ is contained in $P_2$, then since they re prime ideals $P_1 \subset P_2$. this contradicts the fact that they re distinguish and maximal. so this was done.

but next how to use induction for $m$? I need some hint, thanks!

The $m=1$ case furnishes $a\in P_1$ and $b\in P_2$ with $a+b=1$. — Angina Seng, Dec 07 '19 at 12:59
oh from this we then use binomial expansion. yes that's rightI'm a stupid beginner, thank u so much — youknowwho, Dec 07 '19 at 13:03
An integral domain whose prime ideals are maximal is a field. Probably you wanted to say "all non-zero prime ideals are maximal". — user26857, Dec 07 '19 at 18:18

score -1 · Answer 1 · answered Dec 07 '19 at 13:15

consider it to be true for n then multiply by P_2 the RHS is going to be P_2.

see that (P_1)^m*(P_2) + (P_2)^(n+1) is contained in (P_2).

(P_1)^m*(P_2) is contained in (P_1)^m

(P_1)^m is not contained in (P_2)

now let "a" be any such element which is in (P_1)^m but not in (P_2) note that (a) + (P_2) is contained in (a) + (P_1)^m*(P_2) + (P_2)^(n+1) hence (P_1)^m + (P_2)^(n+1)

and from maximality of (P_2) you get (a) + (P_2) is actually A

exercise in commutative algebra

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