Let $R$ be a commutative ring and let $M$ be a nonzero $R$-module. If $m\in M$, define $\operatorname{ord}(m) = \{r \in R \mid rm = 0\}$, and define $F = \{\operatorname{ord}(m) \mid m \in M\ \wedge m \neq 0\}$. Prove that every maximal element in $F$ is a prime ideal.
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Hints:
$ord(m)$ is an ideal for every $m\in M$, so in particular the maximal elements are ideals. This follows from routine verification of the ideal axioms.
Suppose $ord(m)$ is maximal in that set, and that $ab\in ord(m)$. If $b\notin ord(m)$, then what can you say about $bm$?
Then what can you say about $ord(bm)$ in relation to $ord(m)$?
After answering these, one will be able to see why $am=0$.
rschwieb
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If ab ∈ ord(m) then abm=0 . But if b ∉ ord(m) this implies that mb !=0. and a=0. Am i right until now ? – user112127 Dec 18 '13 at 18:06
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The first two claims are right, but the last claim ($a=0$) is wrong. You have not related $ord(bm)$ to $ord(m)$, which is a critical step. – rschwieb Dec 18 '13 at 18:37
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@user112127 No, that reasoning alone is wrong. "Maximal" doesn't mean that it contains everything else. You're on the right track though: determine which one contains the other, and then what that means for $ord(bm)$. Your question mark makes you sound like you are guessing. Don't guess... look at what is actually the case! – rschwieb Dec 18 '13 at 19:10
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this implies that r which is an element of ord(bm) is the form r1*c , where r1 ∈ord(m) – user112127 Dec 18 '13 at 19:35
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@user112127 If you say "$ord(m)$ contains $ord(bm)$" then you are saying that $x(bm)=0$ implies $xm=0$, which is not obviously true. This isn't the right path. How about you look at the opposite containment? – rschwieb Dec 18 '13 at 19:56
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@user112127 $ord(bm)\supseteq ord(m)$ is the right path, but you haven't supplied any proof and it still sounds like you're guessing. In your final result you'll have to carefully explain why these things are true. – rschwieb Dec 18 '13 at 20:53
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we know that ord(bm)⊇ord(m) this implies that an element of ord(m) belongs to ord(bm) .let x be an element of ord(m) . xm=0 and x(bm)=0 but bm!=0 – user112127 Dec 18 '13 at 21:54
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@user112127 First tell me how you know ord(bm) contains ord(m), and then I'll give hints. – rschwieb Dec 18 '13 at 23:37
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We have to prove that an element of ord(m) is in ord(bm). Let be x an element of ord(m). xm=0. Multiplied with b and this results xmb=0=x(mb)=0 . So x is in ord(mb) . So ord(mb) contains ord(m) – user112127 Dec 19 '13 at 07:50
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@user112127 Right! Good work! So you have $ord(m)\subseteq ord(mb)\subsetneq R$. What does ord(m)'s maximality imply? – rschwieb Dec 19 '13 at 13:28
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@user112127 No, that's not possible since $1\in R$ does not annihilate $bm$. What's the other possibility? – rschwieb Dec 19 '13 at 14:51
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@user112127 Correct, that is the only possibility. You already said once before that $a\in ord(bm)$. What else do we know now about $a$? – rschwieb Dec 19 '13 at 15:14
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@user112127 $a\in ord(m)$, Right, and thus we've finished verifying $ord(m)$ is prime! For your second question, yes $a\in ord(bm)$ because $a(bm)=0$. That is the definition of $ord(bm)$ :) Congratulations! I now believe you see all the steps clearly! – rschwieb Dec 19 '13 at 15:52
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$\theta$for $\theta$). Also, please show your working so far :) – Shaun Dec 18 '13 at 17:55