I Need help to prove this:
Let $R$ be a ring and $M$ a maximal ideal of $R$, let $x,y\in R$ such that $xy\in M$, prove that $x\in M$ or $y\in M.$
R is commutative ring with identity I tried to prove M is a prime ideal but i couldn't figured it out
Let R be a ring and M a maximal ideal on R, let x,y∈R such that xy∈M, prove that x∈M or y∈M [closed]
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$\begingroup$ Welcome to MSE. Here is a reference for MathJax. Please keep in mind that this is not a "do my homework for me" site. You will find more help here if you add your thought and/or work. This question, as it stands now, will attract negative votes and will be closed. (Also, don't be discouraged by negative votes; it will help you to understand how to ask a good question). $\endgroup$– KrishCommented Dec 2, 2017 at 13:06
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2$\begingroup$ $R$ is a ring with identity? Commutative or non-commutative? What you have tried? Please add these in the "body" of the question. $\endgroup$– KrishCommented Dec 2, 2017 at 13:08
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2 Answers
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By contrapositive:
Suppose neither $x$ nor $y$ belong to $M$. Then $(x,M)$ and $(y,M)$ strictly contain $M$. $M$ being maximal, these ideals are equal to $R$, which means there exist $a, b\in R$ and $m,n\in M$ such that $ax+m=1$, $by+n=1$. Multiply them: $$1=(ax+m)(by+n)=abxy+axn+bym+mn.$$ In the last sum, the last three terms belong to $M$, so the the first cannot be in $M$ since $1$ doesn't. This implies that $xy$ doesn't belong to $M$.
Added – A shorter-better proof by @egreg:
We prove that if one of $x, y$ doesn't lie in $M$, the other does. Say $y\notin M$. As $M$ is maximal, this means there exists $b\in R,\: n\in M$, such that $by+n=1$, so $$x=x(by+n)=b(xy)+xn\in M.$$
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$\begingroup$ I think the direct way is simpler: if $y\notin M$, then $1=yr+z$ for $r\in R$ and $z\in M$; then $x=xyr+xr\in M$. $\endgroup$– egregCommented Dec 2, 2017 at 15:31
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$\begingroup$ @egreg: Much shorter than the contrapositive indeed. If you don't mind, I'll add it to my answer (with due credits,of course). $\endgroup$– BernardCommented Dec 2, 2017 at 16:30
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$xy\in M \implies $ $x\in M$ or $y\in M$
Hint: This condition defines prime ideals. Prove that every maximal ideal is prime by considering $R/M$.
(I assume that $R$ is a commutative ring with unit.)
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$\begingroup$ Yes, in the assumption that the ring is commutative. $\endgroup$– user370967Commented Dec 2, 2017 at 13:16
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$\begingroup$ @Math_QED, yes, of course. Otherwise: math.stackexchange.com/questions/973066/… $\endgroup$– lhfCommented Dec 2, 2017 at 13:19