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]

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

• 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). Commented Dec 2, 2017 at 13:06
• $R$ is a ring with identity? Commutative or non-commutative? What you have tried? Please add these in the "body" of the question. Commented Dec 2, 2017 at 13:08

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 $$x=x(by+n)=b(xy)+xn\in M.$$

• I think the direct way is simpler: if $y\notin M$, then $1=yr+z$ for $r\in R$ and $z\in M$; then $x=xyr+xr\in M$. Commented Dec 2, 2017 at 15:31
• @egreg: Much shorter than the contrapositive indeed. If you don't mind, I'll add it to my answer (with due credits,of course). Commented Dec 2, 2017 at 16:30
• You're welcome! Commented Dec 2, 2017 at 16:37

$xy\in M \implies$ $x\in M$ 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.)

• Yes, in the assumption that the ring is commutative.
– user370967
Commented Dec 2, 2017 at 13:16
• @Math_QED, yes, of course. Otherwise: math.stackexchange.com/questions/973066/…
– lhf
Commented Dec 2, 2017 at 13:19