0
$\begingroup$

There may be several ways to prove that in a commutative ring with unity, maximal Ideals are prime ideals. I attempted the following proof via contradiction.

Let us consider a commutative Ring $R$ with an unity and let $I$ be a maximal ideal. Suppose there are two elements $r_1,r_2\notin I$ such that $r_1.r_2=i_1\in I$ where "." is the product operation. Consider the two ideals $I_{1,2}=R.r_{1,2}+I=\{r.r_{1,2}+i , r\in R, i\in I\}$. Both $I_{1}$ and $I_{2}$ contain $I$ and $\neq I$. Now $I_{1}.I_{2}=I$, which means that both $I_{1}$ and $I_{2}$ cannot be $R$ itself. So either $I_{1}$ or $I_{2}$ is an ideal which is not $R$ and contains but not equal to $I$, which contradicts the fact that $I$ is a maximal ideal.

If anybody can point out if there are any loop holes or wrong statements above, it will be very helpful.

$\endgroup$
  • $\begingroup$ Looks correct to me. $\endgroup$ – fredgoodman Feb 24 '18 at 4:52
  • $\begingroup$ What is $r_{1,2}$. What are the definitions of $I_1$ and $I_2$. You've only has defined $I_{1,2}$ and $r_{1,2} + I$. $\endgroup$ – Xam Feb 25 '18 at 1:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.