2
$\begingroup$

Is this proof correct?

$M$ is maximal $\iff A/M$ has no proper non trivial ideal

$M$ is prime ideal $\iff A/M$ is a domain

Now, suppose $A/M$ is a commutative ring with no proper non trivial ideal and $\left( A/M \right)^2\neq0$. For every $x\in A/M$, the set $\{y \in A/M |yx=0\}$ is an ideal. Now, let $x$ be a zero divisor. Because $M$ is maximal, $\{y \in A/M |yx=0\}=A/M$. But then, every element is a zero divisor. But then $xy=0 \forall x,y \in A/M$, contradiction

$\endgroup$
2
  • $\begingroup$ Did you notice $(A/M)^2$ is an ideal? So that $(A/M)^2=A/M$? Maybe a better way to look at it is that if $x\notin M$, $(x+M)=A/M$. $\endgroup$ Commented Apr 19, 2023 at 14:22
  • $\begingroup$ Actually, can you clarify what you are asking? Your title says one thing, then you mention a proof followed by two biconditionals, and the question title looks like a question about the paragraph that follows. Are you trying to prove the first biconditoinal? If so, probably omit the second for now. $\endgroup$ Commented Apr 19, 2023 at 14:25

1 Answer 1

Reset to default
1
$\begingroup$

Your reasoning that every element of $A/M$ must be a zero divisor is not correct; taking $x=0$ (note that $0$ is always a zero divisor) in any ring $R$ will yield $\{y\in R\,|\,xy = 0\} = R$.

$\endgroup$

You must log in to answer this question.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.