In a commutative unity ring $R$, if $M$ is a maximal ideal then prove that the ring $R/M^n$ where $n$ is a natural number, has exactly one prime ideal.
My Attempt: first, $M$ is a prime ideal because we know that in a commutative unity ring, every maximal ideal is a prime ideal. After that, I got a prime ideal in $R/M^n$ and called that $I$ but I do not know how to show that this $I$ is unique.
Thank you :)
Lemma. Let $R$ be a ring, and $P$ a prime ideal of $R$. If $A$ is an ideal and $A^n\subseteq P$ for some $n\gt 0$, then $A\subseteq P$.
Proof. Induction on $n$. True if $n=1$. If $n=2$, this is a consequence of the definition of "prime ideal." Assume true for $k$. If $A^{k+1}\subseteq P$, then $AA^k\subseteq P$; because $P$ is prime, this implies either $A\subseteq P$, or $A^k\subseteq P$. In the latter case, induction tells us $A\subseteq P$ holds as well. $\Box$
Let $Q$ be a prime ideal of $R/M^n$. The correspondence theorem tells us that $Q$ is an ideal of the form $P/M^n$ where $P$ is a prime ideal of $R$ that contains $M^n$. By the Lemma, this means that $P$ is a prime ideal of $R$ that contains $M$. Since $M$ is maximal and $P\neq R$, this means $P=M$. So $Q=M/M^n$.
Note that this part does not require $R$ to be commutative or have a unity. If I'm not mistaken, the only place you are using unity is in showing that $M/M^n$ is in fact a prime ideal of $R/M^n$, which is done using the implication that $M$ maximal implies $M$ prime, which holds for rings with unity. In general, a maximal ideal $M$ of a not necessarily commutative ring not necessarily with unity is prime if and only if it does not contain $R^2$. See here and here.
So the result holds in any ring (or rng) for any maximal ideal $M$ that does not contain $R^2$.
You've found that $M/M^n$ is a prime ideal.
Do you know every prime ideal of a commutative ring contains all nilpotent elements?
Everything in $M/M^n$ is nilpotent.
