What are the prime ideals of $K[x]/\langle x^n\rangle$? What are the prime ideals of $K[x]/\langle x^n\rangle$, where $K$ is a field?
I have tried it like this: suppose its prime ideal be $P/\langle x^n\rangle $. Then $x^n$ belongs to $P$, which implies that $x$ is in $P$ since $P$ is a prime ideal of $K[x]$. Now i am confused further how to solve. Is $P= K$?
 A: Consider the general case: $K[x]/\langle f(x)\rangle$.
There is a bijection between the ideals in $K[x]/\langle f(x)\rangle$ and the ideals in $K[x]$ that contain $\langle f(x)\rangle$. This correspondence respects prime ideals.
Since $K[x]$ is a PID, the ideals that contain  $\langle f(x)\rangle$ are exactly those of the form $\langle p(x)\rangle$, where $p$ is an irreducible divisor of $f$.
Therefore, the prime ideals $K[x]/\langle x^n\rangle$ correspond to the prime divisors of $x^n$, which is just $x$.
A: As in this question and many like it, it is easy to show that $(x)$ is the unique maximal ideal of $K[x]/(x)^n=K[x]/(x^n)$. In particular, it is a prime ideal.
But it is also a nilpotent ideal, and nilpotent ideals are contained in all prime ideals. So $(x)$ is contained in all prime ideals. This shows that $(x)$ is the only prime ideal of the ring.
A: Recall that $Spec A/\alpha =\{ p\in Spec A \mid \alpha \subset p\}$.So we can get $Spec k[x]/<x^n> =Spec k[x]/<x>$.we know $<x>$ is a maximal ideal of $k[x]$,so $Spec k[x]/<x^n>$ contains only one element.
