# 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 $P= K$?

• $K$ is not an ideal of $K[x]$, so that is a bad guess. It contains $1$, and any ideal containing $1$ is the entire ring. But it does turn out that $P=(x)$, which is what the duplicate says. There is only that single prime ideal. Commented Jun 8, 2017 at 15:54

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 $x$.

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.

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.