Describe $\operatorname{Spec}(\mathbb{R}[x]/(x^n))$

Question

I want to describe $\operatorname{Spec}(\mathbb{R}[x]/(x^n))$. For starters, I know that since $\mathbb{R}$ is a field, then $\mathbb{R}[x]$ is a PID, and therefore the only prime ideal in $\mathbb{R}[x]$ is $(x)$. Due to the correspondence between the ideals of a ring and a quotient ring, I presume $\operatorname{Spec}(\mathbb{R}[x]/(x^n))$ has only one element, but I don't know how to describe it. Can someone help me?

Answer 1

Let $\pi:\mathbb{R}[x]\rightarrow\mathbb{R}[x]/(x^n)$ be the canonical projection. If $\mathfrak{p}\in{\rm Spec}(\mathbb{R}[x]/(x^n))$ then $\pi^{-1}(\mathfrak{p})\in{\rm Spec}(\mathbb{R}[x])$ therefore there exists $Q\in\mathbb{R}[x]$ irreducible such that $\pi^{-1}(\mathfrak{p})=(Q)$. Since $0\in\mathfrak{p}$, $(x^n)=\pi^{-1}(0)\subseteq\pi^{-1}(\mathfrak{p})=(Q)$ therefore $Q$ divides $x^n$ and since $Q$ is irreducible, $Q=x$. Since $\pi$ is surjective, we have $$ \mathfrak{p}=\pi(\pi^{-1}(\mathfrak{p}))=(x) $$ On the other hand, $(\mathbb{R}[x]/(x^n))/(x)\simeq\mathbb{R}[x]/(x)\simeq\mathbb{R}$ so clearly $(x)\in{\rm Spec}(\mathbb{R}[x]/(x^n))$.