I'm working through professor Vakil's book on algebraic geometry, which I recommend (it can be found at https://math.stanford.edu/~vakil/216blog/FOAGjun1113public.pdf). I'd like to check my answers to Exercises 3.2.A(a) and 3.2.A(b).
3.2.A(a) Describe $\operatorname{Spec}k[\epsilon]/(\epsilon^2)$.
Answer: since the prime ideals of the quotient ring $R/\mathfrak{a}$ are uniquely determined by the prime ideals of $R$ that contain $\mathfrak{a}$, a prime ideal of $k[\epsilon]/(\epsilon^2)$ is given by some $\mathfrak{b} \in \operatorname{Spec}k[\epsilon]$ such that $\mathfrak{b} \supset (\epsilon^2)$. However, since $\mathfrak{b}$ is prime, $\epsilon \in \mathfrak{b}$. Therefore, $\mathfrak{b} \supset (\epsilon)$, and as a consequence, $\operatorname{Spec}k[\epsilon]/(\epsilon^2) \simeq \operatorname{Spec}k[\epsilon]/(\epsilon) \simeq \operatorname{Spec}k = \{pt\}$.
3.2.A(b) Describe $\operatorname{Spec}k[x]_{(x)}$.
Answer: Using corollary 11.20(2) of https://web.mit.edu/18.705/www/13Ed.pdf, the prime ideals of $\operatorname{Spec}k[x]_{(x)}$ are uniquely determined by the prime ideals of $k[x]$ contained in $(x)$. Assume $\mathfrak{p} \in \operatorname{Spec}k[x]$ is such that $(0) \subsetneq \mathfrak{p} \subsetneq (x)$. Since $k[x]$ is a PID, $\mathfrak{p} = (xa)$ for some $a \in k[x]$. However, since $\mathfrak{p}$ is prime and $x \notin \mathfrak{p}$, $a \in \mathfrak{p}$. Then $xa\mid a$ so $x$ is a unit, contradiction. We conclude $\operatorname{Spec}k[x]_{(x)} \simeq \{(x)\}$.
I will appreciate any comments! :)