Is it true that the nilradical is generated by $(x)$ and $(y)$? My intuition says yes because $(ax+by)^2=a^2x^2+2abxy+y^2=y^2$ and thus $(ax+by)^3=axy^2+by^3=0$ so every combination of $x,y$ will be nilpotent but am I forgetting other elements?
Also I have to show $\mathbb{Q}[x,y]/(x^2, xy, y^3)$ is a local ring. So my approach is to take an element of a maximal ideal and showing that it is not a unit. I don't know how to prove that it is not a unit.
I assume that you want to calculate the nilradical of $R = \mathbb{Q}[x,y]/\langle x^2,xy,y^2 \rangle$. I will denote the images of $x,y$ in $R$ also by $x,y$. Then $x,y$ are clearly nilpotent, so that $\langle x,y \rangle \subseteq \mathrm{Nil}(R)$.
I claim that $\langle x,y \rangle \subseteq R$ is a maximal ideal. In fact, the isomorphism theorems imply $ R / \langle x,y \rangle \cong \mathbb{Q}$, which is a field. Hence, $\langle x,y \rangle = \mathrm{Nil}(R)$.
This also shows that $\mathrm{Nil}(R)$ is a maximal ideal of $R$. If $\mathfrak{m}$ is any maximal ideal, then $\mathrm{Nil}(R) \subseteq \mathfrak{m}$, hence $\mathrm{Nil}(R) = \mathfrak{m}$. Therefore, $R$ is local.
