Minimal and maximal primes of the node ring

By [L1], prime ideals of $A$ correspond to prime ideals of $k[x,y]$ containing $(xy)$. If $\mathfrak{p}$ contains $(xy)$, then $xy\in \mathfrak{p}$, so [L2] gives $x\in \mathfrak{p}$ or $y\in \mathfrak{p}$. Therefore every prime of $A$ contains $(\bar{x})$ or $(\bar{y})$. Since $A\mathrm{/}(\bar{x})\cong k[y]$ and $A\mathrm{/}(\bar{y})\cong k[x]$ are integral domains, both $(\bar{x})$ and $(\bar{y})$ are prime, and no smaller prime can contain them. Thus they are the minimal primes of $A$.

If $\mathfrak{m}$ is maximal in $A$, then it is prime, so step 1.1 shows that it contains $(\bar{x})$ or $(\bar{y})$. The maximal ideal $(\bar{x},\bar{y})$ indeed contains both, so "at least one" is the correct boundary statement here.

This gives the minimal-prime picture of the node ring and the maximal-ideal boundary at the singular point.