Assorted Singular Plane Curves

The Union of the Axiis is Singular

Recall that, as a scheme, we can write the union of the axiis as $X := \spec\left[k[x, y]/(xy)\right]$. Consider the local ring $\mc O_{(x, y), X}$, a local ring with maximal ideal $(\bar x, \bar y)$. Consider $(\bar x, \bar y)/(\bar x, \bar y)^2$. Note that $(\bar x, \b y)^2$ is generated by $\b x^2, \b x\b y,$ and $\b y^2$, but $\b x\b y=0$, so $(\b x, \b y)^2 = (\b x^2, \b y^2)$. Then $\dim_k (\b x, \b y)/(\b x^2, \b y^2) = 2$. Suppose $\dim \mc O_{(x, y), X}\neq 1$. We have a nontrivial prime ideal, so the dimension must be positive; thus there must be some tower of prime ideals $0\subsetneq P \subsetneq Q \subsetneq (x, y)$. Elements of $(x, y)$ are of the form $ax + by$ for $a\in k[x]$ and $b\in k[y]$ (if $a$ had any $y$ factor it would vanish, and conversely for $b$). Consider an ideal which contains $ax+by$ for both $a$ and $b$ nonzero; quotienting by such an ideal would yield $\b{x}\b{ax} = \b{x} \b{-by} = 0$ as $xy=0$. Thus such an ideal cannot be prime. No prime ideal can thus contain $ax+by$ for $a, b$ nonzero; moreover, if an ideal contains $ax$ and $by$ it must contain $ax+by$ by closure, so each ideal must contain only multiples of $x$ or multiples of $y$. Moreover, it is clear that $a, b\in k$ for the ideal to be prime, and so any prime ideal in $(x, y)$ is of the form $(x)$ or $(y)$. These ideals don’t contain each other, and so we have found the dimension of $\mc O_{(x, y), X}$ is 1.