The following proof that any artinian ring $R$ has $\text{dim}(R)=0$ is from Theorem 8.1 of Atiyah Macdonald:
Let $\frak{p}$ be a prime ideal of $R$, so that $S=R/\frak{p}$ is an artinian domain (this is because any infinite descending chain of ideals in $S$ could be lifted to an infinite descending chain of ideals in $R$, so because $R$ is artinian, $S$ must be artinian too). For any non-zero $x\in S$, we must have that $(x^n)=(x^{n+1})$ for some $n$ (this is because $S$ is artinian, so we can't have an infinite descending chain $S\supset (x)\supset (x^2)\supset\cdots$), hence $x^n=x^{n+1}y$ for some $y\in S$, but because $S$ is a domain, we can cancel to get $1=xy$, hence $x$ is invertible. Any non-zero element of $S$ is invertible, hence $S=R/\frak{p}$ is a field, hence $\frak{p}$ is maximal. Because any prime ideal of $R$ is maximal, we must have that $\text{dim}(R)=0$.
Theorem 8.5 of Atiyah Macdonald says that $R$ is artinian $\iff$ $R$ is noetherian and $\text{dim}(R)=0$.
So here is an example of a ring $R$ with $\text{dim}(R)=0$ but $R$ not noetherian, and hence not artinian:
$$R=k[x_1,x_2,\ldots]/(x_1,x_2,\ldots)^2\cong k[\epsilon_1,\epsilon_2,\ldots]$$
which is a field $k$ with infinitely many nilpotent elements $\epsilon_i$ added in. There's only one prime ideal of $k$, namely the zero ideal, and nilpotents won't change that, so the ideal $(\epsilon_1,\epsilon_2,\ldots)$ is the only prime ideal of $R$, but $R$ is certainly not noetherian - the chain of ideals
$$(0)\subset (\epsilon_1)\subset(\epsilon_1,\epsilon_2)\subset\cdots$$
is an infinite ascending chain.
Since $\mathbf{C}[\gamma^2,\gamma^3]\subset \mathbf{C}[\gamma]$ is an integral extension we obtain $\dim\mathbf{C}[\gamma^2,\gamma^3]=\dim\mathbf{C}[\gamma]=?$.
$\mathbf{C}[\alpha,\beta,\gamma]/(\gamma-\alpha\beta)\simeq\mathbf{C}[\alpha,\beta]$ while $\mathbf{C}[\alpha,\beta]/(\alpha^2+\beta,\alpha^3\beta^2)\simeq \mathbf{C}[\alpha]/(\alpha^7)$, so the first dimension is ... and the second is ....
Best Answer
The prime ideals of $\mathbb{C}[x,y]/(xy)$ correspond to the prime ideals of $\mathbb{C}[x,y]$ containing $(xy)$. Thus as in the comment of Youngsu, $(x, y-1)$ is a prime ideal, and your conjecture is wrong.
The prime ideal $I$ of $\mathbb{C}[x,y]$ that contains $(xy)$ must contains $x$ or $y$. We assume that $x\in I$, and that $(x)\subsetneq I$. Since $(x)$ is prime, the height of $I$ in $\mathbb{C}[x,y]$ is at least 2. But $\mathrm{dim}\mathbb{C}[x,y]=2$. (It needs some conclusions about dimension of finitely generated algebra over a field. You can find these, for example, in the Ch.11 of Atiyah-Macdonald's book.) Therefore, $I$ must be a maximal ideal of $\mathbb{C}[x,y]$. By Hilbert's Nullstellensatz, $I$ must have the form like $(x, y-b)$. But anyway, the Krull's dimension of $\mathbb{C}[x,y]/(xy)$ is just equal to $1$.