Question on the dimension $\dim 0$. What is the Krull dimension $\dim 0$ of the trivial ring?
Trivial ring is denoted by $0$
 A: Technically speaking, I suppose $-\infty$ is right, being the supremum of an empty set of natural numbers. Realistically speaking though, it's unlikely to make much of a difference: in any case of importance, such as a formula involving dimensions, one should always watch for exceptions with the zero ring, so there should never be (much) confusion.
This question is much like asking what the degree of the $0$ polynomial is, in the following sense: if $R$ is a Noetherian local ring (note: the $0$ ring is not!) then $\dim R = 1 + \deg P_R$, where $P_R$ is the Hilbert polynomial of $R$. In any reasonable sense, the Hilbert polynomial of the $0$ ring ought to be the $0$ polynomial. But whether you take the degree of the $0$ polynomial to be any of $0, -1$, or $-\infty$ depends on context: e.g. if you want the (numerical or regular) derivative of a polynomial to drop the degree by $1$, then $-1$ makes sense (at least in characteristic $0$).
A: Note that $\dim(0):=-\infty$ (which has been already explained in the other answer) is compatible with various dimension formulas, for example $\dim(R \times S)=\sup(\dim(R),\dim(S))$ when $S=0$, and $\dim(R[x])=\dim(R)+1$ for noetherian $R$ when $R=0$. Similarly,  the degree of the zero polynomial is also $-\infty$, which is compatible with $\mathrm{deg}(fg)=\mathrm{deg}(f) +\mathrm{deg}(g)$ (over domains) for $f=0$.
