The polynomial-dimension formula at fields, Artinian rings, and the zero-ring boundary

For the field $k$, one has $\dim k=0$ and [L1] gives $\dim k[x]=1$.

The ring $A=k[t]\mathrm{/}(t^2)$ is Artinian local with the single prime $(\bar{t})$, so $\dim A=0$. Every prime of $A[x]=k[t,x]\mathrm{/}(t^2)$ contains the nilpotent $\bar{t}$, and [L3] identifies its prime chains with those of $A[x]\mathrm{/}(\bar{t})\cong k[x]$. Hence [L1] gives $\dim A[x]=1$. The nilpotent element does not change the one-step dimension jump.

The zero ring is excluded from the statement because it has no prime ideals at all. Its polynomial ring is again the zero ring, so the expression $\dim 0+1$ does not describe its behavior. Thus the theorem is intentionally stated only for ordinary Noetherian rings with the established zero-ring convention left separate.