My question surely is a very basic one, but having just started reading Katz & Mazur's book (Arithmetic Moduli of Elliptic Curves), I would like to make sure I get the first definitions well.
Pages 7-8, the following definition is given:
Let $S$ be an arbitrary scheme. A smooth curve $C/S$ is a smooth morphism $C\rightarrow S$ of relative dimension $1$, which is separated and of finite presentation.
The only definition I know of a smooth morphism of a given relative dimension comes from Hartshorne's book, where these are only defined between schemes of finite type over a field. How does Hartshorne's definition changes in a more general context?
Moreover, I notice that it is nowhere stated that $dim(C)=1$, which is, I assume, an accurate fact when talking about curves. Is it a consequence of the morphism being smooth of relative dimension $1$ and of finite presentation?
Thank you very much for your clarifications.
PS: Here I recall Hartshorne's definition of a smooth morphism.
A morphism $f:X\rightarrow Y$ of schemes of finite type over a field $k$ is smooth of relative dimension $n$ if:
(1) $f$ is flat
(2) if $X' \subseteq X$ and $Y' \subseteq Y$ are irreducible components such that $f(X')\subseteq Y'$, then $\dim (X')=\dim (Y') + n$
(3) for each point $x\in X$, $\dim _{\kappa (x)}(\Omega _{X/Y}\otimes \kappa (x))=n$
You can find the more general definition of smooth morphisms in e.g. Vakil or the Stacks project.
$\dim(C)$ is not equal to $1$ in general! Perhaps a more appropriate name here would be to call $C\rightarrow S$ a family of curves parametrised by the base $S$. In any case, the upshot is that here you want your fibres to be (smooth) curves, hence the condition (smooth of) "relative of dimension $1$".
For example if you consider the subscheme in $\mathbb{P^2} \times \mathbb{A}^1_t$ cut out by $y^2z = x(x-z)(x-tz)$, where $x,y,z$ are coordinates on $\mathbb{P}^2$, and $t$ is your coordinate on $\mathbb{A}^1_t$. Then the projection to $\mathbb{A}^1_t$ defines a (non-smooth!) curve over $\mathbb{A^1}_t$. This is not smooth because your fibre over $t=0,1$ are nodal cubics. However if you restrict to the open subset $\mathbb{A}^1_t - \{0,1\}$, then you do get a smooth morphism.