Geometric intuition behind an $A$-algebra being $0$-smooth / $0$ unramified? Here are the definitions from Matsumura:
An $A$-algebra $B$ is said to be $0$-smooth if for any $A$-algebra $C$ with an ideal $N$ s.t. $N^2 = 0$, any map of $A$-algebras $B \to C / N$ lifts to $C$. It is said to be $0$-etale (or $0$-smooth and $0$-unramified) if this lift is unique. 
These definitions seem pretty unmotivated so far, but their names indicate that there is some geometric meaning behind them. I'm really not seeing it though. Can someone explain where these notions come from?
(I am reading this section in Matsumura in order to read 2.8 in Hartshorne.)
Edit: I think I understand something, which is that this is saying that a map from any closed subscheme of $\operatorname{Spec}C$ to $\operatorname{Spec}A$ with certain conditions on the ideal sheaf extends to a map from $\operatorname{Spec}C$ to $\operatorname{Spec}A$. This reminds me of the idea in differential geometry that shows closed submanifolds manifolds look locally like subspaces in $\mathbb R^n$, so that maps can easily be extended under mild assumptions (ex. compactness of the closed submanifolds.) I'm not sure if this is related, but that was the only idea I had. Of course all regular functions on $\operatorname{Spec}C / N$ can be extended to functions on $C$, so maybe there is no content to this observation. Though it seems reasonable that there are $\operatorname{Spec}B$ that admit functions from $\operatorname{Spec}C / N$ which only extend to rational functions on $\operatorname{Spec}C$, or something like this.)
 A: The example you should have in mind is when $A$ and $B$ are both finite type $k$-algebras (where $k$ is an algebraically closed field), and $C$ is the ring $k[\varepsilon]/(\varepsilon^2)$, with the ideal $N = (\varepsilon)$.

Lemma. A ring homomorphism $\phi \colon A \to C$ corresponds to the choice of a closed point $x \in \operatorname{Spec} A$ (i.e., a maximal ideal $\mathfrak m \subseteq A$) and an element $v \in \operatorname{Hom}_A(\mathfrak m/\mathfrak m^2, k)$. Here, we view $k$ as an $A$-module through the isomorphism $A/\mathfrak m \cong k$.

Proof. Indeed, given such a map $\phi$, we set $\mathfrak m = \phi^{-1}((\varepsilon))$; note that $(\varepsilon) \subseteq k[\varepsilon]/(\varepsilon^2)$ is the unique maximal ideal. Then we define $v$ as the induced morphism
$$\mathfrak m/\mathfrak m^2 \to (\varepsilon)/(\varepsilon)^2 = (\varepsilon) \cong_A k.$$
Conversely, you can check that any element $v \in \operatorname{Hom}_A(\mathfrak m/\mathfrak m^2,k)$ defines a ring homomorphism $\phi$ as above. ∎
The geometric meaning of the set $\operatorname{Hom}_A(\mathfrak m/\mathfrak m^2,k)$ is that it is the tangent space of $\operatorname{Spec} A$ at $x$. More specifically, it is the fibre of the tangent bundle $\operatorname{Hom}_A(\Omega_{A/k},A)$ at the point $x \in \operatorname{Spec} A$. This follows from Proposition II.8.7 in Hartshorne; Eisenbud has a similar but slightly different result in Corollary 16.13, and I haven't found it in Matsumura.
The smoothness condition says the following: the map $A \to B$ is smooth (unramified, étale) if and only if for every ring $C$ and every square zero ideal $N \subseteq C$, the map
$$\operatorname{Hom}_A^{\operatorname{alg}}(B,C) \to \operatorname{Hom}_A^{\operatorname{alg}}(B,C/N)$$
is surjective (injective, bijective, respectively). (The $0$ that Matsumura uses is usually omitted; it refers to the ideal $0$, not the number $0$.) In the example above, $C/N$ is just $k$, so homomorphisms $B \to C/N$ are given by points of $\operatorname{Spec} B$ lying over the given point $x \in \operatorname{Spec} A$.
Then the lemma (applied to both $A$ and $B$, with the specific $C$ above) then implies that if $A \to B$ is smooth (unramified, étale), then for any $x \in \operatorname{Spec} B$, with image $y \in \operatorname{Spec} A$, the map
$$T_x \to T_y $$
is surjective (resp. injective, bijective). That is, we think of the map on spectra being a submersion, an immersion, or a local diffeomorphism, respectively. (Of course, these words have a different or no meaning in the algebraic world.)
Remark. Of course, we don't get quite the converse implication that properties of the map on tangent spaces induce smoothness properties for $A \to B$. However, in many cases (e.g. $A$ and $B$ are both smooth over $k$), it is true that $A \to B$ is smooth if and only if $T_x \to T_y$ is surjective for each $x \in \operatorname{Spec} B$. See for example Hartshorne, Proposition III.10.4. Thus, it is fair to say that smooth morphisms are the submersions of algebraic geometry.
