$I$-smoothness in Algebraic Geometry I was reading in Chapter 10 of Matsumura's book about $I$-smoothness.
In the book, the autor defines this concept by the following universal property:
Let $A$ be a ring, $B$ an $A$-algebra and $I \subset B$ an ideal. We say that $B$ with $I$-topology is $I$-smooth over $A$ if, given


*

*An $A$-algebra $C$

*An ideal $N \subset C$ such that $N^2=0$

*A continuous homomorphism of A-algebras $\phi: B \rightarrow C/N$, where $C/N$ has the discrete topology


then there exist an $A$-algebras homorphism $\varphi:B \rightarrow C$ that lift $\phi$.
Now, a notion of commutative algebra corresponds with something of deeply related with algebraic geometry very often.
There's a link in this case too?
Is the concept of $I$-smoothes somehow related with the concept of classical smoothness of a manifold?
Thank you!
 A: Not a complete answer, but this reminded me of a certain relation with smoothness, which I now try to describe. Smooth varieties $X$ have the infinitesimal lifting property: maps $Y\to X$ (with $Y$ affine) extend to maps from infinitesimal thickenings of $Y$ (this is the square-zero condition). 
Suppose $A=k$ is an algebraically closed field. 
If your $k$-algebra $B$ is smooth over $k$, then the following property holds:

Infinitesimal Lifting Property: For any morphism $f:Y=\textrm{Spec }R\to \textrm{Spec }B$ and every infinitesimal thickening $Y\subset Y'$, there exists a morphism $Y'\to \textrm{Spec }A$ lifting $f$.

Translated at the level of rings, this means: you have $k$-algebra maps $\phi:B\to R$ and a surjection $C\to R=C/N$ with square-zero kernel $N$. Then there is a map $\varphi:B\to C$ lifting $\phi$.
Example (of the converse being false): Let $B=k[x,y]/xy$ (not smooth over $k$), $C=k[t]/t^2$, $N=(t)\subset C$ (so that $R=k$), and consider the small extension
$$0\to (t)\to C\to R\to 0,$$ together with the map $\phi:B\to R$ sending $x\mapsto 0,y\mapsto 0$. Then there is no problem in lifting $\phi$ to a map $B\to C$: just send $x\mapsto at,\,y\mapsto bt$ for some $a,b\in k$. 
This can probably be extended to arbitrary rings $A$, but it is not something I ever faced.
