Smoothness of a variety implies homological smoothness of DbCoh

I have been told that $D^bCoh(X)$ is homologically smooth if $X$ is a smooth variety, and I am trying to construct a proof. My background is not in algebra, so I apologize for elementary questions.

It suffices to construct a finite resolution of the diagonal sheaf $mathcal{O}_Delta$ by the structure sheaf $mathcal{O}_{X times X}$.

1. By naturality of the multiplication map $A otimes A to A$ in the $A$ variable ($A$ is a commutative ring), it seems natural to try and construct such a resolution locally (by covering $X$ by affines).

2. So I believe the statement comes down to: If $A$ is a regular commutative ring, then $A$ is homologically smooth. For this I wanted to invoke two facts: The kernel of the multiplication map $A otimes A to A$ is generated by a regular sequence, and hence the Koszul resolution (by free $A otimes A$-modules) of this map terminates at a finite stage.

I have questions about each stage.

1. While the multiplication map is natural in the $A$ variable, I am not sure that the Koszul resolution is. I would have thought that this naturality depends on the dimension of $A$ (so that each Koszul resolution ends after the same length, hence "glues together" nicely along affine chart overlaps). In topology, I would argue that points in a single "connected component" have uniform dimension to deduce this naturality. I feel like there an algebraic argument to be made for this intuition. What is it?

2. I am not sure why the kernel of the multiplication map is an ideal generated by a regular sequence. (I am only familiar with the Wikipedia definition of regularity.) I am also not sure if this fact is needed; I just want to follow my nose to construct a resolution beginning with $A otimes A to A$, and I thought the Koszul resolution would be the way to go. Is there a standard reference for this, or if I am mistaken, how can I fix my nose?