Let $A \rightarrow B$ and $C \rightarrow B$ be two maps of schemes. How can I compute the derived fiber product $A \otimes^L\_B C$? I'm guessing this is a dg-scheme.


For instance - let $B=\mathbb{A}^1$ and $A = 0 \in \mathbb{A}^1$, and $C \in \mathbb{A}^1$ some point (either $0$ or $1$). 


**Question**: The example I'm really interested in is $\tilde{\mathfrak{g}} \otimes^L\_{\mathfrak{g}} 0$, where $\mathfrak{g}=\mathfrak{sl}\_2$ and $\tilde{\mathfrak{g}}=\{ X,(0 \subset V \subset \mathbb{C}^2) | X \in \mathfrak{sl}\_2, X V \subset V \}$ is the Grothendieck-Springer resolution.



$A \otimes^L_B C$ computing the derived fiber product of schemes
As usual, you replace $A$ (or $C$) with a (sheaf of) DG-algebras which is flat over $B$ and compute the usual tensor product of it with $O\_A$ over $O\_B$. In case when $A$ is a complete intersection subscheme a good choice for such DG-algebra is the Koszul complex. Then the derived fiber product is given by the pullback to $C$ of the Koszul resolution of $O\_A$.


For example, if $A = 0 \subset \mathbf{A}^1$ then $O\_A \cong \{k[t] \stackrel{t}\to k[t]\}$, so the derived fiber product is empty if $C = 1$ and $\{k \stackrel{0}\to k\} = k[\epsilon]/\epsilon^2$ with $\deg\epsilon = -1$, $d\epsilon = 0$ if $C = 0$. The same computation works as well in your second example.


