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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.

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    $\begingroup$ Is there even an ordinary "tensor product" of schemes? $\endgroup$ Sep 19, 2013 at 2:46
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    $\begingroup$ "tensor" = "fiber". $\endgroup$
    – Sasha
    Sep 19, 2013 at 3:05
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    $\begingroup$ The tensor product of commutative rings corresponds to the fiber product of the associated affine schemes, but I think one should be careful with terminology and notation. $\endgroup$ Sep 19, 2013 at 16:39

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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.

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  • $\begingroup$ Thanks - that makes sense! But for the example, $A=\tilde{\mathfrak{g}} \rightarrow \mathfrak{g}=B$ -- what does the corresponding sheaf of dg-algebras (ie. the Koszul complex) look like? $\endgroup$ Sep 21, 2013 at 3:11
  • $\begingroup$ It is a simple exercise, try to do it yourself. $\endgroup$
    – Sasha
    Sep 21, 2013 at 4:03
  • $\begingroup$ When $A$ is not a complete intersection you can rewrite the morphism $A \to B$ as a composition of a closed embedding $A \to \tilde B$ and of a flat morphism $\tilde B \to B$. Then $A\times^L_B C=A\times^L_{\tilde B}(\tilde B\times_BC) = A\times^L_{\tilde B}(\tilde B\times_B C)$. $\endgroup$
    – Sasha
    Sep 23, 2013 at 5:55
  • $\begingroup$ Ah okay. But what is the precise definition of a "complete intersection subscheme"? Hartshorne doesn't have a definition either. $\endgroup$ Sep 23, 2013 at 18:36
  • $\begingroup$ I understand what a flat module over a ring is but -- given a map $A \rightarrow B$ what does it mean for a sheaf of DG-algebras on $A$ to be flat over $B$? Thanks. $\endgroup$ Sep 23, 2013 at 18:48

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