The Stacks Project


Tag 08IB

Chapter 35: Derived Categories of Schemes > Section 35.21: Cohomology and base change, IV

Lemma 35.21.3. Let $g : S' \to S$ be a morphism of schemes. Let $f : X \to S$ be quasi-compact and quasi-separated. Consider the base change diagram $$ \xymatrix{ X' \ar[r]_{g'} \ar[d]_{f'} & X \ar[d]^f \\ S' \ar[r]^g & S } $$ If $X$ and $S'$ are Tor independent over $S$, then for all $E \in D_\textit{QCoh}(\mathcal{O}_X)$ we have $Rf'_*L(g')^*E = Lg^*Rf_*E$.

Proof. For any object $E$ of $D(\mathcal{O}_X)$ we can use Cohomology, Remark 20.29.3 to get a canonical base change map $Lg^*Rf_*E \to Rf'_*L(g')^*E$. To check this is an isomorphism we may work locally on $S'$. Hence we may assume $g : S' \to S$ is a morphism of affine schemes. In particular, $g$ is affine and it suffices to show that $$ Rg_*Lg^*Rf_*E \to Rg_*Rf'_*L(g')^*E = Rf_*(Rg'_* L(g')^* E) $$ is an isomorphism, see Lemma 35.5.1 (and use Lemmas 35.3.8, 35.3.9, and 35.4.1 to see that the objects $Rf'_*L(g')^*E$ and $Lg^*Rf_*E$ have quasi-coherent cohomology sheaves). Note that $g'$ is affine as well (Morphisms, Lemma 28.11.8). By Lemma 35.5.2 the map becomes a map $$ Rf_*E \otimes_{\mathcal{O}_S}^\mathbf{L} g_*\mathcal{O}_{S'} \longrightarrow Rf_*(E \otimes_{\mathcal{O}_X}^\mathbf{L} g'_*\mathcal{O}_{X'}) $$ Observe that $g'_*\mathcal{O}_{X'} = f^*g_*\mathcal{O}_{S'}$. Thus by Lemma 35.21.1 it suffices to prove that $Lf^*g_*\mathcal{O}_{S'} = f^*g_*\mathcal{O}_{S'}$. This follows from our assumption that $X$ and $S'$ are Tor independent over $S$. Namely, to check it we may work locally on $X$, hence we may also assume $X$ is affine. Say $X = \mathop{\rm Spec}(A)$, $S = \mathop{\rm Spec}(R)$ and $S' = \mathop{\rm Spec}(R')$. Our assumption implies that $A$ and $R'$ are Tor independent over $R$ (More on Algebra, Lemma 15.57.5), i.e., $\text{Tor}_i^R(A, R') = 0$ for $i > 0$. In other words $A \otimes_R^\mathbf{L} R' = A \otimes_R R'$ which exactly means that $Lf^*g_*\mathcal{O}_{S'} = f^*g_*\mathcal{O}_{S'}$ (use Lemma 35.3.8). $\square$

    The code snippet corresponding to this tag is a part of the file perfect.tex and is located in lines 4240–4256 (see updates for more information).

    \begin{lemma}
    \label{lemma-compare-base-change}
    Let $g : S' \to S$ be a morphism of schemes.
    Let $f : X \to S$ be quasi-compact and quasi-separated.
    Consider the base change diagram
    $$
    \xymatrix{
    X' \ar[r]_{g'} \ar[d]_{f'} &
    X \ar[d]^f \\
    S' \ar[r]^g &
    S
    }
    $$
    If $X$ and $S'$ are Tor independent over $S$, then for all
    $E \in D_\QCoh(\mathcal{O}_X)$ we have
    $Rf'_*L(g')^*E = Lg^*Rf_*E$.
    \end{lemma}
    
    \begin{proof}
    For any object $E$ of $D(\mathcal{O}_X)$ we can use
    Cohomology, Remark \ref{cohomology-remark-base-change} to get a
    canonical base change map $Lg^*Rf_*E \to Rf'_*L(g')^*E$. To check this
    is an isomorphism we may work locally on $S'$. Hence we may assume
    $g : S' \to S$ is a morphism of affine schemes. In particular, $g$
    is affine and it suffices to show that
    $$
    Rg_*Lg^*Rf_*E \to Rg_*Rf'_*L(g')^*E = Rf_*(Rg'_* L(g')^* E)
    $$
    is an isomorphism, see Lemma \ref{lemma-affine-morphism}
    (and use Lemmas \ref{lemma-quasi-coherence-pullback},
    \ref{lemma-quasi-coherence-tensor-product}, and
    \ref{lemma-quasi-coherence-direct-image}
    to see that the objects $Rf'_*L(g')^*E$ and $Lg^*Rf_*E$
    have quasi-coherent cohomology sheaves). Note that $g'$ is
    affine as well (Morphisms, Lemma \ref{morphisms-lemma-base-change-affine}).
    By Lemma \ref{lemma-affine-morphism-pull-push} the map becomes a map
    $$
    Rf_*E \otimes_{\mathcal{O}_S}^\mathbf{L} g_*\mathcal{O}_{S'}
    \longrightarrow
    Rf_*(E \otimes_{\mathcal{O}_X}^\mathbf{L} g'_*\mathcal{O}_{X'})
    $$
    Observe that $g'_*\mathcal{O}_{X'} = f^*g_*\mathcal{O}_{S'}$. Thus by
    Lemma \ref{lemma-cohomology-base-change} it suffices to prove that
    $Lf^*g_*\mathcal{O}_{S'} = f^*g_*\mathcal{O}_{S'}$. This follows from our
    assumption that $X$ and $S'$ are Tor independent over $S$. Namely, to
    check it we may work locally on $X$, hence we may also assume $X$ is affine.
    Say $X = \Spec(A)$, $S = \Spec(R)$ and $S' = \Spec(R')$. Our assumption
    implies that $A$ and $R'$ are Tor independent over $R$
    (More on Algebra, Lemma \ref{more-algebra-lemma-tor-independent}), i.e.,
    $\text{Tor}_i^R(A, R') = 0$ for $i > 0$. In other words
    $A \otimes_R^\mathbf{L} R' = A \otimes_R R'$ which exactly means
    that $Lf^*g_*\mathcal{O}_{S'} = f^*g_*\mathcal{O}_{S'}$
    (use Lemma \ref{lemma-quasi-coherence-pullback}).
    \end{proof}

    Comments (2)

    Comment #2600 by Kiran Kedlaya on June 5, 2017 a 9:50 pm UTC

    In the first displayed equation in the proof, there is an occurrence of $Rh_*$ which should be $Rg'_*$.

    Comment #2625 by Johan (site) on July 7, 2017 a 12:32 pm UTC

    Thanks, fixed here.

    Add a comment on tag 08IB

    Your email address will not be published. Required fields are marked.

    In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the lower-right corner).

    All contributions are licensed under the GNU Free Documentation License.




    In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following box. So in case this where tag 0321 you just have to write 0321. Beware of the difference between the letter 'O' and the digit 0.

    This captcha seems more appropriate than the usual illegible gibberish, right?