Lemma 30.8.2. The identifications of Equation (30.8.1.1) are compatible with base change w.r.t. ring maps $R \to R'$. Moreover, for any $f \in R[T_0, \ldots , T_ n]$ homogeneous of degree $m$ the map multiplication by $f$

\[ \mathcal{O}_{\mathbf{P}^ n_ R}(d) \longrightarrow \mathcal{O}_{\mathbf{P}^ n_ R}(d + m) \]

induces the map on the cohomology group via the identifications of Equation (30.8.1.1) which is multiplication by $f$ for $H^0$ and the contragredient of multiplication by $f$

\[ (R[T_0, \ldots , T_ n])_{-n - 1 - (d + m)} \longrightarrow (R[T_0, \ldots , T_ n])_{-n - 1 - d} \]

on $H^ n$.

**Proof.**
Suppose that $R \to R'$ is a ring map. Let $\mathcal{U}$ be the standard affine open covering of $\mathbf{P}^ n_ R$, and let $\mathcal{U}'$ be the standard affine open covering of $\mathbf{P}^ n_{R'}$. Note that $\mathcal{U}'$ is the pullback of the covering $\mathcal{U}$ under the canonical morphism $\mathbf{P}^ n_{R'} \to \mathbf{P}^ n_ R$. Hence there is a map of Čech complexes

\[ \gamma : \check{\mathcal{C}}_{ord}^\bullet (\mathcal{U}, \mathcal{O}_{\mathbf{P}_ R}(d)) \longrightarrow \check{\mathcal{C}}_{ord}^\bullet (\mathcal{U}', \mathcal{O}_{\mathbf{P}_{R'}}(d)) \]

which is compatible with the map on cohomology by Cohomology, Lemma 20.15.1. It is clear from the computations in the proof of Lemma 30.8.1 that this map of Čech complexes is compatible with the identifications of the cohomology groups in question. (Namely the basis elements for the Čech complex over $R$ simply map to the corresponding basis elements for the Čech complex over $R'$.) Whence the first statement of the lemma.

Now fix the ring $R$ and consider two homogeneous polynomials $f, g \in R[T_0, \ldots , T_ n]$ both of the same degree $m$. Since cohomology is an additive functor, it is clear that the map induced by multiplication by $f + g$ is the same as the sum of the maps induced by multiplication by $f$ and the map induced by multiplication by $g$. Moreover, since cohomology is a functor, a similar result holds for multiplication by a product $fg$ where $f, g$ are both homogeneous (but not necessarily of the same degree). Hence to verify the second statement of the lemma it suffices to prove this when $f = x \in R$ or when $f = T_ i$. In the case of multiplication by an element $x \in R$ the result follows since every cohomology groups or complex in sight has the structure of an $R$-module or complex of $R$-modules. Finally, we consider the case of multiplication by $T_ i$ as a $\mathcal{O}_{\mathbf{P}^ n_ R}$-linear map

\[ \mathcal{O}_{\mathbf{P}^ n_ R}(d) \longrightarrow \mathcal{O}_{\mathbf{P}^ n_ R}(d + 1) \]

The statement on $H^0$ is clear. For the statement on $H^ n$ consider multiplication by $T_ i$ as a map on Čech complexes

\[ \check{\mathcal{C}}_{ord}^\bullet (\mathcal{U}, \mathcal{O}_{\mathbf{P}_ R}(d)) \longrightarrow \check{\mathcal{C}}_{ord}^\bullet (\mathcal{U}, \mathcal{O}_{\mathbf{P}_ R}(d + 1)) \]

We are going to use the notation introduced in the proof of Lemma 30.8.1. We consider the effect of multiplication by $T_ i$ in terms of the decompositions

\[ \check{\mathcal{C}}_{ord}^\bullet (\mathcal{U}, \mathcal{O}_{\mathbf{P}_ R}(d)) = \bigoplus \nolimits _{\vec{e} \in \mathbf{Z}^{n + 1}, \ \sum e_ i = d} \check{\mathcal{C}}^\bullet (\vec{e}) \]

and

\[ \check{\mathcal{C}}_{ord}^\bullet (\mathcal{U}, \mathcal{O}_{\mathbf{P}_ R}(d + 1)) = \bigoplus \nolimits _{\vec{e} \in \mathbf{Z}^{n + 1}, \ \sum e_ i = d + 1} \check{\mathcal{C}}^\bullet (\vec{e}) \]

It is clear that it maps the subcomplex $\check{\mathcal{C}}^\bullet (\vec{e})$ to the subcomplex $\check{\mathcal{C}}^\bullet (\vec{e} + \vec{b}_ i)$ where $\vec{b}_ i = (0, \ldots , 0, 1, 0, \ldots , 0))$ the $i$th basis vector. In other words, it maps the summand of $H^ n$ corresponding to $\vec{e}$ with $e_ i < 0$ and $\sum e_ i = d$ to the summand of $H^ n$ corresponding to $\vec{e} + \vec{b}_ i$ (which is zero if $e_ i + b_ i \geq 0$). It is easy to see that this corresponds exactly to the action of the contragredient of multiplication by $T_ i$ as a map

\[ (R[T_0, \ldots , T_ n])_{-n - 1 - (d + 1)} \longrightarrow (R[T_0, \ldots , T_ n])_{-n - 1 - d} \]

This proves the lemma.
$\square$

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