Question from $p$-adic HodgeTheory, linear algebra data

Question

I want to understand the definition $2.1.4$ of $\text{Mod}_{\mathscr{O}}^{\varphi}$ category in the pages $4$-$5$ in the $p$-adic Hodge Theory survey paper here, by Brian Conrad.

The map $\varphi_{\mathcal{M}}: \mathcal{M} \to \mathcal{M}$ is $\varphi_{\mathscr{O}}$-semilinear and injective.

In page $5$, it says that " a $\varphi_{\mathscr{O}}$-semilinear operator $\varphi_{\mathcal{M}}: \mathcal{M} \to \mathcal{M}$ is injective if its $\mathscr{O}$-linearisation $$1 \otimes \varphi_{\mathcal{M}}: \varphi_{\mathscr{O}}^{*} \mathcal{M}=\mathscr{O} \otimes_{\mathscr{O},\varphi_{\mathscr{O}}} \mathcal{M} \to \mathcal{M}$$ is injective.

My Question:

$(1)$ What does mean by $\mathscr{O}$-linearisation of $\varphi_{\mathcal{M}}$ ?

$(2)$ The map $1 \otimes \varphi_{\mathcal{M}}$ act as an identity on $\mathscr{O}$, why is so ?

$(3)$ Why we denote $\varphi_{\mathscr{O}}^{*} \mathcal{M}=\mathscr{O} \otimes_{\mathscr{O},\varphi_{\mathscr{O}}} \mathcal{M}$ ?

please help me by explaining those

Answer 1

I always find this notation also a bit confusing and incomplete. Let me spell out, regarding 1. and 2., the construction of linearization explicitly:

$\varphi: M \rightarrow M$ is an $\varphi_O$-semilinear map, meaning that the compatibility with the scalar multiplication is somewhat "twisted", i.e. one has $\varphi(am)=\varphi(a)\cdot \varphi(m), \;a \in O, m \in M$ (rather than the "usual" linearity "$\varphi(am)=a\varphi(m)$"). This can cause some problems, for example, the image $\varphi(M)$ is not an $O$-submodule of $M$ in general.

For that reason, it is good to associate some "actually linear" map to $\varphi,$ hence the name linearization. The way to do it is as follows:

Consider $_O O_\varphi$, that is, $O$ as the $O$-bimodule whose left multiplication is the usual one but whose right multiplication is given by viewing $O$ as an $O$-algebra via the ring homomorphism $\varphi: O \rightarrow O$. That is, one has, for $x\in O$ and a scalar $a \in O,$ $a\cdot x=ax$ but $x \cdot a=x\varphi(a)$. Then there is an actually $O$-bilinear map \begin{align*} _O O_\varphi\times M & \longrightarrow M \\ (a, m) &\longmapsto a \varphi(m), \end{align*}

and so this induces an $O$-linear (with respect to scalars acting on the left, i.e. utilizing the left action of $_O O_{\varphi}$) map $_O O_{\varphi} \otimes_{O}M \rightarrow M$, given by $a \otimes m \mapsto a \varphi(m)$. This is the linearization of $\varphi$. From the explicit description, you can kind of see why the map is described by $1\otimes \varphi_M$.

Answer 2

(1) He means that map $$1 \otimes \varphi_{\mathcal{M}}: \varphi_{\mathscr{O}}^{*} \mathcal{M}=\mathscr{O} \otimes_{\mathscr{O},\varphi_{\mathscr{O}}} \mathcal{M} \to \mathcal{M}$$ which could also be written as $$id_{\mathscr{O}} \otimes \varphi_{\mathcal{M}}: \varphi_{\mathscr{O}}^{*} \mathcal{M}=\mathscr{O} \otimes_{\mathscr{O},\varphi_{\mathscr{O}}} \mathcal{M} \to \mathcal{M}$$

and explicitly is given by

$$\sum_i a_i \otimes m_i \mapsto \sum_i a_i \cdot \varphi_{\mathcal{M}}(m_i).$$

Note that this is well-defined because 1) in the tensor product, we have (EDIT: corrected) $a \otimes b m = a \cdot \varphi_{\mathscr{O}}(b)\otimes m $, and 2) we have $\varphi_{\mathcal{M}}(am) = \varphi_{\mathscr{O}}(a)\cdot \varphi_{\mathcal{M}}(m)$ because that is what $\varphi_{\mathcal{M}}$ being $\varphi_{\mathscr{O}}$-semilinear means.

Note that this map is indeed $\mathscr{O}$-linear in the sense that e.g.

$$(1 \otimes \varphi_{\mathcal{M}}) \left( b\cdot \sum_i a_i \otimes m_i\right) = b \cdot\left((1 \otimes \varphi_{\mathcal{M}}) ( \sum_i a_i \otimes m_i) \right)$$

so it is "a linear version of $\varphi_{\mathcal{M}}$". For example if $\mathcal{M}$ is a free module of finite rank, the matrices representing the semilinear map $\varphi_{\mathcal{M}}$ and the linear map $1 \otimes \varphi_{\mathcal{M}}$ are "the same".

(2) I do not know what you mean by that map "acting as the identity on $\mathscr{O}$" because it does not act on $\mathscr{O}$ or anything that canonically identifies with $\mathscr{O}$. If by $\mathscr{O}$ here you mean the first factor in the tensor product (which, however, is not a subset of the teesor product), then, well, it acts as identity there by definition.

(3) $\varphi_{\mathscr{O}}^{*} \mathcal{M}$ naturally identifies with the extension of scalars of the $\mathscr{O}$-module $\mathcal{M}$ via the ring homomorphism $\varphi_{\mathscr{O}}: \mathscr{O} \rightarrow \mathscr{O}$, and that is often described with that notation with the asterisk. I think this notation, in turn, was chosen to be reminiscent of pullbacks, see https://ncatlab.org/nlab/show/extension+of+scalars. I believe however that understanding the reasons for certain notations is secondary in understanding the actual theory here.