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I have explained the two questions and then showed my effort on question $(1)$ as follows (Please at least check my effort below and suggest to make it perfect):

Let $R, S,T$ be three commutative rings with $R \subset S \subset T$ and the map $ S \to T$ is faithfully flat.

Assumme a functor $F: \text{SMod} \to \text{TMod}$ between two categories, where $\text{SMod}$ is the category of $S$ modules and $\text{TMod}$ is the category of $T$-modules. (We can assume more suitable conditions if needed)

Now consider a finite flat $R$-alebra $A$.

Questions:

$(1)$ I want to show the functor $ \text{SMod} \otimes A \to \text{TMod} \otimes A$ is fully faithful provided the functor $F: \text{SMod} \to \text{TMod}$ is fully faithful.

$(2)$ I want to show if the functor $F: \text{SMod} \to \text{TMod}$ is equivalence then it induces an equivalence functor $\text{SMod} \otimes A \to \text{TMod} \otimes A$.

Note 1: here by notation $\text{SMod} \otimes A$, I want to mean we are taking action of the finite flat $R$-algebra $A$ (or scalar extension by $A$)

Note 2: You can improve my question by adding more conditions if necessary in order to answer question (1).

$$------------------------------------$$ My efforts:

Let for brevity, $\mathscr{C}=\text{SMod}$ and $\mathscr{D}=\text{TMod}$.

$(1)$ Given that $F: \mathscr{C} \to \mathscr{D}$ is fully faithful. So for any $S$-modules $M,N \in \mathscr{C}$, the functor $F$ induces the map $$F_{M,N}: \text{Hom}_{\mathscr{C}}(M,N) \to \text{Hom}_{\mathscr{D}}(F(M),F(N)),$$ which is bijective. So for any $S$-module homomorphism $f: M\to N$ in category $\mathscr{C}$, we get a unique $T$-module homomorphism $F_{M,N}(f): F(M) \to F(N)$ in category $\mathscr{D}$ by the injectiveness property. Also by the surjective property, for every $T$-module homomorphism $F_{M,N}(g): F(M) \to F(N)$ in category $\mathscr{D}$ there exists a $S$-module homomorphism $f: M \to N$ in category $\mathscr{C}$ which corresponds to each other.

Now we consider the action of the finite flat $R$-algebra $A$ on every modules of $\mathscr{C}$ and $\mathscr{D}$ such as $ M \otimes_R A$ or $N \otimes_R A$ or $F(M) \otimes_R A$ or $F(N) \otimes_R A$ etc. We can then think of the functor $F: M\to N$ and the $S$-module homomorphism $f:M \to N$ as follows: $$ {\color{red}{F \otimes 1}}: \mathscr{C} \otimes A \to \mathscr{D} \otimes A, \ \text{and} \ f \otimes 1: M \otimes_R A \to N \otimes_R A,$$ (where $A$ can be thought of as a category of single object (a $R$-module) along identity map $1$).

Thus it seems, by the property of product categories, $${\color{red}{(F \otimes 1)}}_{M,N} (f \otimes 1)=F_{M,N}(f) \otimes F_{M,N}(1),$$ where $1$ denotes the identity map. So it seems that the functor $F \otimes A$ induces the bijective map \begin{align} {\color{red}{(F \otimes 1)}}_{M,N}: &\text{Hom}_{C \otimes A}(M \otimes_R A,N \otimes_R A) \\ & \to \text{Hom}_{D \otimes A}({\color{red}{(F \otimes 1)}}(M \otimes_R A), {\color{red}{(F \otimes 1)}}(N \otimes_RA)), \end{align} which answer question (1), I think. Perhaps we can consider the following diagram for better understanding: \begin{align} \matrix{ \text{SMod} & \overset{F}{ \longrightarrow} & \text{TMod} \cr \downarrow && \downarrow \cr \text{SMod} \otimes A & \overset{{\color{red}{F \otimes 1}}}{\longrightarrow} & \text{TMod} \otimes A} \end{align}

Any comment on this approach.

Edit: The flatness property of $A$ and faithful flatness of the map $S \to T$ is enough to get the above bijective map to my intuition

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    $\begingroup$ The question formulation seems odd to me because $F$ is just any old functor, not connected to the hypotheses like $S \subset T$ being faithfully flat. But a fully faithful functor $F: C \to D$ is an equivalence iff it is essentially surjective on objects, where "essentially surjective" means every object $d$ of $D$ is isomorphic to $F(c)$ for some object $c$ of $C$. If $F$ is essentially surjective and if I have understood your notation, then $F \otimes 1$ is essentially surjective as well, so (1) implies (2). $\endgroup$
    – Todd Trimble
    Feb 1, 2021 at 12:56
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    $\begingroup$ If you're not referring to the extension of scalars functor, tensoring with $T$ over $S$, then I don't know what else you'd be referring to. On the other hand, for that to be fully faithful doesn't seem too common. So I'm not sure. I think maybe you should think about it some more, since you know better what the outlying context is. $\endgroup$
    – Todd Trimble
    Feb 1, 2021 at 14:03
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    $\begingroup$ @Masmath If $SMod \otimes A$ is meant to denote the product category of the category $SMod$ and the category $A$, then (1) and (2) are true in complete generality -- all of the other details are irrelevant. That is, if $F: C \to D$ is a functor and $A$ is a category then the induced functor $F \times 1_A: C \times A \to D \times A$ is is fully faithul (respectively, an equivalence) if and only if $F$ is fully faithful (respectively, an equivalence). You can check this directly from the definitions. $\endgroup$
    – Tim Campion
    Feb 8, 2021 at 20:47
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    $\begingroup$ @Masmath I think your proof is correct (and trivial) if $SMod \otimes A$ is really the product category, except for the part where you say "consider the action of $A$". If $SMod \otimes A$ is really the product category, then the action of $A$ is irrelevant to the discussion. So I think it's much more likely that you have misunderstood what you're supposed to be proving. $\endgroup$
    – Tim Campion
    Feb 9, 2021 at 22:06
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    $\begingroup$ My advice is to go back and reread whatever it is you're reading and make sure you understand all the notation, definitions, etc. I would also suggest you read a bit on basic category theory where it is not mixed up with specific algebra which might be new to you. I've listed a few good books here for example. Once you have a bit more practice these things will become clearer. $\endgroup$
    – Tim Campion
    Feb 9, 2021 at 22:06

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