Why does the isomorphism $A^e\otimes_A M\simeq A\otimes_k M$ imply that $A^e\otimes M$ is $A$-projective? Let $k$ be a field, $A$ an associative $k$-algebra, $M$ a left $A$-module and $A^e := A\otimes_k A^{op}$ the enveloping algebra of $A$. I have an $A^e$-projective resolution of $A$
$$ \cdots \xrightarrow{d_2} P_1 \xrightarrow{d_1} P_0 \xrightarrow{\varepsilon} A \rightarrow 0$$
and then applying the right exact functor $-\otimes_A M$ to this resolution we get
$$ \cdots \xrightarrow{d_2\otimes\boldsymbol{1}} P_1\otimes_A M \xrightarrow{d_1\otimes\boldsymbol{1}} P_0\otimes_A M \xrightarrow{\varepsilon\otimes\boldsymbol{1}} M \rightarrow 0$$
so that $\varepsilon\otimes\boldsymbol{1}$ is surjective.
I'm following along with Hochschild Cohomology for Algebra by Sarah Witherspoon, specifically section 2.5: Actions of Hochschild cohomology and attempting to show that this gives an $A$-projective resolution of $M$. At this point Sarah writes the following paragraph

Each term $P_i$ in the sequence is projective as an $A^e$-module, thus is
a direct summand of a free $A^e$-module. So each term $P_i\otimes_A M$ is projective
as a left $A$-module, where the action is on the left tensor factor only. (It
suffices to prove that $(A\otimes_k A^{op})\otimes_A M$ is projective as a left $A$-module, which
is immediate from the isomorphism $(A \otimes_k A^{op}) \otimes_A M \simeq A \otimes_k M$ given by
$(a\otimes b)\otimes m \mapsto a\otimes bm$ for $a, b \in A, m \in M$.)

But I'm confused by the statement in brackets. As my title states, why does the isomorphism $A^e\otimes_A M\simeq A\otimes_k M$ imply that $A^e\otimes M$ is $A$-projective? The author states that it is immediate so I'm probably just missing something obvious.
From that first sentence I've been focusing on the fact that
$$\bigoplus_{i\in I}A^e \simeq Q\oplus P_i$$
for some $A^e$-module, to try get an idea of what's going on. Applying $-\otimes_A M$ to this gives
$$\bigoplus_{i\in I}(A^e\otimes_A M) \simeq (Q\otimes_A M)\oplus (P_i\otimes_A M) $$
Edit: I thought I understood why showing that $A^e\otimes_A M$ is projective is enough, but I've realised I actually don't. If we're focusing on the fact that $P_i\otimes_A M$ needs to be a summand of a free $A$-module then how is that the $A$-projectivity of $A^e\otimes_A M$ guarantees that $\bigoplus_{i\in I}(A^e\otimes_A M)$ is a free $A$-module? At first I thought that $k$ being a field was what guaranteed it but I don't think I can actually use that structure here can I?
Any help would be greatly appreciated!
 A: Some preparation
For every ring $R$ we have the following properties of free and projective $R$-modules.

*

*Every free module is projective.


*For every family $(P_j)_{j ∈ J}$ of projective $R$-modules, their direct sum $\bigoplus_{j ∈ J} P_j$ is again projective.


*If $P$ is a projective $R$-module, then every direct summand of $P$ is again projective.
More explicitly, if $P'$ and $P''$ are two submodules of $P$ with $P = P' ⊕ P''$, then $P'$ and $P''$ are again projective.
Based on these properties of projective modules we have the following result:
Proposition.
Let $R$ and $S$ be two rings, and let $T \colon R\text{-}\mathrm{Mod} \to S\text{-}\mathrm{Mod}$ be a functor.
Suppose that

*

*$T$ is additive, in the sense that $T( \bigoplus_{i ∈ I} M_i ) ≅ \bigoplus_{i ∈ I} T(M_i)$ for every family $(M_i)_{i ∈ I}$ of $R$-modules, and


*$T(R)$ is projective.
Then $T$ preserves projectives: for every projective $R$-module $P$, the $S$-module $T(S)$ is again projective.
Proof.
First, let $F$ be a free $R$-module.
Then $F ≅ \bigoplus_{i ∈ I} R$ for some index set $I$, and therefore $T(F) ≅ \bigoplus_{i ∈ I} T(R)$ because $T$ is additive.
Each direct summand $T(R)$ is projective, whence $T(F)$ is again projective by property (2).
Now, let $P$ be projective $R$-module.
This means that there exists another $R$-module $P'$ such that the module $P ⊕ P'$ is free.
It follows that the $S$-module $T(P ⊕ P') ≅ T(P) ⊕ T(P')$ is again projective, as shown in the previous paragraph.
Therefore, $T(P)$ is again projective by property (3).
$∎$
First step

I thought I understood why showing that $A^e ⊗_A M$ is projective is enough, but I’ve realised I actually don’t.

A left $A^e$-module is the same as an $A$-$A$-bimodule.
For every left $A^e$-module $P$ and every left $A$-module $M$ we can therefore form the tensor product $P ⊗_A M$, and this is again a left $A$-module via $a ⋅ (p ⊗ m) = (a p) ⊗ m$.
We have in this way a functor
$$
  (-) ⊗_A M
  \colon
  A^e\text{-}\mathrm{Mod} \longrightarrow A\text{-}\mathrm{Mod} \,.
$$
We want this functor to preserve projectives.
It is additive, so according to the above proposition, it suffices to show that $A^e ⊗_A M$ is projective.
Second step

As my title states, why does the isomorphism $A^e ⊗_A M ≃ A ⊗_k M$ imply that $A^e ⊗_A M$ is $A$-projective?

The algebra $A$ has an $A$-$$-bimodule structure and a $$-$A$-bimodule structure.
The tensor product $A ⊗_ A$ has therefore an $A$-$A$-bimodule structure via
$a_1 ⋅ (b_1 ⊗ b_2) ⋅ a_2 = (a_1 b_1) ⊗ (b_2 a_2)$.
When we regard $A^e$ as an $A$-$A$-bimodule, then it is precisely $A ⊗_ A$.
It follows from the associativity of the tensor product that
$$
  A^e ⊗_A M ≅ A ⊗_ A ⊗_A M ≅ A ⊗_ M
$$
as left $A$-modules.
(The left $A$-module structure on $A ⊗_ M$ is given by acting on the first tensor factor.)
We want to show that $A^e ⊗_A M$ is projective, whence we need to show that $A ⊗_ M$ is projective.
We consider therefore the extension of scalars functor
$$
  A ⊗_ (-)
  \colon
  \text{-}\mathrm{Vect} \longrightarrow A\text{-}\mathrm{Mod} \,.
$$
This functor is additive, and the module $A ⊗_  ≅ A$ is free and therefore projective.
We find from the above proposition that the functor $A ⊗_ (-)$ preserves projectives.
But every $$-module is free (since $$ is a field) and therefore also projective by property (1).
We hence find that $A ⊗_ V$ is projective for every $$-module (i.e., vector space) $V$.
This entails that $A ⊗_ M$ is projective.¹

¹ The module $A ⊗_ M$ is not only projective but already free, since the functor $A ⊗_ (-)$ preserves free modules.
