Module theory MOC
Change of ring
Suppose π :π
βπ is a ring homomorphism.
For any π-module π, we have a (left) π
-module πβπ by restriction of scalars. Explicitly, πβπ is the π
-module with the same underlying abelian group as π
and the π
-action defined by π₯π£ =π(π₯)π£.
Since any π-morphism π βπ becomes an π
-morphism of πβπ βπβπ, we have a βforgetfulβ faithful functor πβ :ππ¬ππ½ βπ
π¬ππ½.
In fact, this functor has both left and right adjoints, giving us an adjoint triple

where
Note that is some cases πβ β
π! are naturally isomorphic:
When π is a Frobenius extension.
Extension of scalars
Let π :π
βπ be a ring homomorphism,
and let πππ
denote π regarded as a (π,π
)-bimodule, where the right action is given by π βπ =π π(π).
We define π! from the Tensor product of bimodules as the functor
π!:=πππ
βπ
?.
It follows for an π
-module π, we have the π-action
π 1(π 2βπ£)=π 1π 2βπ£
for π₯ βπ, π¦ βπ
, and π£ βπ.

The unit of the adjunction π! β£πβ is the (not necessarily injective) βinclusion,β with components
ππ:πβπβπ!(π)π£β¦1βπ£.
Proof of adjunction
Suppose π is a π-module and π :π βπβπ is an π
-morphism.
The right diagram says that for π£ βπ,
π(1βπ£)=πβ―ππ(π£)=π(π£).which by π-linearity uniquely defined π.
CoΓ«xtension of scalars
Let π :π
βπ be a ring homomorphism,
and let π
π denote π regarded as a left π
-module via the action π₯ βΉπ¦ =π(π₯)π¦.
We define the functor πβ :π
π¬ππ½ βππ¬ππ½ so that
- the image πβπ of an π
-module π is the π-module with underlying abelian group π
π¬ππ½(π
π,π) and π-action π 1 βΉπ =π 2 β¦π(π 2π 1).1
- the image πβπ of an π
-morphism π :π βπ is the π-morphism with underlying group homomorphism
π
π¬ππ½(π
π,π):π
π¬ππ½(π
π,π)βπ
π¬ππ½(π
π,π)πβ¦πβπ.
Proof of π-linearity
To reduce parentheses, the action βΉ takes precedence over application of functions.
To see that πβπ has a valid π-action:
closure follows from
π 1βΉπ(π1βΉπ 2+π2βΉπ 3)=π 1βΉπ(π(π1)π 2+π(π2)π 3)=π(π(π1)π 2π 1+π(π2)π 3π 1)=π1π(π 2π 1)+π2π(π 3π 1)=π1(π 1βΉπ)(π 2)+π2(π 1βΉπ)(π 3);unitality follows immediately;
multiplicativity follows from
π 1π 2βΉπ(π 3)=π(π 3π 1π 2)=π 2βΉπ(π 3π 1)=π 1βΉπ 2βΉπ(π 3);scalar distributivity follows from
π 1βΉπ1+π2(π 2)=(π1+π2)(π 2π 1)=π1(π 2π 1)+π2(π 2π 1)=π 1βΉπ1(π 2)+π 1βΉπ2(π 2)=(π 1βΉπ1+π 1βΉπ2)(π 2);vector distributivity follows from
(π 1+π 2)βΉπ(π 3)=π(π 3(π 1+π 2))=π(π 2π 1+π 3π 2)=π(π 2π 1)+π(π 3π 2)=π 1βΉπ(π 3)+π 2βΉπ(π 1).That πβπ is well-defined and preserved addition follows from the internal-hom of π π».
To see that πβπ preserves the π-action,
πβπ(π 1βΉπ)(π 2)=πβ(π 1βΉπ)(π 2)=π(π(π 2π 1))=π 1βΉ(πβπ)(π 2)as required.

The coΓΌnit of the adjunction πβ β£πβ is the βprojectionβ with components
ππ:πβπβπβππβ¦π(1)
Proof of π
-linearity and adjunction
To see that ππ is π
-linear, it suffices to note
ππ(π(π)βΉπ1)=π(π)βΉπ(1)=π(π(π))=ππ(1)=π1ππ(π)by the π
-linearity of π.
Next we show that ππ is a coΓΌnit of adjunction.
Suppose that π is a π-module and π :πβπ βπ is an π
-morphism.
The left diagram says that for π€ βπ we have
πβ(π€)(1)=πππβ(π€)=π(π€)βπβπ=π
π¬ππ½(π
π,π)This determines the πβ(π€) completely, as for any π βπ we have
πβ(π π€)(1)=π βΉπβ(π€)(1)=πβ(π€)(π )by π-linearity.
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