Change of Scalars

In this post, we examine how to turn an \(A\)-module into a \(B\)-module, or a \(B\)-module into an \(A\)-module, via a ring homomorphism \(\phi:A \rightarrow B\). Therefore, because abbreviating scalar multiplication and operations as before leaves room for confusion, for multiplication maps we omit \(\cdot\) as before, while denoting actions by \(\cdot\) (or \(\cdot_A\) and \(\cdot_B\)).

Restriction of scalar

Let a \(B\)-module \(\rho_N:B\otimes N \rightarrow N\) be given. Then, considering the following composition

we see that \(\phi^\ast\rho_N:A\otimes N \rightarrow N\) satisfies all the conditions required of an action, and therefore defines an \(A\)-module structure on \(N\). Moreover, considering the following diagram

we know that associating \(A\)-modules in this way is functorial.

Definition 1 For a ring homomorphism \(\phi:A \rightarrow B\), the functor defined in the above manner is written as \(\phi^\ast: \lMod{B} \rightarrow \lMod{A}\), and is called restriction of scalar.

In other words, given an arbitrary \(B\)-module \(\rho_N: B\otimes N \rightarrow N\), this simply defines an action of \(A\) on \(N\) using it via the formula

\[\alpha\cdot_A y:=\phi(\alpha)\cdot_B y\]

The same construction applies to right modules as well, and in this case we also view \(\phi^\ast\) as \(\rMod{B} \rightarrow\rMod{A}\).

Let us consider the special case \(N=B\). Since \(\phi^\ast B\) and \(B\) coincide as sets, we can examine the relationship between the original ring homomorphism \(\phi:A \rightarrow B\) and the action on \(\phi^\ast B\); in this case, one can check that \(\phi\) becomes an \(A\)-linear map. Also, since \(B\) is simultaneously a left \(B\)-module and a right \(B\)-module over itself, \(\phi^\ast B\) becomes a \((B,A)\)-bimodule where \(B\) acts by multiplication on the left and \(A\) acts on the right via \(\beta\cdot_A\alpha=\beta\phi(\alpha)\). Henceforth, when we use \(\phi^\ast B\) as the left factor of a tensor product, we mean this right \(A\)-structure.

Example 2 The forgetful functor \(U: \lMod{B} \rightarrow\Ab\) is induced from the (unique) ring homomorphism \(\mathbb{Z}\rightarrow B\).

Extension of scalar

We now define two functors from \(\lMod{A}\) to \(\lMod{B}\). For convenience, fix an \(A\)-module \(M\).

Consider the tensor product \(\phi^\ast B\otimes_AM\) of the two \(A\)-modules \(\phi^\ast B\) and \(M\). Then on this, through the following formula

\[\beta'\cdot_B(\beta\otimes_A x)=(\beta'\beta)\otimes_A x\]

we can define an action \(\cdot_B\) of \(B\). That this is indeed an action is not difficult to obtain even by a direct computation; or, from the composition

\[B\otimes_\mathbb{Z}(\phi^\ast B\otimes_AM)\cong (B\otimes_\mathbb{Z}\phi^\ast B)\otimes_AM \overset{\mu_B\otimes_A\id_M}{\longrightarrow} \phi^\ast B\otimes_AM\]

it can also be thought of as being obtained.¹ Also, for any \(A\)-linear map \(u:M \rightarrow M'\), one can check that \(\id_{\phi^\ast B}\otimes_A u\) defines a \(B\)-linear map between the two \(B\)-modules defined in this way.

Definition 3 We write the above functor \(\phi^\ast B\otimes_A-:\lMod{A} \rightarrow \lMod{B}\) simply as \(\phi_!\), and call it the extension of scalar.

Coextension of scalar

As before, fix an \(A\)-module \(M\). This time, we consider homomorphisms between the two \(A\)-modules \(\phi^\ast B\) and \(M\). On the following abelian group

\[\Hom_A(\phi^\ast B,M)\]

let us define a \(B\)-module structure by the following:

\[\beta\cdot g: (\beta'\mapsto g(\beta'\beta))\]

Since for any \(\alpha\in A\) and any \(\beta'\in \phi^\ast B\),

\[(\beta\cdot g)(\alpha\cdot \beta')=g(\phi(\alpha)\beta'\beta)=g(\alpha\cdot(\beta'\beta))=\alpha\cdot g(\beta'\beta)=\alpha\cdot (\beta\cdot g)(\beta')\]

\(\beta\cdot g\) is also an \(A\)-linear map. A short computation verifies that this is also functorial, and thus the following is defined.

Definition 4 The functor \(\Hom_A(\phi^\ast B,-): \lMod{A} \rightarrow \lMod{B}\) is called the coextension of scalar and is written \(\phi_\ast\).

Adjoint functors

There are certain adjoint relations between the three functors defined above. ([Category Theory] §Adjoint Functors, ⁋Definition 1) First, we prove the following lemma.

Lemma 5 For a right \(B\)-module \(N_1\) and a left \(B\)-module \(N_2\), consider the two abelian groups \(\phi^\ast N_1\otimes_A \phi^\ast N_2\) and \(N_1\otimes_B N_2\). Then there exists a unique homomorphism \(\Phi:\phi^\ast N_1\otimes_A \phi^\ast N_2 \rightarrow N_1\otimes_BN_2\) such that any \(y_1\otimes_A y_2\in \phi^\ast N_1\otimes_A\phi^\ast N_2\) is sent to \(y_1\otimes_B y_2\in N_1\otimes_BN_2\).

If both \(A\) and \(B\) are commutative rings, then \(\Phi\) is an \(A\)-linear map \(\phi^\ast N_1\otimes_A\phi^\ast N_2 \rightarrow\phi^\ast(N_1\otimes_BN_2)\).

Proof

It suffices to define the map from \(\phi^\ast N_1\times\phi^\ast N_2\) to \(N_1\otimes_B N_2\) by \((y_1,y_2)\mapsto y_1\otimes_B y_2\), and then show that this behaves well with respect to scalar multiplication by \(A\). However, on \(\phi^\ast N_1,\phi^\ast N_2\), the scalar multiplication by \(A\) is defined through \(\phi(\alpha)\) via the \(B\)-action, so for any \(\alpha\in A\)

\[(y_1\cdot_A\alpha,y_2)=(y_1\cdot_B\phi(\alpha), y_2)\mapsto (y_1\cdot_B\phi(\alpha))\otimes_B y_2=y_1\otimes_B(\phi(\alpha)\cdot_B y_2)\]

holds, and since \((y_1\cdot_A\alpha,y_2)\) and \((y_1,\alpha\cdot_Ay_2)\) are sent to the same element, the proof is completed by the universal property of the tensor product.

The following propositions can be proved in the general case as well, but for convenience we assume that \(A, B\) are both commutative rings.

Proposition 6 The adjunction \(\phi_!\dashv\phi^\ast\) exists.

Proof

Fix an arbitrary \(A\)-module \(M\) and a \(B\)-module \(N\). First, for any \(v\in\Hom_B(\phi_!M,N)\), via the composition of functions

we obtain a map \(M \rightarrow N\). Here \(M \rightarrow A\otimes_AM \rightarrow \phi^\ast B\otimes_AM\) is a composition of \(A\)-linear maps, and \(v:\phi^\ast B\otimes_A M \rightarrow N\) is a \(B\)-linear map. First, looking at the composition of the former \(A\)-linear maps for arbitrary \(\alpha\in A\) and \(x\in M\), we have

\[\alpha\cdot_Ax\mapsto \alpha\otimes_A x\mapsto \phi(\alpha)\otimes_A x\]

and for the \(B\)-linear map \(v\), using

\[\phi(\alpha)\otimes_A x=(\phi(\alpha)1)\otimes_A x=\phi(\alpha)\cdot_B(1\otimes_A x)\]

we obtain

\[v(\phi(\alpha)\otimes_A x)=v(\phi(\alpha)\cdot_B(1\otimes_A x))=\phi(\alpha)\cdot_B v(1\otimes_A x)\]

That is, viewing \(N\) as an \(A\)-module via restriction of scalar, the above composition is an \(A\)-linear map.

Conversely, suppose that an arbitrary \(u\in\Hom_A(M, \phi^\ast N)\) is given. Then this time, through the following composition

we obtain a map \(\phi_!M \rightarrow N\). Then for arbitrary \(\beta'\in B\) and \(\beta\otimes_A x\in \phi^\ast B\otimes_AM\),

\[\Phi((\id_{\phi^\ast B}\otimes_A u)(\beta'\cdot_B(\beta\otimes_Ax)))=\Phi((\beta'\beta)\otimes_Au(x))=(\beta'\beta)\otimes_B u(x)\]

and via \(B\otimes_BN\cong N\) this is sent to \((\beta'\beta)\cdot_Bu(x)=\beta'\cdot_B(\beta\cdot_Bu(x))\). That is, the map defined above is a \(B\)-linear map.

Now one can verify that the two maps defined above are inverses of each other, and moreover that they define a natural equivalence.

The following adjoint pair also holds.

Proposition 7 The adjunction \(\phi^\ast\dashv\phi_\ast\) exists.

Proof

Fix a \(B\)-module \(N\) and an \(A\)-module \(M\); it suffices to show \(\Hom_A(\phi^\ast N,M)\cong\Hom_B(N,\phi_\ast M)\). For any \(u\in\Hom_A(\phi^\ast N,M)\), define \(\tilde{u}(y)\) by \(\beta\mapsto u(\beta\cdot_By)\). Then

\[\tilde{u}(y)(\alpha\cdot_A\beta)=u\bigl((\phi(\alpha)\beta)\cdot_By\bigr)=u\bigl(\alpha\cdot_A(\beta\cdot_By)\bigr)=\alpha\cdot_A\tilde{u}(y)(\beta)\]

so \(\tilde{u}(y)\in\phi_\ast M\), and since \(\tilde{u}(\beta'\cdot_By)(\beta)=u((\beta\beta')\cdot_By)=\tilde{u}(y)(\beta\beta')\) equals \((\beta'\cdot_B\tilde{u}(y))(\beta)\) by the \(B\)-action defined just before Definition 4, we have \(\tilde{u}\in\Hom_B(N,\phi_\ast M)\).

Conversely, for any \(w\in\Hom_B(N,\phi_\ast M)\), setting \(u(y)=w(y)(1)\) gives

\[u(\alpha\cdot_Ay)=w(\phi(\alpha)\cdot_By)(1)=w(y)(\phi(\alpha))=w(y)(\alpha\cdot_A1)=\alpha\cdot_Au(y)\]

so \(u\in\Hom_A(\phi^\ast N,M)\). These two correspondences are inverses of each other, because \(\tilde{u}(y)(1)=u(1\cdot_By)=u(y)\), and conversely for the \(u\) obtained from \(w\), we have \(\tilde{u}(y)(\beta)=w(\beta\cdot_By)(1)=w(y)(\beta)\). Finally, since both correspondences are given solely by composition with \(u\) and \(w\), they are natural in both \(M\) and \(N\).

Therefore \(\phi^\ast:\lMod{B} \rightarrow\lMod{A}\) is both a left adjoint and a right adjoint, and hence commutes with all kinds of limits and colimits.

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References

[Bou] Bourbaki, N. Algebra I. Elements of Mathematics. Springer. 1998.

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1. Strictly speaking, for the first isomorphism in this formula to make sense, one must use the fact that \(B\) is a \((\mathbb{Z},A)\)-bimodule. ↩