Change of Scalars

This post was machine-translated from the Korean original by Marvin (via Kimi). It may contain errors or awkward phrasing — the Korean original is the source of truth.

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 \(\phi:A \rightarrow B\). Because abbreviating scalar multiplication and operations as before could cause confusion, we agree to keep omitting \(\cdot\) for multiplication maps, 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 see that this assignment of \(A\)-modules is functorial.

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

In other words, given an arbitrary \(B\)-module \(\rho_N: B\otimes N \rightarrow N\), we use it to define an action of \(A\) on \(N\) by the formula

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

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

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 checks 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 argument 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 of the two \(A\)-modules \(\phi^\ast B\) and \(M\), namely \(\phi^\ast B\otimes_AM\). Then we can define a \(B\)-action \(\cdot_B\) on this by the formula

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

That this is indeed an action is not difficult to verify by direct computation, or alternatively one may view it as arising from the composition

\(B\otimes_\mathbb{Z}(\phi^\ast B\otimes_AM)\cong (B\otimes_\mathbb{Z}\phi^\ast B)\otimes_AM \overset{\mu_B\otimes_A\id_M}{\longrightarrow} \phi^\ast B\otimes_AM\)¹

Also, for any \(A\)-linear map \(u:M \rightarrow M'\), one checks that \(\id_{\phi^\ast B}\otimes_A u\) defines a \(B\)-linear map between the two \(B\)-modules constructed 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 \(M\). We define a \(B\)-module structure on the abelian group

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

by

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

For arbitrary \(\alpha\in A\) and \(\beta'\in \phi^\ast B\), we have

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

so \(\beta\cdot g\) is also an \(A\)-linear map. A short computation shows that this is again 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

The three functors defined above satisfy certain adjoint relationships. ([Category Theory] §Adjoint Functors, ⁋Definition 1) We first 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 \(\Phi:\phi^\ast N_1\otimes_A \phi^\ast N_2 \rightarrow N_1\otimes_BN_2\) sending any \(y_1\otimes_A y_2\in \phi^\ast N_1\otimes_A\phi^\ast N_2\) 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

Define a map \(\phi^\ast N_1\times\phi^\ast N_2 \rightarrow N_1\otimes_B N_2\) 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 \(A\). Since the \(A\)-scalar multiplication on \(\phi^\ast N_1,\phi^\ast N_2\) is defined via \(B\)-action through \(\phi(\alpha)\), for any \(\alpha\in A\) we have

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

and thus \((\alpha\cdot_A y_1,y_2)\) and \((y_1,\alpha\cdot_Ay_2)\) are sent to the same element, so 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 \(A\) and \(B\) 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 \(v\in\Hom_B(\phi_!M,N)\), we obtain a map \(M \rightarrow N\) via the composition of functions

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. 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 an arbitrary \(u\in\Hom_A(M, \phi^\ast N)\) is given. Then this time we obtain a map \(\phi_!M \rightarrow N\) via the following composition

Then for arbitrary \(\beta'\in B\) and \(\beta\otimes_A x\in \phi^\ast B\otimes_AM\), we have

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

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

Now one checks 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, to make the first isomorphism in this formula work, one must use the fact that \(B\) is a \((\mathbb{Z},A)\)-bimodule. ↩