Change of Base Ring

This post was translated from Korean by LLM (Kimi). The translation may contain errors or awkward sentences. The Korean original is the source of truth.

In this article, we examine how to change an \(A\)-module into a \(B\)-module, or a \(B\)-module into an \(A\)-module, via a ring homomorphism \(\phi:A \rightarrow B\). Hence, abbreviating scalar multiplication and operations as we did before may lead to confusion; we therefore continue to omit \(\cdot\) for multiplication maps, and denote actions by \(\cdot\) (or \(\cdot_A\) and \(\cdot_B\)).

Restriction of Scalars

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

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

we see that this assignment of an \(A\)-module is functorial.

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

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

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

In particular, consider the case \(N=B\). Since \(\phi^\ast B\) and \(B\) coincide as sets, we can compare the original ring homomorphism \(\phi:A \rightarrow B\) with the action on \(\phi^\ast B\); here we find that \(\phi\) is an \(A\)-linear map.

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

Extension of Scalars

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 \(M\). We define a \(B\)-action \(\cdot_B\) on it by the formula

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

That this defines an action is readily verified by direct computation, or can be understood as arising from the composition

\(B\otimes_\mathbb{Z}(\phi^\ast B\otimes_AM)\cong (B\otimes_\mathbb{Z}\phi^\ast B)\otimes_AM \overset{\mu_B}{\longrightarrow} \phi^\ast B\otimes_AM\)¹. Furthermore, for any \(A\)-linear map \(u:M \rightarrow M'\), we verify that \(\id_{\phi^\ast B}\otimes_A u\) is a \(B\)-linear map between the \(B\)-modules defined above.

Definition 3 The functor \(\phi^\ast B\otimes_A-:\lMod{A} \rightarrow \lMod{B}\) defined above is simply denoted by \(\phi_!\) and called the extension of scalars.

Coextension of Scalars

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

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

so \(\beta\cdot g\) is also an \(A\)-linear map. A short calculation shows that this is functorial as well, yielding the following definition.

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

Adjoint Functors

The three functors defined above are related by certain adjunctions. We first establish the following lemma.

Lemma 5 Let \(N_1\) be a right \(B\)-module and \(N_2\) a left \(B\)-module, and consider the two abelian groups \(\phi^\ast N_1\otimes_A \phi^\ast N_2\) and \(N_1\otimes_B N_2\). Then there is a unique bilinear map \(\Phi:\phi^\ast N_1\otimes_A \phi^\ast N_2 \rightarrow N_1\otimes_BN_2\) sending each \(y_1\otimes_A y_2\in \phi^\ast N_1\otimes_A\phi^\ast N_2\) to \(y_1\otimes_B y_2\in N_1\otimes_BN_2\).

If \(A\) is a commutative ring, 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 verify that it is balanced with respect to the \(A\)-action. Since the \(A\)-action on \(\phi^\ast N_1,\phi^\ast N_2\) is given by the \(B\)-action through \(\phi(\alpha)\), for any \(\alpha\in A\),

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

holds, and therefore \((\alpha\cdot_A y_1,y_2)\) and \(y_1,\alpha\cdot_Ay_2\) are sent to the same element; the claim follows from the universal property of the tensor product.

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

Proposition 5 An adjunction \(\phi_!\dashv\phi^\ast\) exists.

Proof

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

yields a map \(M \rightarrow N\). Here \(M \rightarrow A\otimes_AM \rightarrow \phi^\ast B\otimes_AM\) is a composite of \(A\)-linear maps, while \(v:\phi^\ast B\otimes M \rightarrow N\) is a \(B\)-linear map.

Now, for any \(\alpha\in A\) and \(x\in M\), the composite of the \(A\)-linear maps is

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

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 have

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

Thus, viewing \(N\) as an \(A\)-module via restriction of scalars, we see that the above composition is an \(A\)-linear map.

Conversely, given any \(u\in\Hom_A(M, \phi^\ast N)\), consider the composition

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

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

and under the isomorphism \(B\otimes_BN\cong N\) this corresponds to \((\beta'\beta)\cdot_Bx=\beta'\cdot_B(\beta\cdot_Bx)\). Hence the map defined above is \(B\)-linear.

We now verify that the two maps defined above are mutually inverse, and moreover that they define a natural equivalence.

The following adjoint pair is proved similarly.

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

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

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