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Extension of scalars is left adjoint to restriction of scalars

Statement

Let $f : R \to S$ be a homomorphism of commutative rings, let $M$ be an $R$ -module, and let $N$ be an $S$ -module. There is a natural bijection

${Hom}_{S} (S \otimes_{R} M, N) ≅ {Hom}_{R} (M, {Res}_{R}^{S} N) .$

It sends $F$ to $m \mapsto F (1_{S} \otimes m)$ . Its inverse sends $u : M \to {Res}_{R}^{S} N$ to the $S$ -linear map determined by

$s \otimes m ⟼ s u (m) .$

Facts & Assumptions

Given: A ring map $f : R \to S$ , an $R$ -module $M$ , and an $S$ -module $N$ .

[L1]

Restriction uses $r n = f (r) n$ , while extension is the $S$ -module $S \otimes_{R} M$ with $s^{'} (s \otimes m) = s^{'} s \otimes m$ (Restriction of scalars and extension of scalars $S \otimes_{R} M$ along a ring homomorphism $R \to S$ ).

[L2]

Balanced pairings induce unique homomorphisms from tensor products (Universal property of the tensor product for balanced maps into abelian groups).

[L3]

An elementary-tensor prescription descends exactly when its underlying pairing is balanced (A formula on elementary tensors defines a homomorphism exactly when its underlying pairing is balanced).

Proof

technique · direct

1.1givenL1algebra

If $F : S \otimes_{R} M \to N$ is $S$ -linear, define $Θ (F) (m) = F (1 \otimes m)$ . Then $Θ (F) (r m) = F (1 \otimes r m) = F (f (r) \otimes m) = f (r) F (1 \otimes m)$ , so $Θ (F)$ is $R$ -linear into the restriction of $N$ .

1.2givenL1L2L3

If $u : M \to {Res}_{R}^{S} N$ is $R$ -linear, the pairing $(s, m) \mapsto s u (m)$ is $R$ -balanced because $(s f (r)) u (m) = s u (r m)$ . By [L2] and [L3] it induces a homomorphism $Λ (u) : S \otimes_{R} M \to N$ .

2.1step 1.2L1algebra

The map $Λ (u)$ is $S$ -linear because $Λ (u) (s^{'} (s \otimes m)) = Λ (u) (s^{'} s \otimes m) = s^{'} s u (m) = s^{'} Λ (u) (s \otimes m)$ .

2.2step 1.1step 1.2

For $u$ , one has $Θ (Λ (u)) (m) = Λ (u) (1 \otimes m) = u (m)$ .

2.3step 1.1step 1.2algebra

Precomposition in $M$ and postcomposition in $N$ commute with both displayed formulas, so the inverse bijections are natural in both modules.

3.1step 1.1step 1.2step 2.1L2

For $F$ , one has $Λ (Θ (F)) (s \otimes m) = s F (1 \otimes m) = F (s \otimes m)$ by $S$ -linearity, so uniqueness on elementary tensors gives $Λ Θ (F) = F$ .

4.1step 2.2step 3.1step 2.3∎

Steps 2.2, 2.3 and 3.1 prove that extension of scalars is left adjoint to restriction of scalars.

Depends on

Used by

Pullback of modules is left adjoint to pushforward Theorem

Dependency tree · two levels

12 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

MIT 18.721 Algebraic Geometry notes, Lemma 2.1.35 (standard reference, not scraped)