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Universal mapping property of the tensor product of commutative algebras

Statement

Let $A, B, C$ be commutative $R$ -algebras. For every pair of $R$ -algebra homomorphisms $f : A \to C$ and $g : B \to C$ , there is a unique $R$ -algebra homomorphism

$h : A \otimes_{R} B ⟶ C$

such that $h (a \otimes 1) = f (a)$ and $h (1 \otimes b) = g (b)$ . It is given by

$h (a \otimes b) = f (a) g (b) .$

Thus $A \otimes_{R} B$ , with its two canonical maps, is the coproduct of $A$ and $B$ among commutative $R$ -algebras.

Facts & Assumptions

Given: Commutative $R$ -algebras $A, B, C$ and $R$ -algebra maps $f : A \to C$ , $g : B \to C$ .

[L1]

The tensor product algebra has multiplication $(a \otimes b) (a^{'} \otimes b^{'}) = a a^{'} \otimes b b^{'}$ and identity $1 \otimes 1$ (The tensor product of $R$ -algebras has multiplication $(a \otimes b) (a^{'} \otimes b^{'}) = a a^{'} \otimes b b^{'}$ ).

[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 formula descends exactly when the corresponding pairing is balanced (A formula on elementary tensors defines a homomorphism exactly when its underlying pairing is balanced).

Proof

technique · direct

1.1givenL1algebra

The canonical maps $j_{A} (a) = a \otimes 1$ and $j_{B} (b) = 1 \otimes b$ are $R$ -algebra homomorphisms: [L1] gives their multiplication and identity laws, and $j_{A} (r a) = r j_{A} (a)$ and $j_{B} (r b) = r j_{B} (b)$ show compatibility with the structure maps.

1.2givenalgebra

The pairing $(a, b) \mapsto f (a) g (b)$ is $R$ -bilinear: additivity is distributivity in $C$ , and $f (r a) g (b) = r f (a) g (b) = f (a) r g (b) = f (a) g (r b)$ because $C$ is commutative and both maps respect $R$ .

2.1step 1.2L2L3

By [L2] and [L3], step 1.2 induces a unique $R$ -linear map $h : A \otimes_{R} B \to C$ satisfying $h (a \otimes b) = f (a) g (b)$ .

3.1step 2.1L1algebra

On pure tensors, [L1] gives $h ((a \otimes b) (a^{'} \otimes b^{'})) = f (a a^{'}) g (b b^{'}) = f (a) g (b) f (a^{'}) g (b^{'})$ , where commutativity of $C$ permits the middle factors to switch; hence $h$ is multiplicative.

3.2step 2.1algebra

One has $h (1 \otimes 1) = 1_{C}$ , $h (a \otimes 1) = f (a)$ , and $h (1 \otimes b) = g (b)$ , so $h$ is an $R$ -algebra homomorphism with the required restrictions.

4.1step 3.2L1L2

If $h^{'}$ has the same restrictions, then $a \otimes b = (a \otimes 1) (1 \otimes b)$ by [L1], so $h^{'} (a \otimes b) = f (a) g (b) = h (a \otimes b)$ . The underlying group homomorphisms consequently induce the same balanced pairing, and uniqueness in [L2] gives $h^{'} = h$ .

5.1step 1.1step 2.1step 3.1step 3.2step 4.1∎

Step 1.1 supplies the two coproduct maps, and steps 2.1 through 4.1 prove the asserted universal mapping property.

Depends on

Used by

Dependency tree · two levels

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Sources

W. Li, Commutative Algebra, Lectures 9-10 (standard reference, not scraped)