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$C \otimes_{R} C ≅ C \times C$ as $R$ -algebras

Example

With complex conjugation defined by $\overline{a + b i} = a - b i$ , the formula

$Φ (z \otimes w) : = (z w, \overline{z} w)$

defines an isomorphism of $R$ -algebras

$C \otimes_{R} C ≅ C \times C .$

Under this isomorphism, the two product idempotents are the images of

$\frac{1}{2} (1 \otimes 1 - i \otimes i) and \frac{1}{2} (1 \otimes 1 + i \otimes i) .$

Facts & Assumptions

Given: The usual real embedding $R \to C$ and $i \in C$ .

[L1]

Every complex number has unique form $a + b i$ , with the usual arithmetic, and $C$ is a field (The complex numbers as $R [x] / (x^{2} + 1)$ , with the real embedding and imaginary unit $i$ , $C = R [x] / (x^{2} + 1)$ is a field, every element is uniquely $a + b i$ , and every nonzero element has inverse $(a - b i) / (a^{2} + b^{2})$ ).

[L2]

The vectors $1, i$ form an $R$ -basis of $C$ ( $C / R$ has power basis $1, i$ and degree $2$ ).

[L3]

Product bases form a basis of a tensor product (The elementary tensors of two bases form the product basis of the tensor product).

[L4]

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

[L5]

$C \times C$ has componentwise ring operations (The product ring $R \times S$ with componentwise operations, its identity $(1_{R}, 1_{S})$ and its units $R^{\times} \times S^{\times}$ ).

Verification

technique · direct

1.1

Conjugation fixes real scalars and is additive and multiplicative by the coordinate formulas in [L1]. Hence $(z, w) \mapsto (z w, \overline{z} w)$ is $R$ -bilinear and induces an $R$ -linear map $Φ$ from the tensor product.

givenL1algebra

1.2

By [L2] and [L3], $1 \otimes 1, i \otimes 1, 1 \otimes i, i \otimes i$ form an $R$ -basis of the source. Their images are $(1, 1), (i, - i), (i, i), (- 1, 1)$ .

L1L2L3

2.1

By [L4] and [L5], $Φ ((z \otimes w) (z^{'} \otimes w^{'})) = (z z^{'} w w^{'}, \overline{z z^{'}} w w^{'}) = Φ (z \otimes w) Φ (z^{'} \otimes w^{'})$ , and $Φ (1 \otimes 1) = (1, 1)$ ; thus $Φ$ is an $R$ -algebra homomorphism.

step 1.1L1L4L5

2.2

Given $(u + v i, x + y i) \in C \times C$ , its unique coordinates in the four images of step 1.2 are $a = (u + x) / 2$ , $b = (v - y) / 2$ , $c = (v + y) / 2$ , and $d = (x - u) / 2$ . Therefore those images form a real basis and $Φ$ is bijective.

step 1.2L1algebra

2.3

Since $Φ (i \otimes i) = (- 1, 1)$ , the two displayed tensors map respectively to $(1, 0)$ and $(0, 1)$ , the standard product idempotents.

step 1.2L5algebra

3.1

Steps 2.1 and 2.2 prove the claimed algebra isomorphism, and step 2.3 identifies its idempotents.

step 2.1step 2.2step 2.3∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 70 results over 15 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources

Wenqi Li, Commutative Algebra, Lecture 9 (standard reference, not scraped)

C⊗RC≅C×CC⊗R​C≅C×C as RR-algebras