# Construction of Direct Product of Fields

## Theorem

Let $\struct {F, +_F, \times_F}$ be a field whose zero is $0$ and whose multiplicative identity is $1$.

Let $E = F \times F$ be the Cartesian product of $F$ with itself.

Let addition be defined on $E$ by:

$\forall a, b, c, d \in F: \tuple {a, b} +_E \tuple {c, d} := \tuple {a +_F c, b +_F d}$

Let multiplication be defined on $E$ by:

$\forall a, b, c, d \in F: \tuple {a, b} \times_E \tuple {c, d} := \tuple {\paren {a \times_F c} +_F \paren {-\paren {b \times_F d} }, \paren {a \times_F d} +_F \paren {b \times_F c} }$

Then $\struct {E, +_E, \times_E}$ is a field.

## Proof

In order to define the structure rigorously, each of the field addition and field multiplication operations were explicitly stated in the above.

However, in order to simplify presentation, the operations will be denoted in the following as:

 $\ds \forall a, b, c, d \in F: \,$ $\ds \tuple {a, b} + \tuple {c, d}$ $=$ $\ds \tuple {a + c, b + d}$ $\ds \tuple {a, b} \tuple {c, d}$ $=$ $\ds \tuple {a c - b d, a d + b c}$

when it is clear from the context which operation is implied.

We check the criteria for $E$ to be a field.

First we note that $\struct {E, +_E}$ is the external group direct product $\struct {F, +_F} \times \struct {F, +_F}$, where $\struct {F, +_F}$ is an Abelian group.

Hence from External Direct Product of Abelian Groups is Abelian Group we have that $\struct {E, +_E}$ is an Abelian group whose identity is $\tuple {0, 0}$.

Now we consider the algebraic structure $\tuple {E, \times_E}$.

We specifically wish to demonstrate that $\tuple {E^*, \times_E}$, where $E^* := E \setminus \tuple {0, 0}$, is an Abelian group.

Taking the group axioms in turn:

### Group Axiom $\text G 0$: Closure

Let $\tuple {a, b}$ and $\tuple {c, d}$ be arbitrary elements of $\tuple {E, \times_E}$.

By definition:

$\tuple {a, b} \times_E \tuple {c, d} := \tuple {\paren {a \times_F c} +_F \paren {-\paren {b \times_F d} }, \paren {a \times_F d} +_F \paren {b \times_F c} }$

Because $\struct {F, +_F}$ is a group, $+_F$ is a closed operation.

Because $\struct {F, \times_F}$ is a group, $\times_F$ is a closed operation.

Hence:

$\paren {a \times_F c} +_F \paren {-\paren {b \times_F d} } \in F$

and:

$\paren {a \times_F d} +_F \paren {b \times_F c} \in F$

That is:

$\tuple {\paren {a \times_F c} +_F \paren {-\paren {b \times_F d} }, \paren {a \times_F d} +_F \paren {b \times_F c} } \in F \times F$

and so by definition:

$\tuple {a, b} \times_E \tuple {c, d} \in E$

Thus $\tuple {a, b} \times_E \tuple {c, d} \in E$ and so $\tuple {E, \times_E}$ is closed.

We specifically note that if $\tuple {a, b} = \tuple {0, 0}$ or $\tuple {c, d} = \tuple {0, 0}$, then:

$\tuple {a, b} \times_E \tuple {c, d} = \tuple {0, 0}$

Suppose that $\tuple {a, b} \tuple {c, d} = \tuple {0, 0}$.

We have:

 $\ds \tuple {a, b} \tuple {c, d}$ $=$ $\ds \tuple {0, 0}$ $\ds \leadsto \ \$ $\ds \tuple {a c - b d, a d + b c}$ $=$ $\ds \tuple {0, 0}$ $\ds \leadsto \ \$ $\ds \tuple {a c - b d}$ $=$ $\ds 0$ $\, \ds \land \,$ $\ds \tuple {a d + b c}$ $=$ $\ds 0$ $\ds \leadsto \ \$ $\ds \tuple {a c d - b d^2}$ $=$ $\ds 0$ where $d^2 := d d$ $\, \ds \land \,$ $\ds \tuple {a c d + b c^2}$ $=$ $\ds 0$ and $c^2 := c c$ $\ds \leadsto \ \$ $\ds b \paren {c^2 + d^2}$ $=$ $\ds 0$

and also:

 $\ds \tuple {a c - b d}$ $=$ $\ds 0$ $\, \ds \land \,$ $\ds \tuple {a d + b c}$ $=$ $\ds 0$ $\ds \leadsto \ \$ $\ds \tuple {a c^2 - b c d}$ $=$ $\ds 0$ $\, \ds \land \,$ $\ds \tuple {a d^2 + b c d}$ $=$ $\ds 0$ $\ds \leadsto \ \$ $\ds a \paren {c^2 + d^2}$ $=$ $\ds 0$

it follows that if either $c \ne 0$ or $d \ne 0$, then $a = b = 0$.

So for $\tuple {a, b} \tuple {c, d} = \tuple {0, 0}$ it must be the case that either $\tuple {a, b} = \tuple {0, 0}$ or $\tuple {c, d} = \tuple {0, 0}$.

Thus it follows that $\tuple {E, \times_E} \setminus \tuple {0, 0}$ is closed.

$\Box$

### Group Axiom $\text G 1$: Associativity

 $\ds \forall a, b, c, d, e, f \in F: \,$ $\ds$  $\ds \paren {\tuple {a, b} \tuple {c, d} } \tuple {e, f}$ $\ds$ $=$ $\ds \tuple {a c - b d, a d + b c} \tuple {e, f}$ Definition of $\times_E$ $\ds$ $=$ $\ds \tuple {\paren {a c - b d} e - \paren {a d + b c} f, \paren {a c - b d} f + \paren {a d + b c} e}$ Definition of $\times_E$ $\text {(1)}: \quad$ $\ds$ $=$ $\ds \tuple {a c e - b d e - a d f - b c f, a c f - b d f + a d e + b c e}$ $F$ is a field: $\times_F$ distributes over $+_F$

 $\ds \forall a, b, c, d, e, f \in F: \,$ $\ds$  $\ds \tuple {a, b} \paren {\tuple {c, d} \tuple {e, f} }$ $\ds$ $=$ $\ds \tuple {a, b} \tuple {c e - d f, c f + d e}$ Definition of $\times_E$ $\ds$ $=$ $\ds \tuple {a \paren {c e - d f} - b \paren {c f + d e}, a \paren {c f + d e} + b \paren {c e - d f} }$ Definition of $\times_E$ $\text {(2)}: \quad$ $\ds$ $=$ $\ds \tuple {a c e - a d f - b c f - b d e, a c f + a d e + b c e - b d f}$ $F$ is a field: $\times_F$ distributes over $+_F$

Because $\struct {F, +_F}$ is an Abelian group, $+_F$ is a commutative operation.

Thus from $(1)$ and $(2)$:

$\paren {\tuple {a, b} \tuple {c, d} } \tuple {e, f} = \tuple {a, b} \paren {\tuple {c, d} \tuple {e, f} }$

Thus $\times_E$ is associative.

$\Box$

### Group Axiom $\text G 2$: Existence of Identity Element

In order for $\tuple {E^*, \times_E}$ to be a group, it must have an identity element

With that in mind, let $\tuple {x, y} \in E$ such that:

$\forall \tuple {a, b} \in E: \tuple {x, y} \tuple {a, b} = \tuple {a, b} = \tuple {a, b} \tuple {x, y}$

We have:

 $\ds \forall a, b \in F: \,$ $\ds \tuple {x, y} \tuple {a, b}$ $=$ $\ds \tuple {a, b}$ $\ds \leadsto \ \$ $\ds \tuple {x a - y b, x b + y a}$ $=$ $\ds \tuple {a, b}$ $\ds \leadsto \ \$ $\ds x a - y b$ $=$ $\ds a$ Equality of Ordered Tuples $\, \ds \land \,$ $\ds x b + y a$ $=$ $\ds b$ $\ds \leadsto \ \$ $\ds x a b - y b b$ $=$ $\ds a b$ $F$ is a field: $\times_F$ distributes over $+_F$ $\, \ds \land \,$ $\ds x a b + y a a$ $=$ $\ds a b$ $\ds \leadsto \ \$ $\ds 2 x a b$ $=$ $\ds 2 a b$ $\, \ds \land \,$ $\ds y \paren {a a - b b}$ $=$ $\ds 0$

Because $F$ is a field, this means that:

$x$ is the multiplicative identity of $F$
$y$ is the zero of $F$.

That is:

$\tuple {x, y} = \tuple {1, 0}$

It suffices to check that:

 $\ds \forall a, b \in F: \,$ $\ds \tuple {a, b} \tuple {1, 0}$ $=$ $\ds \tuple {a 1 - b 0, a 0 + b 1}$ $\ds$ $=$ $\ds \tuple {a, b}$

Thus it has been shown that $\tuple {1, 0}$ is the identity element of $\tuple {E, \times_E}$.

Hence, directly, $\tuple {1, 0}$ is the identity element of $\tuple {E^*, \times_E}$.

$\Box$

### Group Axiom $\text G 3$: Existence of Inverse Element

We have that $\tuple {1, 0}$ is the identity element of $\tuple {E^*, \times_E}$.

We investigate the conditions under which, for any given $\tuple {a, b} \in E$, there exists $\tuple {x, y} \in E$ such that:

$\tuple {x, y} \tuple {a, b} = \tuple {1, 0}$

Hence:

 $\ds \tuple {x, y} \tuple {a, b}$ $=$ $\ds \tuple {1, 0}$ $\ds \leadsto \ \$ $\ds \tuple {x a - y b, x b + y a}$ $=$ $\ds \tuple {1, 0}$ Definition of $+_E$ $\ds \leadsto \ \$ $\ds x a - y b$ $=$ $\ds 1$ $\ds \land \ \$ $\ds x b + y a$ $=$ $\ds 0$ $\ds \leadsto \ \$ $\ds x a b - y b^2$ $=$ $\ds b$ where $b^2 := b b$ $\ds \land \ \$ $\ds x a b + y a^2$ $=$ $\ds 0$ and $a^2 := a a$ $\ds \leadsto \ \$ $\ds y \paren {a^2 + b^2}$ $=$ $\ds -b$ $\ds \leadsto \ \$ $\ds y$ $=$ $\ds \dfrac {-b} {a^2 + b^2}$ $\ds \leadsto \ \$ $\ds x a - \dfrac {-b} {a^2 + b^2} b$ $=$ $\ds 1$ $\ds \leadsto \ \$ $\ds x a$ $=$ $\ds \dfrac {-b^2} {a^2 + b^2} + 1$ $\ds$ $=$ $\ds \dfrac {-b^2 + a^2 + b^2} {a^2 + b^2}$ $\ds$ $=$ $\ds \dfrac {a^2} {a^2 + b^2}$ $\ds \leadsto \ \$ $\ds x$ $=$ $\ds \dfrac a {a^2 + b^2}$ $\ds \leadsto \ \$ $\ds \tuple {x, y}$ $=$ $\ds \paren {\dfrac a {a^2 + b^2}, \dfrac {-b} {a^2 + b^2} }$

Hence, for $\tuple {x, y}$ to be the multiplicative inverse of $\tuple {a, b}$, it is necessary and sufficient that:

$\tuple {x, y} = \paren {\dfrac a {a^2 + b^2}, \dfrac {-b} {a^2 + b^2} }$

This can happen when:

$a^2 + b^2 \ne 0$

that is, when:

$\tuple {a, b} \ne \tuple {0, 0}$

So:

$\forall \tuple {a, b} \in E \setminus \tuple {0, 0}: \tuple {a, b}^{-1} = \paren {\dfrac a {a^2 + b^2}, \dfrac {-b} {a^2 + b^2} }$

Thus every element $\tuple {a, b}$ of $\tuple {E, \times_E} \setminus \tuple {0, 0}$ has an inverse $\paren {\dfrac a {a^2 + b^2}, \dfrac {-b} {a^2 + b^2} }$.

$\Box$

All the group axioms are thus seen to be fulfilled, and so $\tuple {E, \times_E}$ is a group.

### Commutativity

It is sufficient to observe that:

 $\ds \forall a, b, c, d \in F: \,$ $\ds \tuple {a, b} \tuple {c, d}$ $=$ $\ds \tuple {a c - b d, a d + b c}$ Definition of $\times_E$ $\ds$ $=$ $\ds \tuple {c a - d b, c b + d a}$ $\struct {F, +_F, \times_F}$ is a field $\ds$ $=$ $\ds \tuple {c, d} \tuple {a, b}$

and it is seen that $\times_E$ is a commutative operation.

$\Box$

### Distributivity

It remains to be shown that $\times_E$ is distributive over $+_E$.

 $\ds \forall a, b, c, d, e, f \in F: \,$ $\ds$  $\ds \tuple {a, b} \paren {\tuple {c, d} + \tuple {e, f} }$ $\ds$ $=$ $\ds \tuple {a, b} \tuple {c + e, d + f}$ Definition of $+_E$ $\ds$ $=$ $\ds \tuple {a \paren {c + e} - b \paren {d + f}, a \paren {d + f} + b \paren {c + e} }$ Definition of $\times_E$ $\ds$ $=$ $\ds \tuple {a c + a e - b d - b f, a d + a f + b c + b e}$ $F$ is a field: $\times_F$ distributes over $+_F$ $\ds$ $=$ $\ds \tuple {\paren {a c - b d} + \paren {a e - b f}, \paren {a d + b c} + \paren {a f + b e} }$ $\ds$ $=$ $\ds \tuple {a c - b d, a d + b c} + \tuple {a e - b f, a f + b e}$ Definition of $+_E$ $\ds$ $=$ $\ds \tuple {a, b} \tuple {c, d} + \tuple {a, b} \tuple {e, f}$ Definition of $\times_E$

Thus it has been demonstrated that $\times_E$ is distributive over $+_E$.

$\Box$

All the criteria have been checked, and it follows that $\tuple {E, +_E, \times_E}$ is a field.

$\blacksquare$