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 was explicitly stated in the above.

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

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:

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:

and also:

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.

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.

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:

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:

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 $\tuple {E^*, \times_E}$.

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:

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 $\paren {\dfrac a {a^2 + b^2}, \dfrac {-b} {a^2 + b^2} }$.

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:

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

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

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

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