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:

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

So:
 * $\tuple {E, \times_E}$ is closed.

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:
 * $c \ne 0$ \lor $d \ne 0$

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

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 $F$

And:
 * $y$ 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 $\tuple {E^*, \times_E}$.

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

Hence, all the group axioms are fulfilled, and so $\tuple {E, \times_E}$ is a group.

Commutativity
It is sufficient to show 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, $\times_E$ is distributive over $+_E$.

All the axioms have been fulfilled. Hence, $\tuple {E, +_E, \times_E}$ is a field.