Division Algebra has No Zero Divisors

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Theorem

Let $A = \left({A_F, \oplus}\right)$ be an algebra over a field $F$.


Then $A$ is a division algebra if and only if it has no zero divisors.


That is:

$\forall a, b \in A_F: a \oplus b = \mathbf 0_A \implies a = \mathbf 0_A \lor b = \mathbf 0_A$

If the product of two elements of $A$ is zero, then at least one of those elements must itself be zero.


Some sources use this as the definition of a division algebra and from it deduce:

$\forall a, b \in A_F, b \ne \mathbf 0_A: \exists_1 x \in A_F, y \in A_F: a = b \oplus x, a = y \oplus b$


Proof

Let $A$ be a division algebra, in the sense that:

$\forall a, b \in A_F, b \ne \mathbf 0_A: \exists_1 x \in A_F, y \in A_F: a = b \oplus x, a = y \oplus b$

Suppose that $\exists a, b \in A_F \setminus \left\{{\mathbf 0_A}\right\}: \mathbf 0_A = b \oplus a$.

Then by definition of the zero vector we also have that $\mathbf 0_A = b \oplus \mathbf 0_A$.

So there are two elements $x$ of $A_F$ such that $\mathbf 0_A = b \oplus x$, that is, $a$ and $\mathbf 0_A$.


Similarly, suppose that $\exists a, b \in A_F \setminus \left\{{\mathbf 0_A}\right\}: \mathbf 0_A = a \oplus b$.

Then by definition of the zero vector we also have that $\mathbf 0_A = \mathbf 0_A \oplus b$.

So there are two elements $y$ of $A_F$ such that $\mathbf 0_A = y \oplus b$, that is, $a$ and $\mathbf 0_A$.

So $A$ can not be a division algebra.


So it follows that if:

$\forall a, b \in A_F, b \ne \mathbf 0_A: \exists_1 x \in A_F, y \in A_F: a = b \oplus x, a = y \oplus b$

then:

$\forall a, b \in A_F: a \oplus b = \mathbf 0_A \implies a = \mathbf 0_A \lor b = \mathbf 0_A$

$\Box$


Now suppose that:

$\forall a, b \in A_F: a \oplus b = \mathbf 0_A \implies a = \mathbf 0_A \lor b = \mathbf 0_A$


Suppose $\exists x_1, x_2 \in A_F, x_1 \ne x_2$ such that:

$a = b \oplus x_1$
$a = b \oplus x_2$

We have that $x_1 \ne x_2$ and so $x_1 - x_2 = z \ne \mathbf 0_A$.

Thus $b \oplus x_1 - b \oplus x_2 = \mathbf 0_A$.

As $\oplus$ is a bilinear mapping, it follows that $b \oplus \left({x_1 - x_2}\right) = \mathbf 0_A$ and so $b \oplus z = \mathbf 0_A$.

But $z \ne \mathbf 0_A$ and so it is not the case that $\forall a, b \in A_F: a \oplus b = \mathbf 0_A \implies a = \mathbf 0_A \lor b = \mathbf 0_A$.


Similarly it can be shown that if $\exists y_1, y_2 \in A_F, y_1 \ne y_2$ such that:

$a = y_1 \oplus b$
$a = y_2 \oplus b$

then it is not the case that $\forall a, b \in A_F: a \oplus b = \mathbf 0_A \implies a = \mathbf 0_A \lor b = \mathbf 0_A$.


Now suppose that:

$\exists a, b \in A_F, b \ne \mathbf 0_A: \not \exists_1 x \in A_F: a = b \oplus x$

or

$\exists a, b \in A_F, b \ne \mathbf 0_A: \not \exists_1 y \in A_F: a = y \oplus b$



Thus it is not the case that $\forall a, b \in A_F: a \oplus b = \mathbf 0_A \implies a = \mathbf 0_A \lor b = \mathbf 0_A$.

So it follows that if:

$\forall a, b \in A_F: a \oplus b = \mathbf 0_A \implies a = \mathbf 0_A \lor b = \mathbf 0_A$

then:

$\forall a, b \in A_F, b \ne \mathbf 0_A: \exists_1 x \in A_F, y \in A_F: a = b \oplus x, a = y \oplus b$

$\blacksquare$