# Division Algebra has No Zero Divisors

This page has been identified as a candidate for refactoring.Convert this into a standard equivalence proof for Definition:Division Algebra/Definition 1 and Definition:Division Algebra/Definition 2Until this has been finished, please leave
`{{Refactor}}` in the code.
Because of the underlying complexity of the work needed, it is recommended that you do not embark on a refactoring task until you have become familiar with the structural nature of pages of $\mathsf{Pr} \infty \mathsf{fWiki}$.To discuss this page in more detail, feel free to use the talk page.When this work has been completed, you may remove this instance of `{{Refactor}}` from the code. |

## 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

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

This needs considerable tedious hard slog to complete it.Haven't found an approach yet ...To discuss this page in more detail, feel free to use the talk page.When this work has been completed, you may remove this instance of `{{Finish}}` from the code.If you would welcome a second opinion as to whether your work is correct, add a call to `{{Proofread}}` the page. |

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:

$\blacksquare$