Associative Algebra has Multiplicative Inverses iff Unitary Division Algebra

Theorem
Let $\left({A_R, \oplus}\right)$ be an associative algebra over the ring $A_R$.

Then:
 * $\left({A_R, \oplus}\right)$ has a unique multiplicative inverse for every non-zero $a \in A_R$


 * $\left({A_R, \oplus}\right)$ is a unitary division algebra.
 * $\left({A_R, \oplus}\right)$ is a unitary division algebra.

Proof
Let $A = \left({A_R, \oplus}\right)$ be a unitary division algebra whose zero is $\mathbf 0_A$.

From the unitary nature of $A$ we have that $\oplus$ has a unit $1$ such that:
 * $\forall a \in A_R, a \ne \mathbf 0_A: a \oplus 1 = a = 1 \oplus a$

From the division algebra nature of $A$ we have that $R$ is a field and that:
 * $\forall a, b \in A_R, b \ne \mathbf 0_A: \exists_1 x \in A_R, y \in A_R: b \oplus x = a = y \oplus b$

As $1 \in A_R$ it follows that the above holds if $1$ is substituted for $a$:
 * $\exists_1 x \in A_R, y \in A_R: b \oplus x = 1 = y \oplus b$

Hence:

So if we let $b^{-1} := x = y$ in the above, we have:


 * $\exists_1 b^{-1} \in A_R: 1 = b \oplus b^{-1} = 1 = b^{-1} \oplus b$

That is, $b^{-1}$ is a unique multiplicative inverse of $b$.

So every non-zero element of $A_R$ has a unique multiplicative inverse in $A_R$.

Now suppose that $A_R$ has a unique multiplicative inverse in $A_R$.

It follows directly that, by definition, $\left({A_R, \oplus}\right)$ has to be a unitary algebra or there is no $1$ for $a \oplus a^{-1}$ to equal.

First we prove existence, i.e we show that:
 * $\forall a, b \in A_R, b \ne \mathbf 0_A: \exists x \in A_R, y \in A_R: b \oplus x = a = y \oplus b$

Take $a \oplus b^{-1}$ where $b^{-1}$ is the unique multiplicative inverse of $b$.

Then:

Similarly, we show that $\exists y \in A_R: a = y \oplus b$.

Now we prove uniqueness.

Now suppose $\exists x_1, x_2 \in A_R: a = x_1 \oplus b, a = x_2 \oplus b$.

Then:

In exactly the same way, $a = x_2 \oplus b \implies a \oplus b^{-1} = x_2$.

And so $x_1 = x_2$ thus proving uniqueness.

In a similar way we prove that
 * $\forall a, b \in A_R, b \ne \mathbf 0_A: \exists_1 y \in A_R: a = y \oplus b$