Associative Algebra has Multiplicative Inverses iff Unitary Division Algebra
Theorem
Let $\struct {A_R, \oplus}$ be an associative algebra over the ring $A_R$.
Then:
- $\struct {A_R, \oplus}$ has a unique multiplicative inverse for every non-zero $a \in A_R$
- $\struct {A_R, \oplus}$ is a unitary division algebra.
Proof
Let $A = \struct {A_R, \oplus}$ 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:
| \(\ds 1\) | \(=\) | \(\ds b \oplus x\) | for some $x \in A_R$ | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds y \oplus 1\) | \(=\) | \(\ds y \oplus \paren {b \oplus x}\) | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds y \oplus 1\) | \(=\) | \(\ds \paren {y \oplus b} \oplus x\) | Associativity of $\oplus$ | ||||||||||
| \(\ds \leadsto \ \ \) | \(\ds y \oplus 1\) | \(=\) | \(\ds 1 \oplus x\) | as $1 = y \oplus b$ | ||||||||||
| \(\ds \leadsto \ \ \) | \(\ds y\) | \(=\) | \(\ds x\) | Definition of $1$ |
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 $A_R$.
$\Box$
Now suppose that $A_R$ has a unique multiplicative inverse in $A_R$.
It follows directly that, by definition, $\struct {A_R, \oplus}$ 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:
| \(\ds a \oplus b^{-1}\) | \(=\) | \(\ds x\) | ||||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \paren {a \oplus b^{-1} } \oplus b\) | \(=\) | \(\ds x \circ b\) | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds a \oplus \paren {b^{-1} \oplus b}\) | \(=\) | \(\ds x \oplus b\) | Associativity of $\oplus$ | ||||||||||
| \(\ds \leadsto \ \ \) | \(\ds a \oplus 1\) | \(=\) | \(\ds x \oplus b\) | as $b^{-1} \oplus b = 1$ | ||||||||||
| \(\ds a\) | \(=\) | \(\ds x \oplus b\) |
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:
| \(\ds a\) | \(=\) | \(\ds x_1 \oplus b\) | ||||||||||||
| \(\ds \leadsto \ \ \) | \(\ds a \oplus b^{-1}\) | \(=\) | \(\ds \paren {x_1 \oplus b} \oplus b^{-1}\) | $b^{-1}$ being the unique multiplicative inverse of $b$ | ||||||||||
| \(\ds \leadsto \ \ \) | \(\ds a \oplus b^{-1}\) | \(=\) | \(\ds x_1 \oplus \paren {b \oplus b^{-1} }\) | Associativity of $\oplus$ | ||||||||||
| \(\ds \leadsto \ \ \) | \(\ds a \oplus b^{-1}\) | \(=\) | \(\ds x_1 \circ 1\) | as $b \oplus b^{-1} = 1$ | ||||||||||
| \(\ds \leadsto \ \ \) | \(\ds a \oplus b^{-1}\) | \(=\) | \(\ds x_1\) |
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
- $\forall a, b \in A_R, b \ne \mathbf 0_A: \exists_1 y \in A_R: a = y \oplus b$
$\blacksquare$