Theorems of Field

Theorem 1:
The multiplicative inverse of a non-zero element of a field is unique.

Proof:
Let there be two multiplicative inverse ${a^{ - 1}}$ and $a'$ for a non-zero element $a \in F$ . Let $\left( 1 \right)$ be the unity of the field $F$ .
$\therefore \,a \cdot {a^{ - 1}} = 1$ and $a \cdot a' = 1$ so that $a \cdot {a^{ - 1}} = a \cdot a'$ . Since $F - \left\{ 0 \right\}$ is a multiplicative group, applying left cancellation, we get ${a^{ - 1}} = a'$ .

Theorem 2:
A field is necessarily an integral domain.

Proof:
Since a field is a commutative ring with unity, therefore, in order to show that every field is an integral domain we only need to prove that s field is without zero divisors.

Let $F$ be any field and let $a,b \in F$ with $a \ne 0$ such that $ab = 0$ . Let $1$ be the unity of $F$ . Since $a \ne 0$ , ${a^{ - 1}}$ exists in $F$ , therefore

$\begin{gathered} ab = 0 \Rightarrow {a^{ - 1}}\left( {ab} \right) = {a^{ - 1}}0 \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \Rightarrow \left( {{a^{ - 1}}a} \right)b = 0 \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \Rightarrow 1 \cdot b = 0 \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \Rightarrow b = 0 \\ \end{gathered}$

Similarly if $b \ne 0$ then it can be shown that $ab = 0 \Rightarrow a = 0$ .

Thus $ab = 0 \Rightarrow a = 0\,\,\,{\text{or}}\,\,\,b = 0$ . Hence, a field is necessarily an integral domain.

Corollary:
Since the integral domain has no zero divisor and the field is necessarily an integral domain, therefore the field has no zero-divisor.

Theorem 3:
If $a,b$ are any two elements of a field $F$ and $a \ne 0$ , there exists a unique element $x$ such that $a \cdot x = b$ .

Proof:
Let $1$ be the unity of $F$ and ${a^{ - 1}}$ , the inverse of $a$ in $F$ then

$a \cdot \left( {{a^{ - 1}} \cdot b} \right) = \left( {a \cdot {a^{ - 1}}} \right) \cdot b = 1 \cdot b = b$

$\begin{gathered} \therefore \,ax = b \Rightarrow a \cdot x = a \cdot \left( {{a^{ - 1}}b} \right) \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \Rightarrow x = {a^{ - 1}}b \\ \end{gathered}$

Thus, $x = {a^{ - 1}}b \in F$ .

Now, suppose there are two such elements ${x_1},{x_2}$ (say), then $a \cdot {x_1} = b$ and $a \cdot {x_2} = b$ hence $a \cdot {x_1} = a \cdot {x_2}$ . On applying left cancellation, we get ${x_1} = {x_2}$ .