# 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 $${x_1} = {x_2}$$.

Hence the uniqueness is established.

__Theorem 4__**:**

Every finite integral domain is a field, or a finite commutative ring with no zero divisor is a field.