Polynomial Forms over Field form Integral Domain/Formulation 1

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Theorem

Let $\left({F, +, \circ}\right)$ be a field whose zero is $0_F$ and whose unity is $1_F$.

Let $X$ be transcendental in $F$.

Let $F \left[{X}\right]$ be the ring of polynomial forms in $X$ over $F$.


Then $F \left[{X}\right]$ is an integral domain.


Proof 1

We already have from Ring of Polynomial Forms is Commutative Ring with Unity that $F \left[{X}\right]$ is a ring.

Suppose $f$ and $g$ are polynomials in $F \left[{X}\right]$ such that $f \ne 0_F, g \ne 0_F$.

If $\deg \left({f}\right) = \deg \left({g}\right) = 0$ then $f$ and $g$ are elements of $F$.

As $F$ is a field and a field is an integral domain, $f g \ne 0_f$.

Otherwise from Degree of Product of Polynomials over Integral Domain:

$\deg \left({f g}\right) = \deg \left({f}\right) + \deg \left({g}\right)$

and so:

$\deg \left({f g}\right) > 0$

which means $f g \ne 0_F$

Hence the result, by definition of integral domain.

$\blacksquare$


Proof 2

We have from Ring of Polynomial Forms is Commutative Ring with Unity that $F \left[{X}\right]$ is a commutative ring with unity.

The result follows from Ring of Polynomial Forms is Integral Domain.

$\blacksquare$