Polynomial Forms over Field form Integral Domain/Formulation 1/Proof 1

Theorem

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

Let $X$ be transcendental in $F$.

Let $F \sqbrk X$ be the ring of polynomial forms in $X$ over $F$.

Then $F \sqbrk X$ is an integral domain.

Proof

We already have from Ring of Polynomial Forms is Commutative Ring with Unity that $F \sqbrk X$ is a ring.

Suppose $f$ and $g$ are polynomials in $F \sqbrk X$ such that $f \ne 0_F, g \ne 0_F$.

If $\map \deg f = \map \deg g = 0$ 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:

$\map \deg {f g} = \map \deg f + \map \deg g$

and so:

$\map \deg {f g} > 0$

which means:

$f g \ne 0_F$

Hence the result, by definition of integral domain.

$\blacksquare$