Units of Ring of Polynomial Forms over Field

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

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

Then the units of $F \left[{X}\right]$ are all the elements of $F \left[{X}\right]$ whose degree is $0$.

Proof
An element of $F \left[{X}\right]$ whose degree is $0$ is merely an element of $F$.

But note that $0_F$, considered as an element of $F \left[{X}\right]$, has a degree which is not defined, so the null polynomial is seen to be excluded.

Any element $a$ of $F$ has an inverse $1_F / a$.

So all the elements of $F \left[{X}\right]$ whose degree is $0$ are units of $F$ and hence of $F \left[{X}\right]$.

Now suppose $a \left({X}\right) \in F \left[{X}\right]$ is a unit of $F \left[{X}\right]$.

Then $a \left({X}\right) q \left({X}\right) = 1_F$ for some $q \left({X}\right) \in F \left[{X}\right]$.

Neither $a \left({X}\right)$ nor $q \left({X}\right)$ can be null.

So by Properties of Degree $\deg a \left({X}\right) + \deg q \left({X}\right) = 0$.

So $\deg a \left({X}\right) = 0$.