# Units of Ring of Polynomial Forms over Field

## Theorem

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

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

Then the units of $F \sqbrk X$ are all the elements of $F \sqbrk X$ whose degree is $0$.

## Proof

An element of $F \sqbrk X$ whose degree is $0$ is merely an element of $F$.

But note that $0_F$, considered as an element of $F \sqbrk X$, 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 $F \sqbrk X$ whose degree is $0$ are units of $F$ and hence of $F \sqbrk X$.

Now suppose $\map a X \in F \sqbrk X$ is a unit of $F \sqbrk X$.

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

Neither $\map a X$ nor $\map q X$ can be null.

So by Properties of Degree $\map \deg {\map a X} + \map \deg {\map q X} = 0$.

So $\map \deg {\map a X} = 0$.

$\blacksquare$

## Sources

- 1969: C.R.J. Clapham:
*Introduction to Abstract Algebra*... (previous) ... (next): Chapter $6$: Polynomials and Euclidean Rings: $\S 26$. Divisibility: Example $48$