Ring of Polynomial Forms is Integral Domain

Theorem

Let $D$ be an integral domain.

Let $\left\{{X}\right\}$ be a singleton.

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

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

Proof

By Ring of Polynomial Forms we know that $D \left[{X}\right]$ is a commutative ring with unity.

Finally we note that if neither $\displaystyle f \left({x}\right) = \sum_{k=0}^n a_k x^k$ nor $\displaystyle g \left({x}\right) = \sum_{k=0}^m b_k x^k$ are the null polynomial, then their leading coefficients $a_n$ and $b_m$ are non-zero.

Therefore, as $D$ is an integral domain and $a_n,b_m \in D$, so is their product $a_nb_m$.

By the definition of polynomial multiplication, it follows that $fg$ is not the null polynomial.

It follows that $D \left[{X}\right]$ has no proper zero divisors.

Hence $D \left[{X}\right]$ is an integral domain.

$\blacksquare$

Sources

• C.R.J. Clapham: Introduction to Abstract Algebra (1969)... (previous)... (next): $\S 1.5$: Example $5$
• Thomas A. Whitelaw: An Introduction to Abstract Algebra (1978)... (previous)... (next): $\S 64.3$