Set of Polynomials over Integral Domain is Subring

Theorem

Let $\struct {R, +, \circ}$ be a commutative ring.

Let $\struct {D, +, \circ}$ be an integral subdomain of $R$.

Then $\forall x \in R$, the set $D \sqbrk x$ of polynomials in $x$ over $D$ is a subring of $R$.

By application of the Subring Test:

As $D$ is an integral domain, it has a unity $1_D$ and so $x = 1_D x$.

Hence $x \in D \sqbrk x$ and so $D \sqbrk x \ne \O$.

Let $p, q \in D \sqbrk x$.

Then let:

$\ds p = \sum_{k \mathop = 0}^m a_k \circ x^k, q = \sum_{k \mathop = 0}^n b_k \circ x^k$

Thus:

$\ds -q = -\sum_{k \mathop = 0}^n b_k \circ x^k = \sum_{k \mathop = 0}^n \paren {-b_k} \circ x^k$

and so:

$q \in D \sqbrk x$

$p + \paren {-q} \in D \sqbrk x$

Finally, from Polynomials Closed under Ring Product, we have that $p \circ q \in D \sqbrk x$.

All the criteria of the Subring Test are satisfied.

Hence the result.

$\blacksquare$