Set of Polynomials over Integral Domain is Subring

Theorem
Let $\left({R, +, \circ}\right)$ be a commutative ring.

Let $\left({D, +, \circ}\right)$ be an integral domain such that $D$ is a subring of $R$.

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

This subring is denoted $D \left[{x}\right]$.

$D \left[{x}\right]$ contains $D$ as a subring and $x$ as an element.

$D \left[{x}\right]$ is the smallest subring of $R$ with these properties.

Subring
This follows by a straightforward application of the Subring Test:


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

Hence $x \in D \left[{x}\right]$ and so $D \left[{x}\right] \ne \varnothing$.


 * Let $p, q \in D \left[{x}\right]$.

Then let $p = \sum_{k=0}^m a_k \circ x^k, q = \sum_{k=0}^n b_k \circ x^k$.

Thus $-q = -\sum_{k=0}^n b_k \circ x^k = \sum_{k=0}^n \left({-b_k}\right) \circ x^k$ and so $q \in D \left[{x}\right]$.

Thus as Polynomials Closed under Addition‎, it follows that $p + \left({-q}\right) \in D \left[{x}\right]$.


 * Finally, from Polynomials Closed under Ring Product, we have that $p \circ q \in D \left[{x}\right]$.

All the criteria of the Subring Test are satisfied.

Integral Domain Subring
Now we show that $D \left[{x}\right]$ contains $D$ as a subring.

As the expression $\sum_{k=0}^m a_k \circ x^k$ is a polynomial for all $m \in \Z_+$, we can set $m = 0$ and see that $\sum_{k=0}^0 a_k \circ x^k = a_k \circ x^0 = a_k \circ 1_D = a_k$.

Thus $\forall a_k \in D: \sum_{k=0}^0 a_k \circ x^k \in D$.

It follows directly that $D$ is a subring of $D \left[{x}\right]$ by applying the Subring Test on elements of $D$.

Smallest Subring
Now we show that $D \left[{x}\right]$ is the smallest subring of $R$ with these properties.