Subring of Polynomials over Integral Domain Contains that Domain

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

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

Let $x \in R$.

Let $D \sqbrk x$ denote the ring of polynomials in $x$ over $D$.

Then $D \sqbrk x$ contains $D$ as a subring and $x$ as an element.

Proof
We have that $\displaystyle \sum_{k \mathop = 0}^m a_k \circ x^k$ is a polynomial for all $m \in \Z_{\ge 0}$.

Set $m = 0$:
 * $\displaystyle \sum_{k \mathop = 0}^0 a_k \circ x^k = a_k \circ x^0 = a_k \circ 1_D = a_k$

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

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