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.