Polynomials Closed under Addition

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 closed under the operation $+$.

Proof
Let $p, q$ be polynomials in $x$ over $D$.

We can express them as $\displaystyle p = \sum_{k=0}^m a_k \circ x^k, q = \sum_{k=0}^n b_k \circ x^k$ where:
 * 1) $a_k, b_k \in D$ for all $k$;
 * $m, n$ are some non-negative integers.


 * Suppose $m = n$.

Then $\displaystyle p + q = \sum_{k=0}^n a_k \circ x^k + \sum_{k=0}^n b_k \circ x^k$.

Because $\left({R, +, \circ}\right)$ is a commutative ring, it follows that $\displaystyle p + q = \sum_{k=0}^n \left({a_k + b_k}\right) \circ x^k$ which is also a polynomial in $x$ over $D$.


 * Now suppose, with no loss of generality, that $m > n$.

Then we can express $q$ as $\displaystyle \sum_{k=0}^n b_k \circ x^k + \sum_{k=n+1}^m 0_D \circ x^k$.

Thus $\displaystyle p + q = \sum_{k=0}^n \left({a_k + b_k}\right) \circ x^k + \sum_{k=n+1}^m a_k \circ x^k$ which is also a polynomial in $x$ over $D$.


 * Thus the sum of two polynomials in $x$ over $D$ is another polynomial in $x$ over $D$, hence the result.