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 $$k$$;
 * 2) $$m, n$$ 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.