# Polynomials Closed under Addition/Polynomials over Integral Domain/Proof 1

## Theorem

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

Let $\left({D, +, \circ}\right)$ be an integral subdomain of $R$.

Then $\forall x \in R$, the set $D \left[{x}\right]$ of polynomials in $x$ over $D$ is closed under the operation $+$.

## Proof

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

We can express them as:

$\displaystyle p = \sum_{k \mathop = 0}^m a_k \circ x^k$
$\displaystyle q = \sum_{k \mathop = 0}^n b_k \circ x^k$

where:

$(1): \quad a_k, b_k \in D$ for all $k$
$(2): \quad m, n \in \Z_{\ge 0}$, that is, are non-negative integers.

Suppose $m = n$.

Then:

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

Because $\left({R, +, \circ}\right)$ is a commutative ring, it follows that:

$\displaystyle p + q = \sum_{k \mathop = 0}^n \left({a_k + b_k}\right) \circ x^k$

which is also a polynomials in $x$ over $D$.

Without loss of generality, suppose $m > n$.

Then we can express $q$ as:

$\displaystyle \sum_{k \mathop = 0}^n b_k \circ x^k + \sum_{k \mathop = n \mathop + 1}^m 0_D \circ x^k$

Thus:

$\displaystyle p + q = \sum_{k \mathop = 0}^n \left({a_k + b_k}\right) \circ x^k + \sum_{k \mathop = n \mathop + 1}^m a_k \circ x^k$

which is also a polynomials in $x$ over $D$.

Thus the sum of two polynomials in $x$ over $D$ is another polynomial in $x$ over $D$.

Hence the result.

$\blacksquare$