Polynomials Closed under Addition/Polynomials over Integral Domain

Theorem

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

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

Then $\forall x \in R$, the set $D \sqbrk x$ of polynomials in $x$ over $D$ is closed under the operation $+$.

Proof 1

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

We can express them as:

$\ds p = \sum_{k \mathop = 0}^m a_k \circ x^k$

$\ds q = \sum_{k \mathop = 0}^n b_k \circ x^k$

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:

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

Because $\struct {R, +, \circ}$ is a commutative ring, it follows that:

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

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

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