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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Without loss of generality, suppose $m > n$.
Then we can express $q$ as:
- $\ds \sum_{k \mathop = 0}^n b_k \circ x^k + \sum_{k \mathop = n \mathop + 1}^m 0_D \circ x^k$
Thus:
- $\ds p + q = \sum_{k \mathop = 0}^n \paren {a_k + b_k} \circ x^k + \sum_{k \mathop = n \mathop + 1}^m a_k \circ x^k$
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.
$\blacksquare$
Proof 2
A commutative ring with unity is a ring.
An integral subdomain of a commutative ring with unity $R$ is a subring of $R$.
The result then follows as a special case of Polynomials Closed under Addition: Polynomials over Ring
$\blacksquare$