Subring of Polynomials over Integral Domain Contains that Domain
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Theorem
Let $\struct {R, +, \circ}$ be a commutative ring.
Let $\struct {D, +, \circ}$ be an integral subdomain of $R$.
Let $x \in R$.
Let $D \sqbrk x$ denote the ring of polynomials in $x$ over $D$.
Then $D \sqbrk x$ contains $D$ as a subring and $x$ as an element.
Proof
We have that $\ds \sum_{k \mathop = 0}^m a_k \circ x^k$ is a polynomial for all $m \in \Z_{\ge 0}$.
Set $m = 0$:
- $\ds \sum_{k \mathop = 0}^0 a_k \circ x^k = a_k \circ x^0 = a_k \circ 1_D = a_k$
Thus:
- $\ds \forall a_k \in D: \sum_{k \mathop = 0}^0 a_k \circ x^k \in D$
It follows directly that $D$ is a subring of $D \sqbrk x$ by applying the Subring Test on elements of $D$.
$\blacksquare$
Sources
- 1978: Thomas A. Whitelaw: An Introduction to Abstract Algebra ... (previous) ... (next): $\S 64.1$ Polynomial rings over an integral domain