Subring of Polynomials over Integral Domain is Smallest Subring containing Element and Domain

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 $D \sqbrk x$ is the smallest subring of $R$ which contains $D$ as a subring and $x$ as an element.

Proof

This theorem requires a proof. In particular: Whitelaw says "fairly obviously", so should be more or less straightforward. You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by crafting such a proof. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{ProofWanted}} from the code. If you would welcome a second opinion as to whether your work is correct, add a call to {{Proofread}} the page.

Sources

• 1978: Thomas A. Whitelaw: An Introduction to Abstract Algebra ... (previous) ... (next): $\S 64.1$ Polynomial rings over an integral domain