[[Integral domain]]
# The polynomial ring over an integral domain is an integral domain

Let $D$ be a [[ring]] and $D[x]$ be the [[polynomial ring]] in indeterminate $x$.
Then $D[x]$ is an [[integral domain]] iff $D$ is an integral domain. #m/thm/ring

> [!check]- Proof
> Assume $D$ is an integral domain.
> Clearly $D[x]$ is commutative since $D$ is.
> Let $f(x), g(x) \in D[x]$ be nonzero with leading terms $f_{n}x^n$ and $g_{m}x^m$ respectively.
> Then the leading term of $f(x)g(x)$ is $f_{n}g_{m}x^{n+m}$ so $f(x)g(x) \neq 0$. 
> 
> Note if $D[x]$ is an integral domain, then so are its subrings, including $D$. <span class="QED"/>

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