Theorem. If a polynomial ring $D[X]$ over an integral domain $D$ is a principal ideal domain, then coefficient ring $D$ is a field. (Cf. the corollary 4 in the entry polynomial ring over a field.)

Proof. Let $a$ be any non-zero element of $D$ . Then the ideal $(a,\,X)$ of $D[X]$ is a principal ideal $(f(X))$ with $f(X)$ a non-zero polynomial. Therefore, $$a = f(X)g(X), \quad X = f(X)h(X)$$ with $g(X)$ and $h(X)$ certain polynomials in $D[X]$ . From these equations one infers that $f(X)$ is a constant polynomial $c$ and $h(X)$ is a first degree polynomial $b_0\!+\!b_1X$ ($b_1 \neq 0$ ). Thus we obtain the equation $$cb_0+cb_1X = X,$$ which shows that $cb_1$ is the unity 1 of $D$ . Thus $c = f(X)$ is a unit of $D$ , whence $$(a,\,X) = (f(X)) = (1) = D[X].$$ So we can write $$1 = a\!\cdot\!u(X)+X\!\cdot\!v(X),$$ where $u(X),\,v(X) \in D[X]$ . This equation cannot be possible without that $a$ times the constant term of $u(X)$ is the unity. Accordingly, $a$ has a multiplicative inverse in $D$ . Because $a$ was arbitrary non-zero elenent of the integral domain $D$ , $D$ is a field.

Bibliography

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DAVID M. BURTON: A first course in rings and ideals. Addison-Wesley Publishing Company. Reading, Menlo Park, London, Don Mills (1970).