# polynomial ring which is PID

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 (http://planetmath.org/ZeroPolynomial2).  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 polynomial $c$ and $h(X)$ is a first degree polynomial $b_{0}\!+\!b_{1}X$ ($b_{1}\neq 0$).  Thus we obtain the equation

 $cb_{0}+cb_{1}X\;=\;X,$

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.

## References

• 1 David M. Burton: A first course in rings and ideals. Addison-Wesley Publishing Company. Reading, Menlo Park, London, Don Mills (1970).
Title polynomial ring which is PID PolynomialRingWhichIsPID 2013-03-22 17:53:04 2013-03-22 17:53:04 pahio (2872) pahio (2872) 8 pahio (2872) Theorem msc 13P05 PolynomialRingOverFieldIsEuclideanDomain