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# 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. 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 $b_{0}\!+\!b_{1}X$ ($b_{1}\neq 0$). 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).

## Mathematics Subject Classification

13P05*no label found*

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