Show that if $D$ is a domain but not a field then $D[x]$ is not a principal ideal domain.

Question

Show that if $D$ is a domain but not a field then $D[x]$ is not a principal ideal domain.

Sorry for my english....

Dear Cin, this is the third question that you have posted in a span of only 20 minutes, and you just seem to have copied the problems straight out of a textbook. To get better help, it is usually encouraged that people also post what ideas they have or what they're finding hard about the problem. In fact, some people might think that you're just posting your homework problems for others to solve, which is usually frowned upon. — Adrián Barquero, Apr 14 '13 at 03:19
If it is that you're having trouble posting in English, you can try posting your question in Spanish (since I see that you're from Mexico) and I'm pretty sure that another user would gladly provide a translation. — Adrián Barquero, Apr 14 '13 at 03:21

score 5 · Answer 1 · answered Apr 14 '13 at 09:12

Suppose that $D[x]$ is a principal ideal domain. Then $(x)$ is a maximal ideal: if $(x) \subset I$, then there exists $P \in D[x]$ such that $I=(P)$; in particular, $P \mid x$, so either $P=ax$ (where $a$ is invertible) and $(P)=(x)$ or $P$ is invertible and $(P)=D[x]$.

Therefore, $D \simeq D[x]/(x)$ is a field.

score 2 · Answer 2 · answered Apr 14 '13 at 03:18

Hint: If $D$ is a domain but not a field, consider the ideal $$I = \langle a, x \rangle = \{ a f(x) + x g(x) : f(x), g(x) \in D[x] \}$$ of $D[x]$ generated by a nonzero non-unit $a \in D$ and the degree one polynomial $x$. Now prove that $I$ cannot be generated by a single polynomial in $D[x]$.

Show that if $D$ is a domain but not a field then $D[x]$ is not a principal ideal domain.

2 Answers2