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Let $A$ be an integral domain and $a\in A$. When is $I=(a,x)$ a principal ideal of $A[x]$?
My solution:
If $I$ is a principal ideal then $I=(d(x))$ with $d(x) \in A[x]$. $d(x)$ is the greatest common divisor of $a$ and $x$, so $d(x)$ is equal to $1$. So $1=a.f(x)+x.g(x)$ with $f,g \in A[x]$. From there, it follows that $a$ is non-zero and $a$ is invertible.
On the contrary, if $a$ is non-zero and $a$ is invertible, then $I$ is a principal ideal. I haven't been able to prove it this.
Is my solution correct?

walkar
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Ngân Kim
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  • "Greatest common divisor" doesn't really make sense in a ring like $A[x]$. If $A$ is very nice, e.g. a field or UFD, then perhaps more can be said. Also, in the case of $\mathbb{Z}[x]$, the ideal $(2x,x)$ equal to the principal ideal $(x)$, so certainly it doesn't need to be generated by $1$... – walkar Dec 19 '23 at 02:21
  • Also, you should be careful -- "primary ideal" and "principal ideal" mean different things. A principal ideal is one generated by a single element, whereas a primary ideal means something very different, unrelated to the number of generators. – walkar Dec 19 '23 at 02:26
  • oh, i see it's principal ideal – Ngân Kim Dec 19 '23 at 02:47
  • I think d(x) is gcd of (a,x) if A[x] is a principal ring, so my answer is wrong. Can you help me with this question? – Ngân Kim Dec 19 '23 at 02:52
  • See the linked dupe, e.g. this Lemma. – Bill Dubuque Dec 19 '23 at 03:51

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