0

I already demonstrated one side but I don't know how to conclude the proof in the other direction.

Show that in $\mathbb{Q}[x]$, the ideals $ \langle p(x) \rangle$ and $\langle q(x) \rangle$ are equals if and only if $p(x) = a q(x)$, for some $a \in \mathbb{Q^{*}}$

( $\Leftarrow$ it's done). My idea for the other side ($\Rightarrow$) is:

$\langle p(x) \rangle = \langle q(x) \rangle$ implies that $p(x) \in \langle q(x) \rangle$ and $q(x) \in \langle p(x) \rangle$.

Then $p(x) = q(x) \cdot s(x)$, and $q(x) = p(x) \cdot r(x)$ for some $s(x),r(x) \in \mathbb{Q}[x]$

$ \Rightarrow p(x)|q(x)$ and $q(x)|p(x) \Longleftrightarrow p(x) = aq(x)$

This demonstration does not convince me, does anyone know how to do the correct demonstration?

Ted Shifrin
  • 125,228
pucky
  • 41
  • What about degree? Also, what are the units in the polynomial ring? – Ted Shifrin Jun 08 '24 at 21:34
  • Since contains = divides for principal ideals we have $,(p)=(q)\iff (p)\supseteq (q)\supseteq (p)\iff p\mid q\mid p,,$ i.e. $,p,q,$ are associates. Now apply the dupes and you'll get a more general result. – Bill Dubuque Jun 08 '24 at 21:45

0 Answers0