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Here is the proof I am reading from John Beachy, fourth edition:

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I do not understand why assuming $p$ does not divide $b_0$? Is this by the opposite of Euclid's Lemma or what?

Can anyone explain this to me please?

Bill Dubuque
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Brain
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  • By the linked divisibility product rule if $p$ divides both $a_0$ and $b_0$ then $ p^2\mid a_0 b_0,,$ contra hypothesis, hence at least one of $a_0$ or $b_0$ is not divisible by $p.\ \ $ – Bill Dubuque Jan 25 '24 at 01:58
  • The theroem hypotheses states that $p^2\not \mid a_0$. As $a_0=b_0c_0$ and $p^2\not \mid a_0=b_0c_0$ we can not have $p$ dividing BOTH $b_0$ and $c_0$. (That would mean $p^2|a_0$.) So $p$ must not divide one $c_0$ or $b_0$. It could divide $b_0$ but then it wouldn't divide $c_0$. Or it could divide $c_0$ but then it wouldn't divide $b_0$. Or $p$ could divide neither. But since at least one can't be divide and as these are just labels. WOLOG we call the one that is not divisible by $p$, we call it $b_0$. – fleablood Jan 25 '24 at 01:59

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Note the preceding discussion, that $p^2$ does not divide $a_0 = b_0 c_0$. So it cannot be the case that $p$ divides both $b_0$ and $c_0$.

But the roles of $g(x)$ and $h(x)$ are interchangeable, and hence so are $b_0$ and $c_0$. Thus without loss of generality one may assume $p$ does not divide $b_0$.

hardmath
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    @Brain In other words, since we know that $p$ doesn't divide one of the constant terms, we simply choose to label that constant term (the one that $p$ doesn't divide) as $b_0$. We can do that because, as stated in the answer above, the roles of $g(x)$ and $h(x)$ are completely interchangeable. – Robert Shore Jan 25 '24 at 01:27
  • but in order to divide $a_0,$ p needs to divide only one of $b_0$ or $c_0$ not both, right? I do not understand the logic from the beginning @RobertShore – Brain Jan 25 '24 at 01:52
  • Please strive not to post more (dupe) answers to dupes of FAQs. This is enforced site policy, see here. It's best for site health to delete this answer (which also minimizes community time wasted on dupe processing.) – Bill Dubuque Jan 25 '24 at 02:01
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    We are told from the very beginning that $p^2\not \mid a_0=b_0c_0$. That means we can't have $p$ dividing both. So at most $p$ can (but doesn't have to) divide one, but it absolutely can't divide both. So at most it divides one (or none). And the one that isn't divisible by $p$ we call $b_0$. – fleablood Jan 25 '24 at 02:03
  • @fleablood ok I see – Brain Jan 25 '24 at 02:09
  • @fleablood: Note that $p | a_0$ is part of the setup for Eisenstein's criterion, so indeed $p$ divides one of the two constant terms $b_0,c_0$ (just not both of them). – hardmath Jan 25 '24 at 02:53
  • Oh, oops. I wasn't really reading the details. – fleablood Jan 25 '24 at 15:55