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I'm reading Artin's Algebra. In 11.3.24, it is stated

Let $f$ be a monic integer polynomial, and let $g$ be another integer polynomial. If $f$ divides $g$ in $\Bbb Q[x]$, then $f$ divides $g$ in $\Bbb Z[x]$.

The proof being

Since $f$ is monic, we can do division with remainder in $\Bbb Z[x]$: $g = f q + r$. This equation remains true in the ring $\Bbb Q[x]$, and division with remainder in $\Bbb Q[x]$ gives the same result. In $\Bbb Q[x]$, $f $ divides $g$. Therefore $r = 0$, and $f$ divides $g$ in $\Bbb Z[x]$.

I can't understand why division in $\Bbb Q[x]$ gives the same result.

user26857
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DynamoBlaze
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1 Answers1

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As I was writing the question I realized why it must be true.

It hinges on $f$ being monic. So while doing long division, the leading coefficient of $q$ must be the same as the leading coefficient of $g$ (because the leading coefficient of $f$ is $1$), and from then the long division gives the exact same result at every step, thus giving identical $q$ and $r$ in both rings.

DynamoBlaze
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  • Please strive not to post more (dupe) answers to dupes of FAQs, cf. recent site policy announcement here. It is best to delete this Q&A since we already have too many instances of it here. – Bill Dubuque Sep 28 '23 at 16:12
  • Will anything change if we ignore the assumption that f is monic? I'm trying to prove the statement without this... – Stavros Chrysafis Feb 09 '25 at 11:48
  • @StavrosChrysafis Then you might not be able to do division with remainder in $\Bbb Z[x]$. How do you divide $x^2+1$ by $2x+1$? – DynamoBlaze Feb 10 '25 at 07:05