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Artin states "Let $R$ be a ring, $f$ be a monic polynomial and let $g$ be any polynomial, both with coefficients in $R$. There are uniquely determined polynomials $q$, $r$ in $R[x]$ such that $g = fq + r$ with degree $r <$ degree $f$ or $r = 0$."

He does not prove this theorem. The part I am having trouble with is uniqueness.

For the proof in a field, you assume there exists $q_1$, $r_1$ and $q_2$, $r_2$ and show that $$(q_1 - q_2)f = r_1 - r_2$$ In a field this proves that $q_1 = q_2$ and $r_1 = r_2$ due to degree LHS > degree RHS unless both zero. However, in an arbitrary ring I am not certain due to zero divisors.

Thanks for the help.

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  • $\begingroup$ Are your rings commutative? $\endgroup$ – Angina Seng Aug 17 '18 at 17:57
  • $\begingroup$ Yes they are (Artin assumes all rings are commutative) $\endgroup$ – user56628 Aug 17 '18 at 17:57
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I'll assume $f$ is monic. You have $(q_1-q_2)f=r_1-r_2$. The leading term of $f$ is $x^d$ for some $d$. If $q_1-q_2$ is not the zero polynomial, then its leading term is $ax^t$ for some $t$ and the leading term of $(q_1-q_2)f=r_1-r_2$ is $a x^{d+t}$, so this polynomial has degree $d+t$. As $r_1-r_2$ has degree $<d$, we'll get a contradiction.

Notice how $f$ being monic was crucial here. It's also OK if the leading coefficient of $f$ is not a zero-divisor.

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You can use a similar argument as the one for fields. Indeed, if you assume $q_1\neq q_2$, then $\deg (q_1-q_2)f>\deg f$. That's because if you let $aX^m$ be the leading monomial of $q_1-q_2$ and $X^n$ the leading monomial of $f$ (which is monic), then the leading monomial of $(q_1-q_2)f$ is $aX^{m+n}$.

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  • $\begingroup$ This answer is equally good but the above responded 10 seconds faster $\endgroup$ – user56628 Aug 17 '18 at 21:35
  • $\begingroup$ @user56628 It's OK :) $\endgroup$ – Scientifica Aug 17 '18 at 21:41

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