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Let $R$ be a sub-ring of a ring $S$. Let $f,g$ be non-zero polynomials in $R[x]$ and assume that the leading coefficient of $f$ is a unit in $R$. If $f$ divides $g$ in $S[x]$, show that $f$ divides $g$ in $R[x]$.

Thoughts: If $f$ divides $g$ in $S[x]$, then $g(x) = f(x)q(x)$ for some $q(x)$ in $S[x]$. From here I'm not sure what the question is asking. Do I have to show that $q(x)$ is an element of $R[x]$ ? Insights appreciated.

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  • $\begingroup$ Yes, that is precisely what it means for $f$ to divide $g$ in $R[x]$. $\endgroup$ – Servaes Feb 27 at 23:44
  • $\begingroup$ @Servaes how do I know that $q(x)$ is an element of $R[x]$? $\endgroup$ – IntegrateThis Feb 27 at 23:45
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    $\begingroup$ That is the point of the question. Give it some thought. $\endgroup$ – Servaes Feb 27 at 23:46
  • $\begingroup$ Your title and your question are different. The title says that $S$ is a subring of $R$, but the body of your question says that $R$ is a subring of $S$. Which one is true? $\endgroup$ – stressed out Feb 27 at 23:48
  • $\begingroup$ I think I found the solution. Use Euclidean division. Since the leading coefficient of $f$ is invertible, you can do it. In fact, assume that $f$ is a monic polynomial for simplicity. $\endgroup$ – stressed out Feb 27 at 23:50
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Hint $ $ (Euclidean) division with remainder works because the lead coef is a unit, and the quotient and remainder are unique (same proof as when the coef ring is a field).

Thus by uniqueness, the remainder on division in the subring $R$ is the same as in $S$, i.e. $0$.

Remark $ $ Results like this can also be derived more generally from persistence of gcds, e.g. see here.

Or directly $ g = f q = (u x^i\! +\! f')(s x^j\! +\! q') = us\, x^{i+j}\!+\cdots \in\! R[x]\,$ so $\,us\! =\! r\in R$ so $\,s = r/u\in R$. Thus $\, fq' = g - sx^kf\in R[x]\, $ and $\,\deg q' < \deg q\,$ so by induction $\,q'\in R[x]\,$ so $\,q\in R[x]$

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  • $\begingroup$ Ok. I see. So there couldn't be a q(x), r(x) with deg(r(x))> 0 in R[x] such that g = q f + r because that would contradict the uniqueness of euclidean division? $\endgroup$ – IntegrateThis Feb 27 at 23:57
  • $\begingroup$ @IntegrateThis Exactly. I added a remark and link on the gcd perspective. $\endgroup$ – Bill Dubuque Feb 27 at 23:59
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    $\begingroup$ Are $R$ and $S$ assumed to be UFDs? $\endgroup$ – Servaes Feb 28 at 0:05
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    $\begingroup$ @Servaes No, commutative rings $\endgroup$ – Bill Dubuque Feb 28 at 0:08
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    $\begingroup$ @stressedout If $\,\deg r,\deg R < \deg f\,$ and $ q f + r = Q f + R\,$ then $\,(Q-q)f = r-R.\,$ If $Q\neq q$ then, since lead coef of $f$ is a unit, $\deg {\rm LHS} \ge \deg f > \rm \deg RHS \Rightarrow\!\Leftarrow$ Therefore $\,Q = q\,$ so $\,r-R = 0.\ $ $\endgroup$ – Bill Dubuque Feb 28 at 0:34
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Well, as I mentioned in the comments and Bill Dubuque has explained it well in his answer, you can use the Euclidean division algorithm. Note that for the Euclidean division algorithm to work, all that you need is to know that the leading coefficient of the divisor is a unit. To see this, try to divide a polynomial $g(x)$ by another polynomial $f(x)$ of lower degree and you'll see that you can always cancel the term of highest degree in $g(x)$ when the leading coefficient of $f(x)$ is a unit.

For a better insight and seeing what can go wrong when the leading coefficient is not a unit, try to divide $3x^2+1$ by $2x-1$ in $\mathbb{Z}$. You will immediately see that you can't get rid of $3x^2$ because $2 \not\mid 3$.

However, since in this problem your leading coefficient is a unit, you can assume that $f(x)$ is a monic polynomial. Since $1$ divides anything in the ring, no such problem can arise.

Addendum: without using the uniqueness of the divisor and the remainder polynomials, we can argue as follows:

Suppose that $g(x)=f(x)k(x)+r(x)$ in $R[x]$. Since $R[x] \subseteq S[x]$, the same equation holds in $S[x]$. On the other hand, you had assumed that $g(x)=f(x)q(x)$ in $S[x]$; so we get that $f(x)q(x)=f(x)k(x)+r(x)$. Hence, $f(x)\big( q(x) -k(x) \big) = r(x)$.

Since the leading coefficient of $f(x)$ is $1$ and $1$ is never a zero-divisor, unless $q(x)-k(x) = 0$, we have $\deg r(x) \geq \deg f(x)$ which is a contradiction. So, $q(x)-k(x)=0$ and therefore, $r(x)=0$.

Thanks to Bill Dubuque for point out that $f \mid r$ still implies $\deg r \geq \deg f$ because $f$ is monic.

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  • $\begingroup$ For a direct proof we can show by comparing lead coefs that the lead coef of the quotient $q$ is in $R$, and then inductively do the same for the rest of the quotient (rest after subtracting its lead term) - see the Remark in my answer. $\endgroup$ – Bill Dubuque Feb 28 at 1:10
  • $\begingroup$ @BillDubuque Yes, I saw your answer. Your answers are sometimes so spot on and yet comprehensive that voting them up just once is not enough. :) $\endgroup$ – stressed out Feb 28 at 1:17

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