Unit-monic $f\mid g$ in $R[x]$ if $f\mid g$ in $S[x]$ for rings $S\supset R$

Question

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.

Answer 1

Hint $ $ (Euclidean) division with remainder can divide by any monic polynomial (or unit lead coef). The quotient and remainder are unique (same proof as when $R$ is a field, cf. comment below).

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 proved more generally via 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 $\,q\in R[x]$

Answer 2

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.