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.

Yes, that is precisely what it means for $f$ to divide $g$ in $R[x]$. — Servaes, Feb 27 '19 at 23:44
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? — stressed out, Feb 27 '19 at 23:48
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. — stressed out, Feb 27 '19 at 23:50
@stressedout do you mean divide $q(x)$ by something? Not sure what you mean. — IntegrateThis, Feb 27 '19 at 23:51
No. Divide $g(x)$ by $f(x)$. Show that $r(x)=0$ or you'll get a contradiction. ;-) I'm sure it works. I can write it as an answer if you want, but I think you can handle it on your own. — stressed out, Feb 27 '19 at 23:52

Bill Dubuque · Accepted Answer · 2021-04-14T16:38:31.147

6

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 $\,q\in R[x]$

edited Apr 14 '21 at 16:38

answered Feb 27 '19 at 23:52

Bill Dubuque

272,048

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? – IntegrateThis Feb 27 '19 at 23:57
@IntegrateThis Exactly. I added a remark and link on the gcd perspective. – Bill Dubuque Feb 27 '19 at 23:59
1

Are $R$ and $S$ assumed to be UFDs? – Servaes Feb 28 '19 at 00:05
1

@Servaes No, commutative rings – Bill Dubuque Feb 28 '19 at 00:08
@stressedout Yes, you can prove it directly but is more conceptual to use division uniqueness (which is a consequence of the dividend having unit lead coef - same as the classical proof over fields). – Bill Dubuque Feb 28 '19 at 00:13
1

@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.\ $ – Bill Dubuque Feb 28 '19 at 00:34

stressed out · Answer 2 · 2019-02-28T00:42:36.393

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.

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. — Bill Dubuque, Feb 28 '19 at 01:10
@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. :) — stressed out, Feb 28 '19 at 01:17

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

2 Answers2

Linked

Related