Factorizing a polynomial $f$ in $A[x]$ (with $A$ commutative), where $f$ has a zero in its field of fractions

Question

Let $A$ be a commutative ring and $S$ a multiplicative subset of $A$ generated by $s\in A$ which is not a zero-divisor. Consider the polynomial ring $A[x]$.

Given a polynomial $f\in A[x]$, suppose that $f(1/s)=0$ for $s\in S$. Does it follow by some application of the first isomorphism theorem that $f(x)=(sx-1)g(x)$, with $g(x)\in A[x]$?

On the one hand I think this can be proven by constructing an isomorphism from $A[x]$ modulo the relation $x=1/s$ (really the replacement of $x$ with $1/s$). This is evidently an isomorphism from $A[x]/(x=1/s)$ into $S^{-1}A$. It is tempting to conclude (though I'm not certain that this should work), that there must be some surjective homomorphism $\phi: A[X]\to S^{-1}A$ such that $\ker(\phi)=(sx-1)$. [Since generally $(A/B\cong A/C) \nRightarrow (B\cong C)$, it is not sufficient to conclude that, say, the kernel of $\psi: f\to f(1/s)$ is $(sx-1)$.]

I am especially interested in answers as to whether, say, $\psi$ has kernel $(sx-1)$. But for anyone who can follow my own attempt at a proof, I'm also interested in whether my argument breaks down or can be made to work rigorously.

Answer 1

By the the universal property of the polynomial ring there exists an $A$-algebra morphism $\Phi : A[x]\to S^{-1}A$ sending $x$ to $1/s$ .
Since $\Phi$ vanishes on the ideal $(sx-1)$, it induces a morphism $\phi: A[x]/(sx-1)\to S^{-1}A $ which is clearly surjective.
Let' s show $\phi$ is injective by proving that the kernel of $\Phi$ is $(sx-1)$ .

If $p(x)=a_0+a_1x+...+a_nx^n$ satisfies $\Phi (p)=0$ , we have $a_0+a_1\frac{1}{s}+...+a_n (\frac{1}{s})^n=0$ or $\frac{1}{s^n} .( a_0s^n+...+a_n)=0 $
Since $A$ is a domain, this is equivalent to $$ a_0s^n+...+a_n=0 \quad (*) $$

To find a polynomial $q(x)=b_0+b_1x+...+b_{n-1}x^{n-1}$ such that $p(x)=(sx-1).q(x)$ , it suffices to solve the system :

$$ b_i=-a_0s^{i-1}-a_1 s^{i-2}- ...- a_i=0 \quad (0\leq i\leq n-1) \quad (**) $$

$$ b_{n-1}s=a_n \quad (***)$$

The solution of $(*)$ is $b_i=-a_0s^i -a_1 s^{i-1}-... -a_i$ and it satisfies $(***)$ thanks to $(*)$.

Conclusion
We have proved that $\phi: A[x]/(sx-1)\to S^{-1}A $ is an isomorphism of rings.
(Actually this holds also if $A$ has zero divisors, by a small modification of the above reasoning )

Answer 2

Hint $\rm\, \phi\,$ kills $\rm\,f\,$ and $\rm\,sx\!-\!1\,$ so it kills the polynomial obtained by eliminating their constant terms, viz. $\rm\ f + (sx\!-\!1)\, f_0 =: x\, g \in ker\ \phi\ \Rightarrow\ g\in ker\ \phi\,,\,$ since $\rm\,s\,$ is not a zero-divisor. $\,$ Hence we deduce by induction on degree that $\rm\, sx\!-\!1\mid g\,$ thus $\rm\, sx\!-\!1\mid f\, =\, x\, g-(sx\!-\!1)\, f_0.\ \ $ QED

Remark $\ $ This method avoids the messy coefficient recursions in other approaches. It is worth stressing that this can be deduced more conceptually from more general results. First, using the same obvious inductive proof as for the high-school polynomial division algorithm, we deduce that we can divide with remainder by any monic polynomial over any ring. In particular, since $\rm\,f(1/s) = 0\,$ we deduce by the Remainder Theorem that $\rm\ x\!-\!1/s\,\mid f(x)\ $ in $\rm\, A[1/s][x]\, $ so $\rm\,f(x) = (x\!-\!1/s)\ g(x)/s^j\,$ for some $\rm\,g\in A[x].\,$ Thus $\rm\,s^{\,j+1}\ f(x) = (s\,x\!-\!1)\ g(x)\,$ so, by Euclid's lemma, $\rm\ sx\!-\!1\mid f(x)\,$ in $\rm\,A[x]\,$ since $\rm\,s\,$ is coprime to $\rm\,sx\!-\!1\,$ by $\rm\,(s,sx\!-\!1) = (s,1) = 1\, $ (explicitly, if $\rm\, sx\!-\!1\mid s\,f\ $ then it also divides $\rm\ x(sf) - (sx\!-\!1)f = f).$

More generally, monic division in $\rm A[1/s][x]$ yields a nonmonic division algorithm pulled back to $\rm\,A[x],\,$ namely if the divisor $\rm\,g\,$ has leading coefficient $\rm\,s\,$ then we obtain the following $\rm\,s\rm{-Division}$ Algorithm: $\rm\ s^{n} f = q\, g + r\ $ for some $\rm\,q,r\ \in A[x],\ n\in \mathbb N\ $ with $\rm\ deg\ r < deg\ g.$ This has an elementary direct inductive proof, but it is less conceptual than said pullback from $\rm\,A[1/s].$

Answer 3

The homomorphism $\phi: A[X]\rightarrow S^{-1}A,\; X\mapsto \frac{1}{s}$ is surjective, since the elements of $S^{-1}A$ have the form $\frac{a}{s^k} = \phi (aX^k)$, $k\in\mathbb{N}$, $a\in A$.

The kernel of $\phi$ is generated by $(sX-1)$: let $f$ be an element of the kernel; in $S^{-1}A$ it has the unit $\frac{1}{s}$ as a root. Now in general the following holds:

Let $f\in B[X]$ be a polynomial with coefficients in the commutative ring $B$ and such that $f(u)=0$, where $u\in B$ is a unit. Then $f=(X-u)g$ with $g\in B[X]$.

The proof follows easily by looking at the relation between the coefficients $a_k$ of $f$ and $b_k$ of $g$: $a_{k+1}=b_k-ub_{k+1}$.

Hence we get a factorization of the form $f=(X-\frac{1}{s})g$, $g\in S^{-1}A$. Looking again at the relation $a_{k+1}=b_k-ub_{k+1}$ and taking into account that $u=\frac{1}{s}$ it turns out that every $b_k$ has the form $sc_k$ for some $c_k\in A$, which proves the assertion.

Answer 4

Is the suggested idea of the universal property the following?

Since $s$ is a unit in $B$ (because $Xs\equiv 1$ in $B$), the homomorphism $\psi':A\to B$ such that $\psi'(s)=\frac 1s=X$ sends $S=\{s^n\}$ into the units of $B$ so factors through $\pi:A\to S^{-1}A,\;a\mapsto \frac a1$ and induces a unique homomorphism $\psi: S^{-1}A\to B$ with $\psi(\frac a{s^n})=aX^n$ which is surjective.

Then copying the argument above, the surjection $\phi':\;A[X]\to S^{-1}A, \;f(X)\mapsto (\frac 1s)$ factors through the ideal $(sX-1)\subset \ker \phi'$ so there exists a surjection $\phi:B\to S^{-1}A$ defined by $aX^n\mapsto \frac a{s^n}$ with $B=A[X]/(sX-1)$.

Finally, $\phi$ and $\psi$ are inverse each other.