Let $A$ be a domain with field of fractions $K$. Let $f, g \in A[X]$ with $g$ monic. Show that if $f/g \in K[X]$ then $f/g \in A[X]$.

So I tried the direct approach by just assuming $f/g$ has a coefficient $a/b$ and then multiplying out with $g$ with the purpose of getting that $f$ has an "irrational" coefficient; but I can't finisnh.


Theorem. Let $R$ be a commutative ring with unity, let $g(x)\in R[x]$ be a polynomial with leading coefficient a unit, and let $f(x)\in R[x]$. Then there exist polynomials $q(x),r(x)\in R[x]$ such that $f(x)=q(x)g(x) + r(x)$, and either $r=0$ or $\deg(r)\lt \deg(g)$.

Proof. If $f=0$, then there is nothing to do. Otherwise, we do induction on $\deg(f)$. Assume the result holds for any polynomial of degree smaller than $f(x)$. Let $f(x) = a_nx^n + \cdots + a_0$, and write $g(x) = b_mx^m+\cdots + b_0$. If $n\lt m$, then set $q(x)=0$, $r(x)=f(x)$, and we are done. Otherwise, note that since $b_m$ is a unit, then $a_nb_m^{-1}\in R$, so $a_nb_m^{-1}x^{n-m}g(x)\in R[x]$. Note that the leading term of this polynomial equals that of $f(x)$, so $f(x) - a_nb_m^{-1}x^{n-m}g(x)$ is either $0$, or has degree strictly smaller than $f(x)$. Either way, we can write $$f(x) - a_nb_m^{-1}x^{n-m}g(x) = q'(x)g(x) + r(x)$$ with $q'(x),r(x)\in R[x]$, and $r=0$ or $\deg(r)\lt \deg(g)$, by the induction hypothesis. Setting $q(x) = a_nb_m^{-1}x^{n-m}+q'(x)$ gives the desired conclusion. $\Box$

Because the leading coefficient of $g(x)$ is a unit in $A[x]$, you can perform long division and write $f(x) = q(x)g(x)+r(x)$ with $q(x)\in A[x]$, $r(x)\in A[x]$, and either $r(x)=0$ or $\deg(r)\lt\deg(g)$.

In $A[X]$ this expression may not be unique; however, since $K$ is a field, the division algorithm together with its uniqueness clause does hold; since you have found an expression for $f(x)$ as $f(x) = q(x)g(x)+r(x)$ with $q(x),r(x)\in A[x]\subseteq K[x]$ and $r(x)$ satisfies the degree conditions to be the remainder when dividing $f$ by $g$, then...


HINT $\ $ Let $\rm\:h = f/g\:.\:$ $\rm\ f = g\:h = (x^k +\cdots\:)\:(h_n\:x^n + h')\in A[x]\ \Rightarrow h_n \in A\ \Rightarrow\ g\:h' \in A[x]\:$ hence $\rm\:h'\in A[x]\:$ by induction, so $\rm\: h\in A[x]\:.\:$

Alternatively, equivalently, suppose $\rm\:g\not\in A[x]\:.$ Then $\rm\: g\:h = f \equiv 0\ \ (mod\ A)\:$ contra $\rm\:g\:h\:$ has nonzero leading coefficient $\rm(mod\ A)$, viz. the leading coefficient of $\rm\ g\ mod\ A\:.$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.