TITLE: Different version of Gauss's Lemma
QUESTION [6 upvotes]: 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.

REPLY [2 votes]: Hint $ $ Let $\rm\:h = f/g.\:$ $\rm\ f = gh = (x^k +\cdots\:)(h_nx^n + h')\in A[x]\ \Rightarrow h_n \in A\ \Rightarrow\ gh' \in A[x]\:$ hence $\rm\:h'\in A[x]\:$ by induction, so $\rm\: h\in A[x]$.
Or, equivalently, suppose $\rm\:g\not\in A[x].\,$ Then $\rm\bmod A\!:\,\ f=gh \equiv 0\,$ contra $\rm\,\ell(gh)\equiv \ell(g)\not\equiv 0$ where $\,\ell$ = lead coefficient.