quotient of integral polynomials not being integral - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-18T13:02:25Z http://mathoverflow.net/feeds/question/76212 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/76212/quotient-of-integral-polynomials-not-being-integral quotient of integral polynomials not being integral Hugo Chapdelaine 2011-09-23T15:54:33Z 2011-09-23T20:06:09Z <p>So let $R$ be an integral domain and $K$ its fraction field. Let $f,g,h\in K[x]$ be 3 monic polynomials such that $f=gh$. So I would like to have a simple example of a ring $R$ for which one has that $f,g\in R[x]$ but $h\notin R[x]$. </p> <p>P.S. May be working with an non-maximal order of a Dedekind ring is good enough. Nevertheless I could not come up with such an example </p> http://mathoverflow.net/questions/76212/quotient-of-integral-polynomials-not-being-integral/76215#76215 Answer by Angelo for quotient of integral polynomials not being integral Angelo 2011-09-23T16:03:06Z 2011-09-23T16:03:06Z <p>I don't think this can happen. The division algorithm works over arbitrary rings, with unique quotient and remainder, as long as you are dividing by a monic polynomial. This implies that if $g$ is monic and divides $f$ in $K[x]$, it also divides it in $R[x]$, and $h$ must be the quotient in both rings.</p> http://mathoverflow.net/questions/76212/quotient-of-integral-polynomials-not-being-integral/76219#76219 Answer by Pham Hung Quy for quotient of integral polynomials not being integral Pham Hung Quy 2011-09-23T16:43:04Z 2011-09-23T20:06:09Z <p>We show that $h$ must be in $R[x]$. Suppose that we have polynomials</p> <p>$$g = x^n + a_1x^{n-1} + ... + a_n \in R[x]$$ $$h = x^m +b_1x^{m-1} + ... + b_m \in K[x]$$ such that $f = gh \in R[x]$ but $h \notin R[x]$. Let <code>$r:= \min \{i \mid b_i \notin R\}$</code>. Since $f \in R[x]$ we have $b_r + a_1b_{r-1} + ... + a_{r-1}b_1 + a_r \in R$, so $b_r \in R$. It is a contradiction. </p>