0

Let $R_1 \subset R_2$ be two integral domains. If $f \in R_1[x]$ is a polynomial which factors as $gh$ with $g,h \in R_2[x]$, then the coefficients of $g$ and $h$ are integral over $R_1$.

I tried a bunch of stuff, but didn't really get anywhere without using the extra assumption of it being a UFD/PID/Dedekind domain (which is not provided)

Mohan
  • 17,980
katana_0
  • 1,862
  • This is probably a consequence of the Dedekind Prague lemma, but as it stands it's not stated correctly -- you need to assume something about $g$ and $h$, lest you could have $R_1 = \mathbb{Z}$, $R_2 = \mathbb{Q}$, $f = x$, $g = 1/2$ and $h = 2x$. – darij grinberg Jun 10 '20 at 11:17

1 Answers1

0

If $g$ and $h$ are not assumed to be monic (leading coefficient 1) this fails, as Darij's example shows.

When they are monic, the result holds even when $R_1$ and $R_2$ are not domains: let $R_3\supseteq R_2$ be a splitting ring for $g$ and $h$ (so that both split as products of monic linear factors in $R_3[x]$). The zeroes of $g$ and $h$ in $R_3$ are zeroes of $f$, so they are integral over $R_1$. As the coefficients of $g$ and $h$ are elementary symmetric polynomials in these roots, they too must be integral over $R_1$ (as elements of $R_3$, hence as elements of $R_2$).

See here for details on constructing such an $R_3$.

Matthé van der Lee
  • 764