Suppose $f(x)$ is a monic polynomial in $R[x]$ ($R$ an integral domain) of degree $n$, and all coefficients (except the leading one) are elements of a prime ideal $P$. If we suppose that $f(x) = a(x)b(x)$, then if bar denotes reduction of coefficients modulo $P$ we get $x^n = \overline{a(x)b(x)}$ in $(R/P)[x]$. Why does this imply that both $a(x)$ and $b(x)$ have constant term in $P$? I know that $R/P$ is an integral domain, and so it has no zero divisors, but why can it not be the case that one of the constant terms, say the constant term of $a(x)$, is not in $P$?
Coefficients of product of polynomials divisible by $p$ implies constant terms divisible by $p$?
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abstract-algebra
polynomials
ring-theory
1 Answers
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You'll need to assume neither $a$ nor $b$ is constant, hence they both have degree strictly between $0$ and $n$. Furthermore, their leading coefficients are invertible, hence not in $P$.
The ring $S = R/P$ is an integral domain. Over $S$, the polynomial $x^n$ is factored as $\bar{a}\bar{b}$. But any such factorization can only be as the product of two monomials. (Consider the product of the two terms of highest degree, and the product of the two terms of lowest degree.)
Since $\bar{a}$ and $\bar{b}$ are monomials of degree $>0$, their constant terms are zero.