$p$ is irreducible and $p\mid bc$ follow that $p\mid b$ or $p\mid c$

Question

Any explanation why $d$ must be a monic polynomial? ($p,b,c$ are not monic). Thanks.

Answer 1

In any commutative ring $R$, two elements $r_1, r_2 \in R$ may have more than one greatest common divisor.

For example:

• If $R = \mathbb Z$, $r_1 = 4$, $r_2 = 6$, then $2$ is a greatest common divisor of $r_1$ and $r_2$, but so is $-2$.
• If $R = \mathbb Q[x]$, $r_1 = (x-1)(x-2)$ and $r_2 = (x-2)^3(x-7)$, then $(x-2)$ is a greatest common divisor of $r_1$ and $r_2$, but so is $12345(x-2)$.

In general if $d$ is a greatest common divisor of $r_1, r_2$, then so is $du$, where $u$ is any unit in $R$.

In $F[x]$, the units are the non-zero elements of $F$. If $d$ is a greatest common divisor of $f, g \in F[x]$, then we can construct another greatest common divisor of $f, g \in F[x]$ that is a monic polynomial. Indeed, if $c \in F^\times$ is the leading coefficient of $d$, then $c$ is a unit in $F[x]$, so $d' := c^{-1}d$ is another greatest common divisor of $f$ and $g$ that also happens to be a monic polynomial

Presumably your textbook adopts the following convention: Whenever your textbook talks about "the" greatest common divisor of $f, g \in F[x]$, it is really referring the (unique) monic greatest common divisor $f, g \in F[x]$.

Whether $f$ and $g$ are themselves monic is irrelevant.