Prove that if $f$ and $m$ are polynomials over $K$, and $m$ is irreducible, then hcf$(f,m)$ is either $1$ or $m$.

Question

Proof that if $f$ and $m$ are polynomials over the field $K$, and $m$ is irreducible, then hcf$(f,m)$ is either $1$ or $m$.

I would really appreciate any help.

Answer 1

Hint:

If hcf$(f,m)=1$ then there is nothing to prove. Now suppose hcf$(f,m)\neq 1$ and then show $m|f $.

Let $m=\alpha_n x^n+\alpha_{n-1}x^{n-1}+ \cdot \cdot +\alpha_1x+C $. Since $m$ is irreducible(it has monic divisors $1$ and ${\alpha_n}^{-1}m$) and since hcf$(f,m)\neq 1$ this implies hcf$(f,m)= {\alpha_n}^{-1} m $. Now since hcf$(f,m)= {\alpha_n}^{-1} m$ we have $({\alpha_n}^{-1} m) |f$ and thus $m|\alpha_nf$. Now you should be able to conclude.

Answer 2

First, I use the notation $\gcd$ instead of $\text{hcf}$.

Second, we observe that the conclusion presented in the problem as stated, that $\gcd(f, m) = 1$ or $\gcd(f, m) = m$--is not entirely correct, since $\gcd$ is in fact only defined up to a unit factor, as is explained in this wikipedia entry. Therefore we might consider a modified version of the original statement, which reads (my interpretation):

Proposition: Let $f(x), m(x) \in K[x]$ for some field $K$; then either $\gcd(f(x), m(x))$ is a unit in $K[x]$ or $\gcd(f(x), m(x))$ is an associate of $m(x)$, that is, $\gcd(f(x), m(x)) = um(x)$ for some unit $u \in K[x]$.

Proof of Proposition: Let

$d(x) = \gcd(f(x), m(x)); \tag 1$

then we have

$d(x) \mid f(x), \; d(x) \mid m(x); \tag 2$

since $d(x) \mid m(x)$, we have $q(x) \in K[x]$ with

$m(x) = q(x) d(x); \tag 3$

since $m(x)$ is irreducible in $K[x]$, it cannot be the product of two non-units of $K[x]$; if $d(x)$ is a unit, we are done; if $d(x)$ is not a unit, $q(x)$ must be a unit $u \in K[x]$, whence

$m(x) = ud(x), \tag 4$

i.e. $m(x)$ and $d(x)$ are associates in $K[x]$. End: Proof of Proposition.

In light of the way this question was stated, I feel I should point out that $d(x)$ a unit means that

$d(x) \mid 1, \tag 5$

whence

$v d(x) = 1 \tag 6$

for some unit $v$ in $K[x]$; thus we have

$d(x) = v^{-1}1, \tag 7$

where $v^{-1}$ is a unit since $vv^{-1} = v^{-1}v = 1$.

If $d(x)$ is not a unit, i.e., if (4) binds, then we have

$d(x) = u^{-1}m(x) \tag 8$

with $u^{-1}$ a unit as is $v^{-1}$ above. We see from these considerations that $d(x)$ is in any event within a unit multiple of either $1$ or $m(x)$.

Of course, if we stipulate that $d(x) = \gcd(f(x), m(x))$ is monic, then by (6)-(7) we in fact do have $d(x) = 1$; and if we postulate that $m(x)$ is monic as well, then we also have $d(x) = m(x)$ as is clear from (8).

We close by examining the nature of the units in $K[x]$. If $k(x) \in K[x]$ is such a unit, we see that $\deg k(x) = 0$, since

$uk(x) = 1 \tag 9$

for some unit $u$. It follows that the units of $K[x]$ are precisely the non-zero elements of $K$ considered as a subfield of $K[x]$ in the usual manner. So $d(x)$ and $1$ or $d(x)$ and $m(x)$ are simply related via multiplication by an invertible field element $k \in K$.