I'm trying to prove the statement:
Any two distinct irreducible monic polynomials are relatively prime (coprime).
Attempt: Let $\phi_1(x)$ and $\phi_2(x)$ be two distinct irreducible polynomials. Assume they are not relatively prime. Then there exists a polynomial, $h(x)$, such that
$\phi_1(x) = h(x)h'(x) \quad \text{and} \quad \phi_2(x) = h(x)h''(x) $,
for polynomials $h'(x)$ and $h''(x)$ of appropriate degree. But this implies that $\phi_1(x)$ and $\phi_2(x)$ can be expressed as a product of two polynomials. Hence, they can't be irreducible.
Comment: I'm not too sure if the proof is correct. I haven't seemed to have used the conditions that the polynomials are both distinct and monic. What am I doing wrong?
I don't have much background in abstract algebra. I came across this thereom in an appendix on polynomials in a book on liner algebra, so I am going through it to cover the unit on minimal polynomials.
Recall that constants count as polynomials - so, for example, $x^2 + 2x + 2$ can be written as $(x^2 + 2x + 2) \cdot 1$. Thus, $x^2 + 2x + 2$ and $2x^2 + 4x + 4$ aren't relatively prime, because $2x^2 + 4x + 4 = (x^2 + 2x + 2) \cdot 2$. That's where the monic condition comes in.
The distinct condition comes from the fact that if $\phi_1$ and $\phi_2$ are not distinct, then $\phi_2 = \phi_1 \cdot 1$.