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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.

Bill Dubuque
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Junaid Aftab
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  • The appendix doesn't commit to a field. I think it is implicit the fields is either $\mathbb{R}$ or $\mathbb{C}$, as is the case with most linear algebra books. – Junaid Aftab Feb 13 '17 at 17:35

2 Answers2

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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 $\phi_2 = \phi_1 \cdot 1$.

Reese Johnston
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  • So as of yet, my proof is incomplete; but the part of it that it stated is right? – Junaid Aftab Feb 13 '17 at 17:34
  • @JunaidAftab Well, almost. The statement "$\phi_1(x)$ cannot be expressed as the product of two polynomials" is false; it can't be expressed as the product of two nonconstant polynomials. So your proof so far just shows that either $h(x)$ is a constant or both $h'(x)$ and $h''(x)$ are. – Reese Johnston Feb 13 '17 at 19:03
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Hint $ $ If $\,p,q\,$ are irreducible then nonunit $\,d\mid p,q\,\Rightarrow\, up = d = vq\ $ for units (invertibles) $u,v,\,$ so $\, q = v^{-1}u p,\,$ so comparing lead coef's using $\,p,q\,$ monic yields $\,v^{-1}u = 1,\,$ hence $\,q = p$.

Remark $ $ The sketched proof (implicitly) uses the fact that only constants can be units, which need not hold true in rings that are not integral domains, e.g. see here. Generally factorization theory is more complicated in non-domains, e.g. $\rm\:x = (3+2x)(2-3x)\in \Bbb Z_6[x].\:$ Basic notions such as associate and irreducible bifurcate into a few inequivalent notions, e.g. see here.

Bill Dubuque
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