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I am in abstract algebra and am currently reading on a theorem which describes equivalent conditions relating to polynomial irreducibly. Here is the theorem:

Let $\mathbb{F}$ be a field and $p(x)$ a non-constant polynomial in $\mathbb{F}[x]$. Then the following conditions are equivalent:

  1. $p(x)$ is irreducible
  2. if $b(x)$ and $c(x)$ are any polynomials such that $p(x)|b(x)c(x)$, then $p(x)|b(x)$ or $p(x)|c(x)$
  3. if $r(x)$ and $s(x)$ are any polynomials such that $p(x) = r(x)s(x)$, then $r(x)$ or $s(x)$ is a nonzero constant polynomial.

When the textbook proves condition $3$ by using condition $2$, they say the following:

"If p(x) = r(x)s(x), then p(x)|r(x) or p(x)|s(x), by 2." It goes on to show that this will lead to either r(x) or s(x) being a unit, and thus a constant polynomial. What I'm not getting, though, is where it says that if p(x) = r(x)s(x), then p(x)|r(x) or p(x)|s(x). Condition 2 hinges on p(x)|r(x)s(x), meaning r(x)s(x) = t(x)p(x) for some t(x) in F[x]. Wouldn't p(x)=r(x)s(x) mean that r(x)|p(x) and s(x)|p(x), rather than the other way around?

I had the idea that perhaps p(x)|r(x) or p(x)|s(x) when t(x)=1 in r(x)s(x)=p(x)t(x), but I'm not sure if this is the correct interpretation. Any bit of insight would be appreciated.

vitamin d
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floridaguy
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1 Answers1

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Your thinking is right. We can see that if $1_F$ is the unit of $F$, then we have $1_Fp(x)=p(x)=r(x)s(x)$ so, for $1_F$ is a constant polynomial (that's the crucial part), we do have that there exists a polynomial $g(x)\in F[x]$ such that $r(x)s(x)=p(x)g(x)$. So that means, by definition, that $p(x)|r(x)s(x)$.

Just_a_student
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