Recently my lecture introduced a criterion for the reducibility polynomials:

Let $R$ be an UFD, $p \in R$ prime element (notice that in an UFD $p$ prime $\Leftrightarrow p$ irreducible), $f \in R[X],\,f \neq 0, p \nmid a_n$ ($f$'s leading coefficient).

$$ \varphi: R[X] \rightarrow R/(p)[X]\,\,\,\,\sum_{i=1}^{n} a_iX^i \mapsto \sum_{i=1}^{n} \bar{a_i}X^i $$

  • $f$ is irreducible in Quot($R$)[X] if $\varphi(f)$ is irreducible in $R/(p)[X]$
  • $f$ is irreducible in $R[X]$ if $\varphi(f)$ is irreducible in $R/(p)[X]$ and $f$ is primitive

Now I'm wondering why more requirements are necessary to show that $f$ is irreducible in $R[X]$. For me it would be more intuitive, if more requirements would be necessary to show that f is irreducible in Quot($R$)[X].

Isn't it true, that every reducible polynomial in $R[X]$ is also reducible in Quot($R$)$[X]$ and also that every irreducible polynomial in Quot($R$)[X] is also irreducible in $R[X]$? I think I misunderstand something crucial here.

  • $\begingroup$ See this question. $\endgroup$ Jan 3, 2018 at 15:04
  • $\begingroup$ @DietrichBurde What do you mean specifically? Should I look at the proof? Also why do people downvote my question without leaving feedback at all? $\endgroup$ Jan 3, 2018 at 17:20

1 Answer 1


Notice that $f = 2 \cdot T \in \mathbb{Z}[T]$ is reducible (because $2$,$T$ are no units),

but $f = 2 \cdot T \in \mathbb{Q}[T]$ is irreducible (because $T$ is irreducible in $\mathbb{Q}[T]$ and $2$ is a unit in $\mathbb{Q}[T]$).

So you can see that not every factorization of a polynomial in not-units in $R[T]$ is a factorization in not-units in $Quot(R)[T]$.


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