1
$\begingroup$

Let $R$ be an unique factorization domain (UFD) and $K$ its quotient field, i.e. $Q(R)=K$. Further, let $P\in R[X]$ be a primitive polynomial, that is if $$P(X)=\sum_{i=0}^{n}a_iX^i,$$ then $\text{gcd}(a_1,...,a_n)=1$.

Under these assumptions, the following statement holds:

$P(X)$ is irreducible in $K[X]\Longleftrightarrow P(X)$ is irreducible in $R[X]$

My lecture notes prove this statement by proving that the polynomial is reducible in $K[X]$ iff it is reducible in $R[X]$.

I understand the proof of one direction of the implication, but the other one confuses me. I will quote that part of the proof:

Suppose $P(X)$ is reducible in $R[X]$. Since it is primitive it factors into two factors of lower degree than $P(X)$. Hence $P(X)$ is reducible in $K[X]$.

First of all, I don't understand why it is stated like this at all. Isn't this direction of the biconditional trivially true? Also, I don't understand why the author explictly states that $P$ is primitive. The way I see it, if $P$ is reducible, then by definition it can be factored into two non-units. Since the units of $R[X]$ are precisely those of $R$ (is this actually true?), it follows that the factors must both be at least linear, therefore $P(X)=S(X)T(X)$ for some $S,T\in R[X]$ with $\text{deg}(S),\text{deg}(T)\ge1$. Further, in general we have $\text{deg}(ST)=\text{deg}(S)+\text{deg}(T)$ for any polynomials over any domain, so naturally both $S$ and $T$ have lower degree than $P$. As you can see, I haven't used the notion of primitiveness anywhere, so am I missing something here or does the author simply uses some shortcut (which I'm clearly missing then) to aviod putting up with a similar proof?

$\endgroup$
2
$\begingroup$

We don't have that any non-unit in $R[X]$ has degree $\geq 1$, for example, in $\mathbb Z[X]$ $2$ is not a unit, thus $2x+2$ is reducible in $\mathbb Z [X]$, but not in $\mathbb Q[X]$. Of course this kind of behaviour cannot happen for primitive polynomials, that's we need this assumption.

$\endgroup$
  • $\begingroup$ I see, thanks. So it's not true that $R[X]^*=R^*$. Is there a condition on $R$ which does make this statement true? Because I seem to remember a case in which it did hold. EDIT: Found it. Apparently this is true whenever $R$ is a field. $\endgroup$ – Tyron Jul 5 '17 at 20:58
  • 1
    $\begingroup$ @Tyron it is indeed true that $R[X]^* = R^*$, if $R$ is a reduced commutative ring, so in particular for domains. In this case $\mathbb Z[X]^*= \mathbb Z^* = \{\pm 1\}$. The problem is that if $R$ is not a field, then $R^*$ is smaller than $R \setminus \{0\}$ which gives rise to non-unit polynomials of degree $0$. $\endgroup$ – MatheinBoulomenos Jul 5 '17 at 21:02
  • 1
    $\begingroup$ @Tyron - You write "the units of $R[X]$ are precisely those of $R$". This is ok. You add "it follows that the factors must both be at least linear". This is not ok. $\endgroup$ – Pierre-Yves Gaillard Jul 5 '17 at 21:09
  • 1
    $\begingroup$ @MatheiBoulomenos Ah! So if $R$ is not a field, then I can pick an $a\in ((R-\{0\})-R^*$), which is a polynomial of degree $0$ in $R[X]$, but not a unit in $R$, thus not a unit in $R[X]$. $\endgroup$ – Tyron Jul 5 '17 at 21:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.