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My book states, without a proof, that

If $q(X)$ is reducible in $\mathbb Z[X]$, then it's reducible in $\mathbb Z_p[X]$ for every prime $p$.

The contrapositive of the above result is more useful: if one finds a prime $p$ such that $q(X)$ is irreducible in $\mathbb Z_p[X]$, then it's irreducible in $\mathbb Z[X]$.

I was wondering how one might prove the result, but then read this question: Irreducible polynomial which is reducible modulo every prime, which seems to be in contrast with the above statement. How is that possible?

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  • $\begingroup$ I would quibble with this statement in the case that $q(X)$ is not monic. It's a matter of semantics whether one considers $5X^2+6X+1$ to be reducible in $\mathbb Z_5[X]$. $\endgroup$ – Erick Wong Jun 17 '16 at 19:32
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The following implication is true (as you mention): $$q(X) \in \Bbb Z[X] \text{ monic reducible } \implies \bar q(X) \in \Bbb Z/p\Bbb Z[X] \text{ reducible for all prime } p$$ The contrapositive of this implication is: $$\bar q(X) \in \Bbb Z/p\Bbb Z[X] \text{ irreducible for some prime } p \implies q(X) \in \Bbb Z[X] \text{ irreducible (or non monic) }$$

However, the implication $$q(X) \in \Bbb Z[X] \text{ irreducible monic } \implies \bar q(X) \in \Bbb Z/p\Bbb Z[X] \text{ irreducible for some prime } p$$

is the converse of the first implication, which doesn't hold in general (as you provide a link to give a counter-example).


The first implication can be proven as follows: if you assume $q= fg$ with $f,g \in \Bbb Z[X] \setminus \{±1\}$, then $\bar q = \bar f \bar g \in \Bbb Z/p\Bbb Z[X]$. If you suppose that the dominant coefficient of $q$ is not a multiple of $p$ (e.g. $q$ monic), then both $\bar f$ and $\bar g$ have positive degree, hence $\bar q $ is reducible.


More generally, if $A$ is a UFD, $a \in A$ is an element, $f \in A[X]$ is a polynomial with content $1$ and with dominant coefficient invertible in $A/(a)$, then the irreducibility of $\bar f$ in $A/(a)[X]$ gives the irreducibility of $f$ in $A[X]$. [Berhuy, theorem IV.2.3].

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  • $\begingroup$ Ah, I see where I went wrong, thanks for pointing out the difference between the contrapositive and the converse. I still don't understand why $p$ has to be prime though. If $a,b,c \in \mathbb Z$ and $a = bc$, then it's true that $\bar a = \bar b \bar c$ in $\mathbb Z_n$ even if $n$ is not prime. What am I missing? $\endgroup$ – rubik Jun 18 '16 at 7:44
  • $\begingroup$ @rubik : you are right. However, we look at $\Bbb Z/p\Bbb Z$ when $p$ is prime because it is a field, while $\Bbb Z/n\Bbb Z$ is not a field (it's not even an integral domain) if $n$ is composite. $\endgroup$ – Watson Jun 18 '16 at 9:48
  • $\begingroup$ Ok, that's clear, thanks. But the important thing here is that the coefficients are in a PID (like $\mathbb Z$), not necessarily in a field, right? $\endgroup$ – rubik Jun 18 '16 at 9:56
  • $\begingroup$ Yes, you can even generalize to UFD's. A PID is always a UFD. I will correct the end of my answer, since we can pick an element $p$ in our UFD $A$ even if it is not irreducible in $A$. So in our case, we can consider $\Bbb Z/n\Bbb Z$ even if $n$ is not prime (in $A=\Bbb Z$), if you wish. $\endgroup$ – Watson Jun 18 '16 at 10:02
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The idea of the proof

Suppose $q(X)$ reducible over $\mathbb Z[X]$. Then, $$q(X)=R(X)P(X)$$ where $\deg q,\deg R>0$. In particular, in $\mathbb Z_p[X]$, $$\bar q(X)=\bar R(X)\bar P(X).$$

Now, you have to manage with primitive polynomials, since in $\mathbb Z_p[X]$, some polynomial can be null (that is not possible with primitive roots) (for example $2X^2+2$ in $\mathbb Z_2[X]$. Be careful, since $2$ is not a unit in $\mathbb Z$, you have that $2X^2+2$ is a reducible over $\mathbb Z$).

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