Relations between irreducibility on $\mathbb{Q}[x]$, and on $\mathbb{Q}_p[x]$ ($p$-adic numbers) I'm reading "$p-$adic numbers: An introduction" by Fernando Q.Gouvêa, and I'm currently on page 79 of the book.

Problem 121.
Show that the equation $(X^2 - 2)(X^2 - 17)(X^2 - 34) = 0$ has a root in $\mathbb{Q}_p$, for all $p \le \infty$, but has no root in $\mathbb{Q}$.

Ok, I'm fine with this question. We just need to consider 3 cases:

*

*$p = 2$, then $X^2 - 17 = 0$ has a root, because $17 \equiv 1 (\mbox{ mod }8)$

*$p = 17$, then $X^2 - 2 = 0$ has a root, because $6^2 \equiv 2 (\mbox{ mod }7)$, so it's an application of Helsel's lemma.

*$p \notin \{ 2; 17 \}$, then if 2, and 17 are both not perfect square in $\mathbb{Z}/p\mathbb{Z}$, then 34 must be. So by Hensel's lemma, one of the 3 factors must have a root.


I'm typing problem 121 in because I think the following problem, which I'm stuck is kind of related to it.

Problem 123.
Decide whether it's true that a polynomial in one variable with coefficient in $\mathbb{Z}$ is irreducible in $\mathbb{Q}[X]$ iff it's irreducible in all $\mathbb{Q}_p[X]$, for all $p \le \infty$.

I'm fine with this problem too, it's false, the $\Rightarrow$ is incorrect, since $X^2 - 2$ is irreducible in $\mathbb{Q}[X]$, but not in $\mathbb{Q}_{17}[X]$ as shown above.

Problem 123 (Modified).
Decide whether it's true that a polynomial in one variable with coefficient in $\mathbb{Z}$ is irreducible in $\mathbb{Q}[X]$ iff it's irreducible for some $\mathbb{Q}_p[X]$.

I think this is false too, so I have tried to find a counter example (i.e some polynomial that is irreducible in $\mathbb{Q}[X]$, and then point out it's reducible for all $\mathbb{Q}_p[X]$), but without any luck.
Am I on the right track? Should it be true?
Thanks guys so much,
And have a good day,
 A: For Problem123(modified) I would produce a counterexample as follows.
Let $f(x)\in\mathbb{Z}[x]$ be the minimal polynomial of 56th roots of unity, 
$$f(x)=\phi_{56}(x)=x^{24}-x^{20}+x^{16}-x^{12}+x^8-x^4+1=\phi_7(-x^4).$$ 
The idea is to show that for all primes $p$ the reduction of the polynomial $f(x)$ in $\mathbb{F}_p[x]$ has at least two distinct irreducible factors. Then we can apply Hensel to prove that it factors in $\mathbb{Q}_p[x].$
I picked a cyclotomic polynomial, because then we can use properties of finite fields (=the residue class fields of unramified extensions of the $p$-adic fields). If $q=p^n$ is a power of a prime, the multiplicative group of the finite field $\mathbb{F}_q$ is cyclic of order $q-1$. Therefore this field contains a primitive root of unity of order $d$, iff $d\mid p-1$. We need this.
In order to apply Hensel, we study the factors of $f(x)$ in polynomial rings $\mathbb{F}_p[x]$, $p$ any prime.
A key consequence of the preceding paragraph is that, if $56\mid p^m-1$ for some integer $m$, then the polynomial $f(x)$ splits into factors of degree $m$. This is because any of its roots in an extension field of $\mathbb{F}_p$, say $\alpha$, then generates a degree $m$ extension, namely $K=\mathbb{F}_{p^m}=\mathbb{F}_p[\alpha]$. Therefore the minimal polynomial of $\alpha$ over $\mathbb{F}_p$ must be of degree $m$. Furthermore, $K^*$ is cyclic and thus then contains all the 56 zeros of $x^{56}-1$. The 24 primitive ones have degree $m$ minimal polynomials, the others may (or may not) have lower degree minimal polynomials.
The details vary a bit depending on $p$. The prime factors of $56$, namely $p=2$ and $p=7$ are special, because then $f(x)$ has repeated factors.
If $p=2$, then in the ring $\mathbb{F}_2[x]$ we have
$$
f(x)=\phi_7(x^4)=\phi_7(x)^4.
$$
This is no good for Hensel, but I picked $7$ as a factor for a reason. Here $7\mid 2^3-1$, so $\phi_7(x)$ factors modulo $2$ to a product of cubic factors. It is easy to check that we have
$$
\phi_7(x)=x^6+x^5+x^4+x^3+x^2+x+1\equiv(x^3+x+1)(x^3+x^2+1)\pmod2.
$$
Therefore we also have
$$
f(x)\equiv(x^3+x+1)^4(x^3+x^2+1)^4\pmod2,
$$
and Hensel's lemma tells us that we can find factors congruent to $(x^3+x+1)^4$ and $(x^3+x^2+1)^4$ modulo two in $\mathbb{Q}_2[x]$.
Next we deal with the case $p=7$. Modulo $7$ we have $\phi_7(x)\equiv(x-1)^6$, so
$$
f(x)=\phi_7(-x^4)\equiv(-x^4-1)^6=(x^4+1)^6\pmod7.
$$
But $x^4+1\equiv(x^2+3x+1)(x^2-3x+1)\pmod7$ (see the general $p$ for a reason as to why I knew in advance that $x^4+1$ factors modulo $7$). Therefore
$$
f(x)\equiv(x^2+3x+1)^6(x^2-3x+1)^6\pmod7,
$$
and again Hensel's Lemma bites.
Then the general case $p\neq2,7$.
If $p>2$, then $p^2\equiv1\pmod 8$, so the finite field $\mathbb{F}_{p^2}$ contains all the primitive eighth roots of unity. If furthermore $p\neq7$, then also $p^6\equiv1\pmod7$. The Chinese Remainder Theorem then says that $p^6\equiv1\pmod{56}$. Therefore the field  $\mathbb{F}_{p^6}$ contains primitive roots of unity of order $56$. Hence modulo $p$ our polynomial $f(x)$ has four factors, each of degree six. This time ($p\nmid 56$) the polynomial $f(x)$ has no repeated factors modulo $p$, so all these four irreducible sextic factors are distinct, and we have many ways of applying Hensel's Lemma.
