Question on the irreducibility of $x^4 - 10x^2 +1$ I am currently doing some work in Galois theory, and the following situation has me perplexed.
The polynomial $~x^4 -10x^2 + 1 $ is irreducible over $\mathbb Q, $ which can be shown in a variety of ways.
In the text Galois Theory by Joseph Rotman, Exercise 66 states the following:
"If $\sigma : R \to S $ is a ring map, then $~\sigma^*: R[x] \to S[x]~$ defined by $~\sum r_i x^i \to \sum \sigma(r_i) x^ i ~$ is also a ring map.
Prove that if $R$ and $S$ are domains, and if $~\sigma^*\big( p(x)\big) \in S[x]~$ is irreducible and has the same degree as $~p(x),~$ then $~p(x) ~$ is irreducible over $~R.$"
Now, Exercise 67 from the same text states the following:
"Let $~\sigma:\mathbb Z \to \mathbb Z_p ~$ be the natural map.  Use the preceding exercise with a suitable choice of prime $~p~$ to show that $~f(x) = x^4 -10x^2 +1 ~$ is irreducible over $~\mathbb Z.$"
Yet, I have encountered multiple other sources, including Proving that $x^4 - 10x^2 + 1$ is not irreducible over $\mathbb{Z}_p$ for any prime $p$. $~~$ and Prove that the irreducible polynomial for $\sqrt2+\sqrt3$ over $\mathbb{Q}$ is reducible modulo $p$ for every prime $p$ $~~$ that say otherwise.
I am hoping that someone can help clarify what I am missing in this picture.
Thanks
(Edit: I believe I have a copy of the 1st edition of Rotman's book)
 A: I suppose that that's an error from Rotman's textbook. To be more precise, from the first edition of the textbook, since the statement of that exercise in the second edition is:

Prove that $f(x)=x^4-10x^2+1$ is irreducible in $\Bbb Q[x]$. (Hint. Use
Exercise 63 to show that $f(x)$ has no rational roots; then show that there
are no rationals $a$, $b$ and $c$ with$$x^4-10x^2+1=(x^2+ax+b)(x^2-ax+c).\text)$$

A: It is easy to see that the splitting field of this polynomial is $\Bbb{Q}(\sqrt{2},\sqrt{3})$.
So you can compute the Galois group of $\Bbb{Q}(\sqrt{2},\sqrt{3})/\Bbb{Q}$ to be $$\langle \sigma,\tau:\sigma^{2}=\tau^{2}=1\rangle\cong \Bbb{Z}_{2}\times\Bbb{Z}_{2}\cong K_{4}$$.
Where $\sigma=\begin{cases} \sqrt{2}\mapsto -\sqrt{2}\\ \sqrt{3}\mapsto \sqrt{3}\end{cases}$ , $\tau=\begin{cases} \sqrt{2}\mapsto \sqrt{2}\\ \sqrt{3}\mapsto -\sqrt{3}\end{cases}$
.
Now let $f(x)=x^{4}-10x^{2}+1$ be reducible. Then if it is the product of two irreducible factors $g(x)$ and $h(x)$ which are distinct as $f$ is separable . Then if $\alpha,\beta$ be roots of $g(x)$ and $h(x)$.
Then for any $\zeta\in \text{Gal}(\Bbb{Q(\sqrt{2},\sqrt{3})}/\Bbb{Q})$ , $\zeta(\alpha)$ is a root of $g(x)$ and $\zeta(\beta)$ is a root of $h(x)$. and neither $\zeta(\alpha)$ nor $\zeta(\beta)$ can simultaneously be roots of $g$ and $h$ as the Galois group takes roots of irreducible polynomials to itself and $g$ and $h$ are distinct.
But this is a contradiction . Because if we assume WLOG that $\alpha=\sqrt{2}+\sqrt{3}$. We can get to any root of $h(x)$ which is of the form $(\pm\sqrt{2}\pm\sqrt{3})$ by applying $\sigma$ and $\tau$ in the correct order.
This means that $f(x)$ is irreducible.
Note 1: that we actually have sort of provided a way to prove the fact that if the Galois group acts transitively on the roots of a polynomial, then it is irreducible. In fact the converse is also true.
Note 2: This is essentially the same thing as that of Jose's answer but phrased in the language of Galois Theory as you asked in the question.
