irreducible polynomial over $Q(t_5)$

Question

Consider $f(x)=x^5-5 \in \mathbb{Q}[x]$. I have already shown that f is irreducible. Now I want to prove that f is irreducible in $Q(t_5)[x]$, with $t_5=e^{\frac{i2\pi}{5}}$ but I have no idea how to start. Can anybody give me hint? Best, Redrose

Answer 1

Consider the action of the Galois group on the possible irreducible factors of $f(x)$.

If $\sigma \in G = Gal_\Bbb Q(\Bbb Q(t_5))$ and $g(x)$ is a factor of $f(x)$ in $\Bbb Q(t_5)[x]$, then $\sigma(g(x))$ is also a factor of $\sigma(f(x)) = f(x)$.
Since $G$ is cyclic of order $4$ (say $G = \{id,\sigma,\sigma^2,\sigma^3\}$), there are three ways that the group can act on a factor, according to what $Stab_G(g)$ is :

• if $Stab_G(g) = G$, then $g$ is invariant by $G$.
• if $Stab_G(g) = \{id,\sigma^2\}$, then $g$ comes with one conjugate $\sigma(g)$.
• if $Stab_G(g) = \{id\}$, then $g$ comes with three other conjugates $\sigma(g),\sigma^2(g),\sigma^3(g)$.

Since conjugate factors have the same degree and $f(x)$ has degree $5$ which is odd, there must be an irreducible factor $g(x)$ of $f(x)$ that is invariant by $G$. Since $g$ is invariant by $G$, it must have rational coefficients, and is actually a factor of $f$ over $\Bbb Q$. Since $f$ is irreducible over $\Bbb Q$, this factor must be $f$, and so $f$ is still irreducible over $\Bbb Q(t_5)$

Answer 2

Use some Galois theory. Changing your notation $t_5$ to standard $\zeta$ note that $\zeta$ generates a Galois extension with cyclic Galois group. And it can be shown that if $f(x)$ is irreducible over the rationals and reducible over a Galois extension it has to split into factors of equal degree. For a proof see here:

Why does an irreducible polynomial split into irreducible factors of equal degree over a Galois extension?

Now $5$ being prime $f(x)$ can only split into linear factors, if at all. So all the roots, in particular the real positive root, of $x^5-5$ is in $\mathbf{Q}[\zeta]$, and consequently the root has to be in its real subfield viz $\mathbf{Q}[\zeta +\bar \zeta]$. But that real subfield is quadratic and cannot contain a degree 5 algebraic number. This contradiction proves irreducibilty over $\mathbf{Q}[\zeta]$.