$-3x^{2m}+7x^m-3$ is irreducible for all $m\geq 1$

Question

I have heard that $p(x)=-3x^2+7x-3$ is the simplest polynomial for which $p(1)=1$, $p(x)=x^{\deg p}\cdot p(x^{-1})$ and $p(x^m)$ is irreducible for all $m\geq 1$. I have tried to show the last part, i.e., the irreducibility of $p(x^m)$ but wasn't able to.

My attempt: The case $m=1$ is trivial, so assume $m\geq 2$. Let $\alpha=\frac{7+\sqrt{13}}{6}$ be a root of $p(x)$ and let $\beta=\sqrt[m]{\alpha}$. It suffices to show that $$[\mathbb{Q}(\beta):\mathbb{Q}]=[\mathbb{Q}(\beta):\mathbb{Q}(\alpha)][\mathbb{Q}(\alpha):\mathbb{Q}]=2m,$$ or equivalently $[\mathbb{Q}(\beta):\mathbb{Q}(\alpha)]=m$, since this would imply that $p(x^m)$, being a polynomial of degree $2m$, is a minimal polynomial of $\beta$ over $\mathbb{Q}$.

Now we are left to show $[\mathbb{Q}(\beta):\mathbb{Q}(\alpha)]=m$. Since $\beta$ is a root of $x^m-\alpha\in\mathbb{Q}(\alpha)[x]$, this is equivalent to $x^m-\alpha$ being irreducible over $\mathbb{Q}(\alpha)$. In light of the answer given here, we need to show that $\sqrt[k]{\alpha}\not\in\mathbb{Q}(\alpha,\zeta_m)$ for all $1<k\leq m$, where $\zeta_m=\exp(\frac{2\pi i}{m})$. Here is where I'm stuck, however.

My another attempt was to simply assume $-3x^{2m}+7x^m-3=f(x)g(x)$ and reach contradiction via different methods (e.g., by reducing modulo 3) but I was not successful here either.

Any help or suggestion would be appreciated. Thank you in advance.

Answer 1

I was able to solve this in Eisenstein's fashion.

Suppose $-3x^{2m}+7x^m-3=f(x)g(x)$ for $f,g\in\mathbb{Z}[x]$. Reducing mod $3$, we obtain $\overline{f(x)}\cdot\overline{g(x)}=\bar{x}^m$ in $\mathbb{Z}_3[x]$. Since $\mathbb{Z}_3[x]$ is a PID, up to signs we obtain $\overline{f(x)}=\bar{x}^k$ and $\overline{g(x)}=\bar{x}^l$ for $k+l=m$. Since $f(0)g(0)=-3$, WLOG we may assume $g(0)=1$ so we have $(k,l)=(m,0)$.

So far we have $$ f(x)=-3+a_1x+\cdots+a_mx^m+\cdots+a_nx^n,\\ g(x)=1+b_1x+\cdots+b_{2m-n}x^{2m-n}$$ for $a_i$ and $b_j$ all divisible by $3$ except for $a_m\equiv1$ mod $3$ and $b_0=1$. Observe that the leading coefficient $a_nb_{2m-n}=-3$ of $fg$ is not divisible by $9=3\cdot 3$. If $n>m$ then $a_n$ is divisible by $3$ so $b_{2m-n}$ is not divisible by $3$, which can only occur when $g$ is the constant polynomial $1$. Thus we may assume $f$ and $g$ have degree $m$ and $a_m=1$, $b_m=-3$. If we investigate the coefficient of $x^m$, however, $$7=a_0b_m+a_1b_{m-1}+\cdots+a_{m-1}b_1+a_mb_0\equiv10\;\mathrm{mod}\;9$$ since $a_0b_m+a_mb_0=(-3)^2+1^2=10$ and all other $a_i$ and $b_j$ are divisible by $3$. A contradiction.