Irreducibility of some polynomials

Question

Consider the polynomials $$p_1(t)= 3t^2-7t+3$$ and $$p_2(t)= 4t^2-9t+4.$$ I have tried to show that $p_1(t^s)$ and $p_2(t^s)$ are irreducible over $\mathbb Q[t]$ for any positive integer $s$.

I'm considerably sure for the first one and cannot produce a counterexample for the second one.

Any help will be appreciated.

Answer 1

Handling the polynomial $p_1(t^s)$ using this result. I could not locate an on-site proof of that except in the case where $s$ is a prime. The claim is false for $p_2(t^s)$ unless $s$ is odd. I highlight the differences in the bottom part of this post.

In addition to that result, my argument needs a few facts about algebraic number theory, particularly in the field $F=\Bbb{Q}(\sqrt{13})$. Our interest in the field $F$ comes, of course, from the fact that by quadratic formula the zeros of $p_1(t)$ are $$ t_1=\frac16(7-\sqrt{13})\quad\text{and}\quad t_2=\frac16(7+\sqrt{13}). $$

I intend to apply the cited resul to the field $F$ and the element $t_1$ to prove that the polynomial $$ Q_s(t):=t^s-t_1 $$ is irreducible in the polynomial ring $F[t]$.

We will be working with the ring of algebraic integers $\mathcal{O}_F=\Bbb{Z}[(1+\sqrt{13})/2]$ of the field $F$. The norms of the numerator $7-\sqrt{13}$ and the denominator $6$ are both equal to $$ N_{F/\Bbb{Q}}(7-\sqrt{13})=36=N_{F/\Bbb{Q}}(6). $$ This tells us to first take a census of the prime ideals of $\mathcal{O}_F$ lying above the rational primes $2$ and $3$.

Because $13\equiv5\pmod8$ the law of quadratic reciprocity implies that the prime $2$ is inert. On the other hand the prime $p$ splits as $$ 3=(4-\sqrt{13})(4+\sqrt{13}). $$ A bit of testing with the potential factors then reveals that $$ t_1=u\cdot\frac{4-\sqrt{13}}{4+\sqrt{13}}, $$ where $u=(11+3\sqrt{13})/2$ is a unit of $\mathcal{O}_F$.

This means that the fractional ideal generated by $t_1$ is $\mathfrak{p}_1\mathfrak{p}_2^{-1}$, where $\mathfrak{p}_i,i=1,2,$ are the prime ideals of $F$ lying above the rational prime $3$.

The multiplicity of $\mathfrak{p}_1$ in this fractional ideal is one, implying that $t_1\neq v^s$ for any $v\in F$ and any integer $s>1$. Similarly, the multiplicity of the prime ideal $(2)$ in the fractional ideal generated by $-4t_1^4$ is equal to two, implying that $-4t_1^4$ is not a fourth power of an element of $F$.

By the cited result this gives us the irreducibility of the polynomial $Q_s(t)=t^s-t_1$ over $F$ for all integers $s>1$.

This implies that the extension $F(t_1^{1/s})/F$ is of degree $s$. By the tower law then $[F(t_1^{1/s}):\Bbb{Q}]=2s$.

But, $t_1^{1/s}$ is a zero of $p_1(t^s)$, a polynomial of degree $2s$. Therefore $p_1(t^s)$ is the minimal polynomial of $t_1^{1/s}$ over $\Bbb{Q}$. In particular it is irreducible.

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A similar study of $p_2(t)$ lead to the observation that (this time in the field $\Bbb{Q}(\sqrt{17})$ the zeros of $p_2$ generate a fractional ideal with multiplicities equal to two (both primes lying above $2$ play a role). It is then no longer a surprise that $$ p_2(t^2)=4t^4-9t^2+4=(2t^2-t-2)(2t^2+t-2) $$ fails to be irreducible. However, it follows in the same way that $p_2(t^s)$ is irreducible for all odd integers $s>1$. The main points are:

• The zeros of $p_2(t)$ are $z_{\pm}/4$ with $z_{\pm}=(9\pm\sqrt{17})/2$. The norm of $z_{\pm}$ is $16=N(4)$.
• The only relevant prime $p=2$ splits as $$-2=\left(\frac{3+\sqrt{17}}2\right)\left(\frac{3-\sqrt{17}}2\right).$$ Let's denote by $\mathfrak{q}_1$ the prime ideal generated by the left factor and by $\mathfrak{q}_2$ the prime ideal generated by the right factor.
• We have $$z_-=u \left(\frac{3+\sqrt{17}}2\right)^4$$ with the unit factor $u=33-8\sqrt{17}$.
• The fractional ideal generated by $z_-$ is thus $\mathfrak{q}_1^2\mathfrak{q}_2^{-2}$ implying that $z_-$ is not of the form $v^s$ for any odd $s>1$ and $v\in\Bbb{Q}(\sqrt{17})$.