If $p(x) = a_3 + a_2x + a_1x^2 + a_0x^3$ is irreducible in $\mathbb{Q}[x]$, then $q(x) = a_0 + a_1x + a_2x^2 + a_3x^3$ is also irreducible?

Question

I have this doubt because what I want to prove is that, with the given hypothesis,$\frac{\mathbb{Q}[x]}{\langle q(x) \rangle}$ is an integral domain. I have the same problem but for polynomials of degree 4, it occurs to me to give an isomorphism $T: \mathbb{Q}[x] \longrightarrow \mathbb{Q}[x]$ in general but the proof becomes tedious when it comes to proving that the function $T$ is a ring homomorphism. My idea is to suppose that $q(x)$ is reducible and by means of the isomorphism to show that $T(q(x(x)) = p(x)$ showing that $p(x)$ is also reducible reaching a contradiction, but I would like to know if there is an alternative way that does not involve constructing the isomorphism.

Answer 1

$p(x) = a_3 + a_2x + a_1x^2 + a_0x^3\Rightarrow p\left(\dfrac 1x\right)=\dfrac{a_3x^3 + a_2x^2 + a_1x + a_0}{x^3}=\dfrac{q(x)}{x^3}$

If $q(x)$ were reducible then for some $\beta\ne0$ (because $a_0\ne0$) we would have $q(\beta)=0$ with the degree of $\beta$ equal to $1$ or $2$ over $\Bbb Q$. But in this case $\dfrac {1}{\beta}$ is also of degree $1$ or $2$ over $\Bbb Q$ and $p\left(\dfrac {1}{\beta}\right)=0$. This is a contradiction because each root of $p(x)$ must be of degree $3$ since $p(x)$ is irreducible in $\mathbb{Q}[x]$.

Answer 2

It's easier to prove the contrapositive: If $q(x)=a_0+a_1x+a_2x^2+a_3x^3$ is reducible, then $p(x)=a_3+a_2x+a_1x^2+a_0x^3$ is reducible. I'm actually just going to prove a generalized version of this, not just for degree $4$, so we are proving that, for any integer $d > 0$, if $$ q(x)=\sum_{i=0}^d a_ix^i $$ is reducible, then $$ p(x)=\sum_{i=0}^d a_{d-i}x^i $$ is also reducible.

Unfortunately, I do have to add an extra assumption, which is $a_0\neq 0$. Otherwise, you could have a situation where $q(x)=x^3+x^2$, which is reducible, and $p(x)=x+1$, which is irreducible, which clearly contradicts what we are trying to prove.

To make the proof simpler, I will split it into two cases: The first case is $a_d=0$, which is very simple because $a_d$ is the constant term of $p(x)$ and thus if $a_d=0$, $p(x)$ is clearly reducible. The second case is $a_d\neq 0$, which is what follows. In this case, I can safely assume that the degree of $q$ is $d$, because the leading coefficient $a_d$ is nonzero.

Suppose that $q(x)=b(x)c(x)$ for some $b,c\in\mathbb{Q}[x]$ where $b$ and $c$ both have degree at least $1$. Now, let $d_1$ and $d_2$ be the degrees of $b$ and $c$ respectively. Then, there exist rational numbers $b_0,\dots,b_{d_1},c_0,\dots,c_{d_2}$ such that: $$b(x)=\sum_{i=0}^{d_1} b_ix^i$$ $$c(x)=\sum_{i=0}^{d_2} c_ix^i$$ For purpose of generality, we also define that $b_i=0$ for all $i < 0$ or $i > d_1$ and $c_i=0$ for all $i < 0$ or $i > d_2$.

Thus, by polynomial multiplication, we have: $$q(x)=\sum_{i=0}^d \sum_{j=0}^i b_jc_{i-j}x^i$$

Thus, for any $0\leq i\leq d=d_1+d_2$, since $a_i$ is the coefficient of $x^i$ in $q(x)$, $$ a_i=\sum_{j=0}^i b_jc_{i-j} $$ Notice that, since we defined $b_i=0$ and $c_i=0$ for $i < 0$, we can actually decrease the lower limit and increase the upper limit of the sum to whatever we would like. For convenience, we decrease the lower limit to $-d$ and increase the upper limit to $d$: $$ a_i=\sum_{j=-d}^d b_jc_{i-j} $$

Now, we define two new polynomials: $$m(x)=\sum_{i=0}^{d_1} b_{d_1-i}x^i$$ $$n(x)=\sum_{i=0}^{d_2} c_{d_2-i}x^i$$ Notice that $m$ and $n$ have degrees $d_1$ and $d_2$, respectively, because their leading coefficients are $b_0$ and $c_0$, respectively. We know that these leading coefficients are both nonzero because, from the above equation, we have $a_0=b_0c_0$ and $a_0$ is nonzero, which implies $b_0$ and $c_0$ must be nonzero.

Then, we have: $$ \begin{align*} m(x)n(x) &= \sum_{i=0}^{d_1+d_2} \sum_{j=0}^i b_{d_1-j}c_{d_2-(i-j)}x^i \\ &= \sum_{i=0}^d \sum_{j=0}^i b_{d_1-j}c_{d_2+j-i}x^i \\ &= \sum_{i=0}^d \sum_{j=d_1-i}^{d_1} b_jc_{d_2+d_1-j-i}x^i & &[\text{replace }j\text{ with }d_1-j] \\ &= \sum_{i=0}^d \sum_{j=d_1-i}^{d_1} b_jc_{(d-i)-j}x^i \\ &= \sum_{i=0}^d \sum_{j=-d}^d b_jc_{(d-i)-j}x^i & &[\text{changing limits only adds terms equal to }0] \\ &= \sum_{i=0}^d a_{d-i}x^i \\ &= p(x) \end{align*} $$ Thus, $p(x)=m(x)n(x)$ and, since $m$ and $n$ have the same degree as $b$ and $c$, respectively, this suffices to show that $p$ is reducible.