Can you prove polynomial irreducibility via substitution to a quadratic polynomial?

Question

Suppose I want to find the minimal polynomial of $\alpha \in \mathbb{C}$ over $\mathbb{Q}$; by this, I mean that I want to find a polynomial that $\alpha$ satisfies and then prove its minimality, usually by contradiction. For the sake of argument, suppose such a polynomial has the form

$$m_\alpha(x) = x^{2n} +bx^n + c \hspace{1cm} b,c \in \mathbb{Q}$$

Then by making the substitution $u = x^n$, I obtain a new polynomial

$$m_\alpha(u) = u^2+bu+c$$

Then the roots of $m_\alpha(u)$ are given by

$$u = - \frac{b}{2} \pm \frac{\sqrt{b^2-4c}}{2}$$

Supposing that $\sqrt{b^2-4ac} \not\in \mathbb{Q}$, we could then conclude that $m_\alpha(u)$ is irreducible over $\mathbb{Q}$ since irreducibility of quadratic polynomials is characterized by its roots. Am I then allowed to undo the substitution to declare that $m_\alpha(x)$ is also irreducible over $\mathbb{Q}$? It seems as though my substitution causes us to lose some information about $m_\alpha(\dots)$ which is the source of my skepticism.

Answer 1

We have $x^2+4$ is irreducible over $\Bbb Q$ but $x^4+4 =(x^2+2x+2)(x^2-2x+2)$

Answer 2

These kinds of substitutions can work and the keyword is Capelli's Lemma:

Let $p,q\in K[x]$ be polynomials over a field $K$. Then the composition of $p\circ q$ is irreducible if and only if $p$ is irreducible and $q-\alpha\in K(\alpha)[x]$ is irreducible, whereby $\alpha$ is any root of $p$.

Answer 3

The answer is NO. Consider the case $b=-21,c=4$. Then $\sqrt{b^2-4ac}=\sqrt{425}=5\sqrt{17}\not\in{\mathbb Q}$, so $P(X)=X^2+bX+c$ is irreducible; however

$$ P(X^2)=X^4-21X^2+4=(X^4+4X^2+4)-25X^2=(X^2+2-5X)(X^2+2+5X) $$

Answer 4

Given a polynomial with the form

$P(x)=ax^{2n}+bx^n+c; a,b,c\in\mathbb{Z};b^2-4ac\not\in\mathbb{Z}^2$

we have that $P(x)$ is orreducible over $\mathbb{Q}[x^n]$. But $\mathbb{Q}[x]$ is a larger set than $\mathbb{Q}[x^n]$ for $n\ge2$, and the adfitional elements of the latter set may admit factors of $P(x)$ that do not belong to the former.

Thus $x^4+x^2+1$ is irreducible in $\mathbb{Q}[x^2]$, but it has factors $x^2+x+1,x^2-x+1$ that are in $\mathbb{Q}[x]$ and not in $\mathbb{Q}[x^2]$.