Some basic question about obtain minimal polynomial of algebraic element.

Question

For example, let $c=p^{1/2}+q^{1/3}$ for some prime integers $p$ is not $q$. To find minimal polynomial of $c$ over $\mathbb{Q}(p^{1/2})$, use $(c-p^{1/2})^3=q$, then we can obtain $f=x^3-3p^{1/2}x^2+3px-p^{3/2}-q$. It is minimal polynomial of $c$ over $\mathbb{Q}(p^{1/2})$, but I don't know how to sure that this polynomial is minimal polynomial.

If I prove the polynomials which degree 0, 1, 2 cannot satisfy $g(c)=0$, then can I say that $f$ is minimal polynomial? I guess I also have to show that $f$ is irreducible.. I already know there are unique monic irreducible polynomial, so if I prove $f$ is irreducible, then we can say that minimal polynomial is $f$ which have degree 3?

Answer 1

Either of the following statements would show that $f$ is the minimal polynomial of $c$ in $\mathbb Q(p^{1/2})$:

1. There is no polynomial $g\in\mathbb Q(p^{1/2})[x]$ with $\deg g<\deg f$ and $g(c)=0$.
2. The polynomial $f$ is irreducible.

It is clear that the first condition implies the second since, if $f=gh$ with $\deg g,\deg h>0$, then either $g(c)=0$ or $h(c)=0$, but both have strictly smaller degree. The second also implies the first because we can take the greatest common divisor: if $g$ is a polynomial for which $g(c)=0$, then $\gcd(f,g)$ is also such a polynomial (this essentially follows from the fact that the minimal polynomial exists). This polynomial has degree strictly smaller than that of $f$, and must also divide $f$, so its existence would contradict the irreducibility. In general, the minimal polynomial of $c$ must divide such an $f$ -- this can actually be taken to be a definition of the term "minimal polynomial."

Condition 1 is useful if you want to show something has minimal polynomial of high degree, but can't be sure what its minimal polynomial should actually be. Condition 2 requires a lot less to be verified; in particular, $f$ is of degree $3$, so, if it is not irreducible, it has to have a linear factor. Proving that your $f$ has no linear factor is much easier than dealing with an arbitrary polynomial of degree $\leq 2$, so it's a much better path forward.