# Some basic question about obtain minimal polynomial of algebraic element.

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?

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.