Theoretical justification of method to find the minimal polynomial of $\alpha=\sqrt p + \sqrt q$ over $\mathbb Q$ with $p,q $ distinct prime numbers

Question

Find the minimal polynomial of $\alpha=\sqrt p + \sqrt q$ over $\mathbb Q$ with $p,q $ distinct prime numbers

My approach when I saw this problem was to just use some high-school algebra. I recalled a theorem that says that if a polynomial with rational coefficientes has a root of the form $\sqrt p + \sqrt q$ then the other roots are $\sqrt p - \sqrt q, -\sqrt p - \sqrt q, -\sqrt p + \sqrt q$, and from that I get the polynomial. Another way is also to isolate one radical, square it and then isolate the new radical, square again and get the same result. The answer is $x^4 -2(p+q)x^2 +(p-q)^2$.

In my field theory course the T.A said that was kind of guessing, and that I first needed to find the degree of the minimal polynomial.He argued in this terms:

$Q[\alpha]$ is generated by $1, \alpha, \alpha^2,...$ if $p(x)$ is the minimal polynomial of $\alpha$, then
$p(\alpha)=\sum_{i=0}^{n}c_i\alpha^i=0$ for some $n$, which is a linear relation on the $c_i$. Now solve the equations that result from $\sum_{i=0}^{n}c_i\alpha^i=0$ for $n=1,2,3...$ considering that they must be rational numbers until you find and $n=N$ for wich the $c_i$ are not all zero, that is that they are l.d., then $N$ is the degree of the minimal polynomial and you also get the polynomial at the same time, since you have computed the $c_i$'s

so I did:

$n=1: 0=c_0 +c_1(\sqrt p + \sqrt q) \implies c_0=0, c_1=0$

$n=2: 0=c_2 +c_1(\sqrt p + \sqrt q) + c_2(\sqrt p + \sqrt q)^2=c_2(p+q)+c_0+c_1\sqrt p +c_1\sqrt q + 2c_2\sqrt{pq} \implies c_0=0, c_1=0, c_2=0$

analogously for $n=3$, all $c_i$'s are $=0$

and for $n=4$ on the other hand, I got the following:

$c_0=c_4(p-q)^2, c_2=-2c_4(p+q), c_3=c_1=0$. Since the coefficients are l.d, the degree of the minimal polynomial is N=4 and the minimal polynomial is, setting $c_4=1$ to get a monic polynomial :$x^4 -2(p+q)x^2 +(p-q)^2$

Same solution as with my "primitive" methods but but with considerable more effort.

My questions are

1 what is the rationale behind this method? I don't understand why this works. For instance why is he talking about a $Q[\alpha]$ being generated by $1, \alpha, \alpha^2,...$? This exercise is not about a field extension, we are just in $\mathbb Q$, right?And why are my methods not rigorous?

2 I have also seen other people working with $Q(\sqrt p + \sqrt q)$ and $Q(\sqrt p)(\sqrt q)$, proving that they are equal and then doing some reasoning with the tower rule regarding also $Q(\sqrt p)$ . Is that another rigourous method. Does it matter $Q(.)$ or $Q[.]$ in this method? I saw them mixing them up (maybe just a distraction)

Answer 1

Much of the answers to your question relies on the fundamental properties of minimal polynomials and field extensions. The minimal polynomial for an algebraic number $\alpha$ over $\Bbb{Q}$ is the unique rational monic polynomial $p$ of minimal degree satisfying $p(\alpha)=0$. This immediately shows why the method mentioned by your TA works, since you are checking whether a polynomial exist satisfying $p(\alpha)=0$ for each degree, until you find one - and this must then have minimal degree.

Your method does in fact work, if the result you are referencing is indeed shown rigorously, since having $4$ distinct roots means you must have at least degree $4$, and you show that the resulting degree $4$ polynomial has rational coefficients. But using this high school result might not be in the spirit of the field theory course.

Algebraic numbers and their minimal polynomials are intimately tied to field extensions. For example, the degree of the extension $\Bbb{Q}(\alpha)/\Bbb{Q}$ is exactly the degree of the minimal polynomial of $\alpha$ over $\Bbb{Q}$. Note that $\Bbb{Q}(\alpha)$ are usually reserved for extending $\Bbb{Q}$ with a number, and allowing multiplicative inverses, while $\Bbb{Q}[x]$ usually means forming the polynomial ring over $\Bbb{Q}$, where you consider $x$ as a variable, rather than a number.

The above identity of degrees is the basis of the method in your second question. It is easy to find the polynomial you find, but without knowing your high school result it is not immediate that this is minimal in degree. But if we know that $\Bbb{Q}(\sqrt{p}+\sqrt{q})$ has degree $4$ over $\Bbb{Q}$, we would be done. You can do this by establishing that $\Bbb{Q}(\sqrt{p}+\sqrt{q}) = \Bbb{Q}(\sqrt{p})(\sqrt{q})$, and then arguing that the RHS is a degree $4$ extension, which is not too hard (indeed, you use the tower rule). You can find an answer doing it here: Minimal polynomial of $\sqrt{2}+\sqrt{3}$ over $\mathbb Q$

Answer 2

1. He notes the generates since this allows you to write an arbitrary polynomial $p \in \mathbb{Q}$ in the way that he did. From there, he offers a very constructive proof of the degree of $\sqrt{p} + \sqrt{q}$ (abuse of notation here to use the same $p$ for a polynomial and a prime, but just copying what's in the question). As you note, you get that it's 4. Your method shows the degree is at most 4 but it doesn't rigorously show it is exactly 4, i.e. You correctly found a polynomial, but you did not prove it was minimal. His method does this.
2. Short answer is yes but maybe this will come later in your course. Also $\mathbb{Q}(\alpha)$ is a field, while $\mathbb{Q}[\alpha]$ is a ring. One has inverses, the other doesn't.