Degree of minimal polynomial of the sum of two algebraic elements over $\mathbb Q$

Question

The minimal polynomial of $a$ over $\mathbb{Q}$ is quadratic.

The minimal polynomial of $b$ over $\mathbb{Q}$ is cubic.

Is the minimal polynomial of $a+b$ necessarily of degree $6$?

If so, what is the shortest/most elegant proof of this?

Answer 1

Yes, the degrees of sums in separable field extensions multiply, if they are relatively prime, see here. Since $\gcd(2,3)=1$ we obtain, that the degree of $a+b$ is equal to $2\cdot 3=6$, and $\mathbb{Q}(a,b)=\mathbb{Q}(a+b)$.

Answer 2

Denote $c= a+b$. It is enough to show that $\mathbb{Q}(c) = \mathbb{Q}(a,b)$, the last one easily seen to be of degree $6$. The proof involves the standard trick, but we need a bit of work.

Let $P$ the minimal polynomial of $a$, with root $a=a_1$, $a_2$, and $Q$ the minimal polynomial of $b$, with roots $b=b_1$,$b_2$, $b_3$. Consider now the polynomial $P(c - X)$. It has the root $b_1$. Let's show that it doesn't have any other commons roots with $Q$. It it did, we would have $P(c-b_2) = 0$, so say $c-b_2= a_2$, or $$a_1 + b_1= a_2 + b_2$$ or $$b_2 = b_1 + (a_1 - a_2)$$ Now, $(a_1-a_2) = \sqrt{d}$. So $Q(b + \sqrt{d}) = 0$. Now $$Q(b+ \sqrt{d}) = Q(b) + Q'(b) \sqrt{d} + \frac{1}{2}Q''(b)d + \frac{1}{6}Q^{(3)}(b) d \sqrt{d}=\\ \left(Q'(b) + \frac{1}{6}Q^{(3)}(b)\right) \sqrt{d} + \frac{1}{2}Q''(b)d$$ Now $Q'(b) + \frac{1}{6}Q^{(3)}(b)\ne 0$, since $b$ is of degree $3$. It follows that $\sqrt{d}\in \mathbb{Q}(b)$, contradiction.

Therefore, the GCD of the polynomials $P(c-X)$ $Q$ is $X-b$. Therefore, this polynomial must be in $\mathbb{Q}(c)[X]$, so $b \in \mathbb{Q}(c)$. This also gives an effective way to find the expression of $b$ in terms of $c$.

Observation: From the above, we conclude: if $b$, $e$ are irrationals of degree $3$, then the only possibilities for the degrees of $b+e$ are $1$, $3$, $6$, $9$ ( never $2$).

Note that the proof works for $\deg a = 2$ and $\deg b$ odd.