degree of a field extension

Solution 1:

If you are one of those people who have trouble proving that a nonic polynomial is irreducible, you can try the following sketch instead. This depends on a bit of algebraic number theory.

Consider the prime ideal generated $p=2$. Because $x^3+3x-1$ is irreducible modulo $2$, this prime is inert in the field $K=\mathbb{Q}(\alpha)$. OTOH the polynomial $x^3+x-2$ factors modulo $2$ as $x(x+1)^2$. It is a bit tedious but straightforward to show that $1,\beta,\beta^2$ is an integral basis of $L=\mathbb{Q}(\beta)$. Therefore in its ring of integers two factors as $(2)={\frak{p}_1\frak{p}_2^2}$. In the compositum $F=LK=\mathbb{Q}(\alpha,\beta)$ all the prime ideals lying above two will have inertia degree at least three (because they lie above the inert prime of $K$), so the only possibility is that the prime ideals $\frak{p}_1,p_2$ are both totally inert in $F/L$, and we have $[F:L]=3$. Therefore $[F:\mathbb{Q}]=[F:L]\cdot [L:\mathbb{Q}]=9$ and $L$ and $K$ are linearly disjoint.

The discriminants of the two cubic polynomials are $-135$ and $-104$ respectively. Neither of these is a square in the rationals, so the Galois groups of both of them are $S_3$. Furthermore, the discriminants are coprime, so even the splitting fields of the two cubics intersect trivially, and as Galois extensions are thus linearly disjoint. So if we let $M$ to be the splitting field of $(x^3+3x-1)(x^3+x-2)$, then we can identify the Galois group $$G=\operatorname{Gal}(M/\mathbb{Q})=\operatorname{Sym}(\{1,2,3\})\times \operatorname{Sym}(\{4,5,6\})\simeq S_3\times S_3\le S_6$$ as permutations on the six roots - each copy of $S_3$ acting on the roots of its factor and keeping the roots of the other factor fixed. If we let $\alpha$ and $\beta$ be the roots number 1 and 4 respectively, then $F$ will be the fixed field of $H=\langle (23)\rangle \times \langle (56)\rangle$. It is easy to convince yourself that the only subgroups of $G$ strictly between $H$ and $G$ are $\langle(23)\rangle\times \operatorname{Sym}(\{4,5,6\})$ and $\operatorname{Sym}(\{1,2,3\})\times\langle(56)\rangle$. Clearly the Galois correspondence takes these subgroups to $K$ and $L$ respectively. So $K$ and $L$ are the only non-rational proper subfields of $F$. By linear disjointness the element $\alpha^2+\beta$ does not belong to either, so it must generate all of $F$.

There may be some holes in the argument (a sketch only), but I think that this approach works. It is probably also overkill in its use of algebraic number theory. I am more than willing to be educated about a cleaner approach.

Solution 2:

If you have found such a polynomial, then this suffices.

The reason is that the degree $\lvert \mathbb{Q}(\alpha^2+\beta):\mathbb{Q} \rvert$ equals the degree of the minimal polynomial $f \in \Bbb{Q}[x]$ of $\alpha^{2} + \beta$ over $\Bbb{Q}$.

Now you have found a monic polynomial $g$ of degree $9$ in $\Bbb{Z}[x] \subseteq \Bbb{Q}[x]$ which is irreducible in $\Bbb{Q}[x]$ and has $\alpha^{2} + \beta$ as a root.

Now the minimal polynomial $f$ must be a non-constant, monic divisor of $g$. Since $g$ is irreducible in $\Bbb{Q}[x]$, we obtain $f = g$.