Computations with Galois Theory (14.2.1-3)

Dummit and Foote Abstract Algebra, section , exercise :

1. Determine the minimal polynomial over $\mathbb{Q}$ for the element $\sqrt{2}+\sqrt{5}$.
2. Determine the minimal polynomial over $\mathbb{Q}$ for the element $1+\sqrt[3]{2}+\sqrt[3]{4}$.
3. Determine the Galois group of $(x^2-2)(x^2-3)(x^2-5)$.

Proof: Lemma: Let $K/F$ be Galois over a perfect field and let $α∈K$. Then the minimal polynomial of $α$ over $F$ is the squarefree part of the polynomial $$\prod_{σ∈\text{Gal}(K/F)}(x-σα)$$ Proof: We show $\{σα~|~σ∈\text{Gal}(K/F)\}$ is the full set of zeros for the minimal polynomial $p(x)$. They are all zeros as the automorphisms fix the coefficients of $p(x)$, and as well for any other root $β$ of $p(x)$ we have the isomorphism $F(α)→F(β)$ and extending automorphism $K$ preserving this map, since being Galois $K$ is a mutual splitting field for $F$, $F(α)$, and $F(β)$.

As well, $F$ being perfect, $p(x)$ is separable and thus has no repeated roots, and we may now say $p(x)~|~\prod_{σ∈\text{Gal}(K/F)}(x-σα)$. Since the zeros of the latter are precisely the (nonrepeated) zeros of the former, we have the squarefree part is indeed $p(x)$.$~\square$ This solves the problem of determining minimal polynomials (over perfect fields) when the Galois group of a containment Galois extension is known (and still provides much information when $F$ isn't perfect).

(1) We see $\sqrt{2}+\sqrt{5}∈\mathbb{Q}(\sqrt{2},\sqrt{5})=\mathbb{Q}(\sqrt{2})(\sqrt{5})$, where the latter is computed to be Galois (splitting field of $(x^2-2)(x^2-5)$) of degree $4$. The four automorphisms must be the four uniquely defined by the identity, $α~:~\sqrt{2}↦-\sqrt{2}$, $β~:~\sqrt{5}↦-\sqrt{5}$, and the composite $αβ$. Thus the four roots of the minimal polynomial are $\pm \sqrt{2} \pm \sqrt{5}$ and the minimal polynomial is calculated to be $x^4-14x^2+9$.

(2) We put the element in the Galois extension $\mathbb{Q}(ρ,\sqrt[3]{2})$ and observe the effect of the elements of the Galois group previously determined. The result is the polynomial $$(x-1-\sqrt[3]{2}-\sqrt[3]{4})(x-1-ρ\sqrt[3]{2}-ρ^2\sqrt[3]{4})(x-1-ρ\sqrt[3]{4}-ρ^2\sqrt[3]{2})=$$ $$x^3-3x^2-3x-1$$ (3) With some simple algebra we may see $x^2-5$ is irreducible over the field $\mathbb{Q}(\sqrt{2},\sqrt{3})$ of degree $4$ over $\mathbb{Q}$, so the splitting field in question $\mathbb{Q}(\sqrt{2},\sqrt{3},\sqrt{5})$ is Galois of degree $8$. We see the automorphisms are precisely the automorphisms generated by the maps $α~:~\sqrt{2}↦-\sqrt{2}$, $β~:~\sqrt{3}↦-\sqrt{3}$, $γ~:~\sqrt{5}↦-\sqrt{5}$. By noting that these automorphisms all commute and are all of order $2$, we conclude the Galois group is $Z_2^3$.$~\square$