Show that $x^3-x+1$ is irreducible over $ℚ$. Let $K=ℚ(α)$, where $α$ is a root of $x^3-x+1$. Show that $\{1, α, α^2\}$ is an integral basis for $𝒪_K$.

Any $x ∈ ℚ$ satisfying $x^3-x+1=0$ would have to satisfy $x ∈ ℤ$ (since monic and in $ℤ[x]$) and would divide the constant term 1; but $± 1$ are not roots. Hence there is no linear factor over $ℚ$ and so it is irreducible (since any reducible cubic would have a linear factor). If $\{1, α, α^2\}$ were not an integral basis, there would have to be a prime $p$ such that $p^2 ∣ Δ^2(\{1, α, α^2\})$ such that $\frac{1}{p}(c_1+c_2 α+c_3 α^2) ∈𝒪_K$ for $c_1, c_2, c_3 ∈ ℤ$ not all divisible by $p$. But $Δ^2(\{1, α, α^2\})=-4 a^3-27 b^2=-23$ is square-free, so no such $p$ exists.

Show that $x^3-x+2$ is irreducible over $ℚ$. Let $K=ℚ(β)$, where $β$ is a root of $x^3-x+2$. For any $r, s, t ∈ ℚ$ find $\operatorname{Norm}_{K / ℚ}(r-β)$ and $\operatorname{Norm}_{K / ℚ}(r+β)$, and show that $\operatorname{Tr}_{K / ℚ}(r+s β+t β^2)=3 r+2 t$. Show that $\{1, β, β^2\}$ is an integral basis for $𝒪_K$.

Any $x ∈ ℚ$ satisfying $x^3-x+2=0$ would have to satisfy $x ∈ ℤ$ (since monic and in $ℤ[x]$) and would divide the constant term $2$; but $± 1, ± 2$ are not roots. Hence there is no linear factor over $ℚ$ and so it is irreducible (since any reducible cubic would have a linear factor). Let $β_1, β_2, β_3$ be conjugates of $β$, so that $x^3-x+2=(x-β_1)(x-β_2)(x-β_3)$. Then $\operatorname{Norm}_{K / ℚ}(r-β)=(r-β_1)(r-β_2)(r-β_3)=r^3-r+2$, and $\operatorname{Norm}_{K / ℚ}(r+β)=-\operatorname{Norm}_{K / ℚ}(-r-β)=r^3-r-2$. Also, $\operatorname{Tr}_{K / ℚ}(1)=1+1+1=3, \operatorname{Tr}_{K / ℚ}(β)=β_1+β_2+β_3=0$, and $\operatorname{Tr}_{K / ℚ}(β^2)=β_1^2+β_2^2+β_3^2=(β_1+β_2+β_3)^2-2(β_1 β_2+β_1 β_3+β_2 β_3)=0-2(-1)=2$. Hence $\operatorname{Tr}_{K / ℚ}(r+s β+t β^2)=3 r+2 t$.$$Δ^2(\{1, β, β^2\})=-4 a^3-27 b^2=-104=-8 ⋅ 13$$so if $\{1, β, β^2\}$ were not an integral basis, there would have to exist $\frac{1}{2}(c_1+c_2 β+c_3 β^2) ∈ 𝒪_K$ for $c_1, c_2, c_3 ∈ ℤ$ not all divisible by 2. We only need to check for $c_i ∈\{0,1\}$. But by the above, this has trace $\frac{3}{2} c_1+c_3 ∈ ℤ$ so that $c_1=0$. This leaves only to check: $\frac{1}{2} β, \frac{1}{2} β^2, \frac{1}{2} β(β+1)$, which have norms $-\frac14,\frac12,\frac12 ∉ ℤ$, so these are not in $𝒪_K$. Hence no such element exists, and $\{1, β, β^2\}$ is an integral basis.

Show that $x^3-x+8$ is irreducible over $ℚ$. Let $K=ℚ(γ)$, where $γ$ is a root of $x^3-x+8$. Let $e_1=1, e_2=γ, e_3=\frac{γ^2+γ}{2}$.

[Throughout this question, you may use the fact that, if $K=ℚ(δ)$, where $δ$ is a root of $x^3+a x+b$ (irreducible over ℚ), then $Δ^2(\{1, δ, δ^2\})=-4 a^3-27 b^2$; you may also assume that 431 is a prime number.]