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.

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 $e_3=\frac{4}{1-γ}$ and that $e_3 ∈ 𝒪_K$.
- Show that $\{e_1, e_2, e_3\}$ is an integral basis for $𝒪_K$.
- Express $e_1^2, e_2^2, e_3^2$ as $ℤ$-linear combinations of $e_1, e_2, e_3$ and show that there does not exist $θ ∈ 𝒪_K$ such that $θ ∉(2)$ and $θ^2 ∈(2)$.
- Compute $N((2+γ))$ and $N((1-γ))$.
- Show that (2) is the product of three distinct prime ideals. Deduce that there does not exist $ρ ∈ 𝒪_K$ such that $𝒪_K=ℤ[ρ]$.

- Any $x ∈ ℚ$ satisfying $x^3-x+8=0$ would have to satisfy $x ∈ ℤ$ (since monic and in $ℤ[x]$) and would divide the constant term 8; but $± 1, ± 2, ± 4, ± 8$ 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). Note that $\{γ^2+γ\}\{1-γ\}=-γ^3+γ=8$, so that $e_3=\frac{γ^2+γ}{2}=\frac{4}{1-γ}$. Then $γ=1-\frac{4}{e_3}$, so that $e_3$ satisfies $(1-\frac{4}{e_3})^3-(1-\frac{4}{e_3})+8=0$ and so $e_3^3-e_3^2+6 e_3-8=0$, which is monic, so $e_3 ∈ 𝒪_K$.
- $Δ^2(\{1, γ, γ^2\})=-4 a^3-27 b^2=-2^2 ⋅ 431$, change of basis matrix $\pmatrix{1\\&1\\&\frac12&\frac12}$ has determinant $\frac12$, so $Δ^2(\{e_1, e_2, e_3\})=-431$, which is prime, square-free, and so $\{e_1, e_2, e_3\}$ is an integral basis.
- We compute: $e_1^2=e_1, e_2^2=γ^2=-e_2+2 e_3$ and $e_3^2=\frac{1}{2} γ^2-\frac{3}{2} γ-4=-4e_1-2 e_2+e_3$. Now suppose that $θ ∉(2)$; then $θ=a_1 e_1+a_2 e_2+a_3 e_3$, for $a_1, a_2, a_3 ∈ ℤ$ not all even. Then $θ^2-\{a_1^2 e_1^2+a_2^2 e_2^2+a_3^2 e_3^2\}=2\{a_1 a_2 e_1 e_2+a_1 a_3 e_1 e_3+a_2 a_3 e_2 e_3\} ∈(2)$ and $e_i^2-e_i∈(2)$ so: $θ^2-\{a_1^2 e_1+a_2^2 e_2+a_3^2 e_3\} ∈(2)$. But $a_1^2 e_1+a_2^2 e_2+a_3^2 e_3 ∉(2)$, since $a_1^2, a_2^2, a_3^2$ are not all even. Hence $θ^2 ∉(2)$.
- $N((2+γ))=|\operatorname{Norm}_{K / ℚ}(2+γ)|=|-\operatorname{Norm}_{K / ℚ}(-2-γ)|=|\{-2\}^3-\{-2\}+8|=2$, and $N((1-γ))=|1^2-1+8|=8$.
- Note that $N((2))=|\operatorname{Norm}_{K / ℚ}(2)|=8$ and that $P=(2+γ)$ satisfies $N(P)=2$ (which forces $P$ to be a prime ideal, since it has prime norm), so that $P ∣(N(P))=(2)$. We cannot have $(2)=P^2 Q$ since then $P Q$ would strictly contain (2), so there would exist $θ ∈ P Q$ such that $θ ∉(2)$; but then $θ^2 ∈ P^2 Q^2 ⊆(2)$, which we have shown impossible. We can similarly exclude $(2)=P^3$ and $(2)=P Q^2$. We also cannot have $(2)=P Q$, a product of two distinct prime ideals, since then $N(Q)=4$, and $(1-γ) ∣(8)$, so the prime ideal factors of $(1-γ)$ come from $P, Q$; but $P ∤(1-γ)$ since that would mean $1-γ, 2+γ, 2 ∈ P$, giving $1 ∈ P$, incompatible with $P$ being prime; this would force $(1-γ)$ of norm 8 to be a power of $Q$ of norm 4, which is impossible. The only remaining possibility is that (2) must be the product of three distinct prime ideals.

We finally note that there cannot exist $ρ$ such that $𝒪_K=ℤ[ρ]$ since the Dedekind's Theorem would be applicable at the prime 2 to the minimal polynomial of $ρ$, forcing it to be the product of three distinct linear factors mod 2, which is impossible as there are only 2 possible distinct roots $\bmod 2$.#### Alternate proof

Since $γ$ is a root of $x^3-x+8$. $$-8=γ(γ-1)(γ+1)$$ Considering ideals $N((2))=N(γ)=N(γ-1)=N(γ+1)=8$ $$(2)^3=(γ)(γ-1)(γ+1)$$ since $γ,γ-1$ are coprime, and $γ,γ+1$ are coprime, and $γ-1,γ+1$ are coprime, (2) must split into three distinct prime ideals.

$x^3-x+8=x(x-1)^2\pmod2$, by Dedekind's criterion, $(2)=𝔭_1𝔭_2^2$.

By Chinese Remainder Theorem, $𝒪_K/(2)=𝒪_K/𝔭_1⊕𝒪_K/𝔭_2^2$ then $∃θ∈𝒪_K/(2),θ≠0,θ^2=0$ contradicting (iii).