Dedekind's example of a cubic field $K$ for which $O_K$ does not have the form $\mathbb Z[\alpha]$

Question

Let $\alpha$ be a root of $f(X)=X^3+X^2-2X+8$ and $\beta = \frac{4}{\alpha}$. It can be shown that $O_K=\mathbb{Z}[\alpha, \beta]$.

How does one then establish the following ring isomorphism $$\frac{O_K}{2O_K} \cong \mathbb{F}_2 \times \mathbb{F}_2 \times\mathbb{F}_2\ ?$$

We can't use the Dedekind Criterion here, since $2=|O_K : \mathbb{Z}[\alpha]|$ so the index and prime $2$ are not coprime.

This result is important, as it shows that $(2)$ is totally split and the result $O_K \neq \mathbb{Z}[\nu]$ for any $\nu$ would then follow by Dedekind.

I would be able to prove the existence of this isomorphism indirectly (after some calculation/manipulation of norms of ideals) if I know that $2$ is unramified, but I don't see why this should hold.

Answer 1

If all you need is that $2$ is unramified, and you have that $\mathcal{O}_K=\mathbb{Z}[\alpha,\beta]$, then you should be able to show that $\Delta_K=-503$. The ramified primes are precisely those primes that divide the discriminant.

Answer 2

Let $\alpha$ be a root of $f(X)=X^3+X^2-2X+8$ and $\beta = \frac{4}{\alpha}$. It can be shown that $O_K=\mathbb{Z}[\alpha, \beta]$.

How does one then establish the following ring isomorphism $$\frac{O_K}{2O_K} \cong \mathbb{F}_2 \times \mathbb{F}_2 \times\mathbb{F}_2\ ?$$

You know that $α$ is a root of $f$ and $K=(α)$ then $e_1, e_2, e_3$ is an integral basis for $_K$, where $e_1=1$, $e_2=α$ and $e_3=\frac{1}{2} α(α+1)$.

We will show that the linear maps $ψ_v: _K → _2$ defined by $ψ_v(e_i)=v_i$ are ring homomorphisms, for the following values of $v$:

• $v=(1,0,0)$;
• $v=(1,1,0)$;
• $v=(1,0,1)$.

then we define $ψ:_K→_2^3$ by $ψ=(ψ_{(1,0,0)},ψ_{(1,1,0)},ψ_{(1,0,1)})$

and we'll conclude that $_K/(2) ≅ _2^3$.
$_2^3$ is a product of fields, so doesn't have nilpotent element, so $(2)$ doesn't ramify.

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To prove ψ is a ring homomorphism.
$\begin{array}{c|ccc} &e_1&e_2&e_3\\\hline e_1&e_1&e_2&e_3\\ e_2&e_2&2e_3-e_2&2e_3+4e_1\\ e_3&e_3&2e_3+4e_1&3e_3-2(e_2+3e_1) \end{array}$ this table mod 2 is the table $\begin{array}{c|ccc} &v_1&v_2&v_3\\\hline v_1&v_1&v_2&v_3\\ v_2&v_2&v_2&0\\ v_3&v_3&0&v_3 \end{array}$
Three values of $v$ all satisfy the table, so $ψ(e_i)ψ(e_j)=ψ(e_ie_j)$ for $1≤i≤j≤3$.

To prove ψ is surjective.
For any $(x_1,x_2,x_3)∈_2^3$ we find $$(x_1,x_2,x_3)=x_1(1,1,1)+(x_2-x_1)(0,1,0)+(x_3-x_1)(0,0,1)$$ so $$(x_1,x_2,x_3)=ψ(x_1e_1+(x_2-x_1)e_2+(x_3-x_1)e_3)$$ To prove ker ψ=(2).
Clearly $(2)⊂\kerψ$. On the other hand $N((2))=2^3=8$, $N(\kerψ)={|_K/\kerψ|}=|_2^3|=8$, so $N((2))=N(\kerψ)$, so $\kerψ=(2)$.

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we conclude that $_K/(2) ≅ _2^3$.
$_2^3$ is a product of fields, so doesn't have nilpotent element, so $(2)$ doesn't ramify.

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To verify the above, $(2)=\kerψ_{(1,0,0)}\kerψ_{(1,1,0)}\kerψ_{(1,0,1)}$, \begin{array}{lll} \kerψ_{(1,0,0)}=(e_2,e_3)=(α,α(α+1)/2)\\ \kerψ_{(1,1,0)}=(e_1-e_2,e_3)=(1-α,α(α+1)/2)\\ \kerψ_{(1,0,1)}=(e_1-e_3,e_2)=(1-α(α+1)/2,α) \end{array}

in SageMath we can compute

X = polygen(ZZ)
K.<a> = NumberField(X^3-X^2-2*X-8)
ideal(2+0*a)==ideal (a,a*(a+1)/2)*ideal(1-a,a*(a+1)/2)*ideal(1-a*(a+1)/2,a)

result is True.