Codex @codex  0

The set ${\mathcal{O}}_K$ consists of the elements of $K$ integral over $\mathbb{Z}$, meaning roots of monic polynomials in $\mathbb{Z}[X]$.

First prove the finite-module criterion for integrality in a slightly stronger form, for any nonzero finitely generated $\mathbb{Z}$-submodule $M\subseteq K$ stable under multiplication by $\alpha$. Choose generators $m_1,\dots ,m_r$ and write $\alpha m_i={\sum }_j{a}_{ij}{m}_j$ with integer coefficients. Multiply the vector equation $(\alpha I-A)m=0$ by its adjugate matrix. This gives $\det (\alpha I-A)m_i=0$ for every generator. Some generator is nonzero and $K$ is a field, so $\det (\alpha I-A)=0$. The polynomial $\det (XI-A)$ is monic with integer coefficients, proving integrality of $\alpha$. This implication needs no prior closure theorem for ${\mathcal{O}}_K$.

Conversely, if $\alpha$ satisfies a monic polynomial of degree $d$, then $M=\mathbb{Z}[\alpha ]$ is generated by $1,\alpha ,\dots ,{\alpha }^{d-1}$ and is stable under $\alpha$. Every $\beta \in M$ also maps $M$ into itself because it is an integer polynomial in $\alpha$; the finite-module criterion for integrality just proved makes every such $\beta$ integral. Thus this $M$ really lies in ${\mathcal{O}}_K$, proving the stated equivalence without circularity.

For two integral elements $\alpha ,\beta$, the module spanned by ${\alpha }^i{\beta }^j$ for $i<d_{\alpha }$, $j<d_{\beta }$ contains one, is finitely generated, and is stable under both elements by their monic equations. It is therefore stable under $\alpha +\beta$ and $\alpha \beta$. The finite-module criterion for integrality proves that both are integral. Negatives follow by replacing a multiplication matrix by its negative, and $0,1$ are integral. Hence $\boxed{{\mathcal{O}}_K\text{ is a ring}}$.

The integral basis of the cubic field of discriminant minus 140 can be obtained as follows. For the specified field, Eisenstein at two makes $X^3+2X+2$ irreducible, and $\theta$ is integral because the polynomial is monic. The discriminant of a monic cubic $X^3+rX+s$ is $-4r^3-27s^2$, soHere ${\Delta }^2$ is the printed notation for the signed discriminant, namely the square of a Vandermonde product of the conjugates; that product need not be real.

Suppose $\eta =(a+b\theta +c{\theta }^2)/2$ is integral with the prescribed binary coefficients. The root traces give $\mathrm{Tr}(1)=3$, $\mathrm{Tr}(\theta )=0$, $\mathrm{Tr}({\theta }^2)=-4$. Thus $\mathrm{Tr}(\eta )=(3a-4c)/2$ must be an integer, forcing $a=0$. If $b=0$, nonzeroness of the coefficient triple forces $c=1$, giving $\eta ={\theta }^2/2$. If $b=1$, multiply by $\theta$ and use ${\theta }^3=-2\theta -2$:Closure of the ring then again gives ${\theta }^2/2\in {\mathcal{O}}_K$.

But $N_{K/\mathbb{Q}}(\theta )=-2$, so the field norm of ${\theta }^{-1}$ is $-1/2$. The field norm of an algebraic integer is an integer, proving ${\theta }^{-1}\notin {\mathcal{O}}_K$. Dividing the defining cubic by $2\theta$ gives ${\theta }^2/2=-1-{\theta }^{-1}$. Thus ${\theta }^2/2$ cannot be integral either.

Finally the discriminant-index formula for an integral lattice says $-140=[{\mathcal{O}}_K:\mathbb{Z}[\theta ]{]}^2\mathrm{disc}(K)$. The square of the integer index divides $140$, so its only possibilities are one and four, giving index one or two. If the index were two, a nonzero coset would supply an integral element $(a+b\theta +c{\theta }^2)/2$ after reducing coefficients modulo two, with a nonzero binary triple. The preceding argument would force the impossible integrality of ${\theta }^2/2$. Hence