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Minkowski's lemma states that if $\Lambda$ is a full lattice in ${\mathbb{R}}^n$ and $C\subset {\mathbb{R}}^n$ is measurable, convex, and centrally symmetric, thenimplies that $C$ contains a nonzero point of $\Lambda$.

Let ${\sigma }_1,{\sigma }_2:K\to \mathbb{R}$ be the two real embeddings. The Minkowski embedding of a real quadratic field identifies ${\mathcal{O}}_K$ with a latticeof covolume $\sqrt{|D_K|}$. Choose a constant $c$ satisfyingFor $R>0$, consider the closed diamondIt is convex and centrally symmetric, and its area isMinkowski's lemma supplies a nonzero ${\alpha }_R\in {\mathcal{O}}_K$ whose two embeddings lie in $C_R$.

For $(x,y)\in C_R$, the arithmetic-geometric mean inequality givesSince the coordinate product is the field norm,Finally, $|{\sigma }_2({\alpha }_R)|\le c/R\to 0$ as $R\to \infty$. No fixed nonzero algebraic integer can occur for arbitrarily large $R$, because its second embedding is nonzero. Consequently the elements ${\alpha }_R$ obtained along an unbounded sequence of $R$ contain infinitely many distinct values. Here $N(\alpha )$ denotes the absolute norm, as usual in this inequality.

Since $2\not\equiv 1(\mathrm{mod}4)$, the ring of integers of $\mathbb{Q}(\sqrt{2})$ iswith integral basis $1,\sqrt{2}$ and field discriminant eight. Multiplication is ordinary polynomial multiplication followed by $(\sqrt{2}{)}^2=2$:For example, $(1+\sqrt{2}{)}^2=3+2\sqrt{2}$ and $(1+\sqrt{2}{)}^3=7+5\sqrt{2}$. The integer coefficients stay integral, as required for a ring.

Under the two field embeddings, the element $a+b\sqrt{2}$ becomes the point $(a+b\sqrt{2},a-b\sqrt{2})$. The integral basis vectors become $(1,1)$ and $(\sqrt{2},-\sqrt{2})$, giving a Minkowski embedding of a real quadratic field. Their determinant has absolute value $2\sqrt{2}=\sqrt{8}$, consistent with the field discriminant.