For K=Q(i)K=\mathbb Q(i), the is C/Z[i]\mathbb C/\mathbb Z[i]. Write ω=u+iv\omega=u+iv. Translation by ω\omega is ergodic exactly when 1,u,v1,u,v are linearly independent over Q\mathbb Q.

Fourier calculation

Here OK∨=12Z[i]\mathcal O_K^\vee=\tfrac12\mathbb Z[i] and Tr⁡∞z=2Re⁡z\operatorname{Tr}_\infty z=2\operatorname{Re}z. With a=(p−iq)/2a=(p-iq)/2,

Tr⁡∞(aω)=pu+qv,\operatorname{Tr}_\infty(a\omega)=pu+qv,

recovering the ordinary two-torus criterion. The increment 2+i3\sqrt2+i\sqrt3 gives an ergodic translation.

One translation can fail while the joint action succeeds

For ω=2\omega=\sqrt2, translation fixes the imaginary coordinate modulo one, so it is not ergodic. Nevertheless k2∉Z[i]k\sqrt2\notin\mathbb Z[i] for every nonzero k∈Z[i]k\in\mathbb Z[i].

The joint action includes translation by 2\sqrt2 and by i2i\sqrt2. A character e2πi(pRe⁡z+qIm⁡z)e^{2\pi i(p\operatorname{Re}z+q\operatorname{Im}z)} is fixed by both only if p2,q2∈Zp\sqrt2,q\sqrt2\in\mathbb Z, forcing p=q=0p=q=0. The action is therefore ergodic although neither generator is individually ergodic.