Let $k$ be a number field of absolute degree $d$. I consider the absolute codifferent $\mathfrak{d}^{-1}$, which is the dual of the maximal order of $k$ for the form induced by the trace map. I am interested in bounding the minimum of the algebraic norm over $\mathfrak{d}^{-1}$.
For instance when $k$ is a cyclotomic field, we have $\min_{x\in\mathfrak{d}^{-1}} N_{k/\mathbb{Q}}(x) = \Delta^{-1}_{k/\mathbb{Q}}$ the discriminant of the field, since we can show that the codifferent is principal (as the maximal order is also principal, meaning that the inverse of the different of any of its generator spans the codifferent).
In general, we thus have the bound $\min_{x\in\mathfrak{d}^{-1}} N_{k/\mathbb{Q}}(x) \geq \Delta^{-1}_{k/\mathbb{Q}}$ with equality iff the maximal order is principal.
Seeing the codifferent as a $\mathbb{Z}$-lattice for the norm induced by the trace (namely $\|x\|^2 = \textrm{Tr}_{k/\mathbb{Q}}(x\bar{x})$), Hermite-Minkowski theorem asserts that there exists an element $x$ of norm bounded by $$\sqrt{\gamma_d}\textrm{covol}(\mathfrak{d}^{-1}) = \sqrt{\gamma_d}N(\mathfrak{d}^{-1})^\frac1d\Delta_k^\frac1{2d} = \sqrt{\gamma_d}\Delta_k^\frac{-1}{2d}$$ since the norm of the codifferent is $\Delta_k^{-1}$ and the covolume of a fractional ideal saw as a lattice is its norm times the square root of the discriminant of the field. By arithmetic / geometric inequality, the norm of $x$ is then bounded by $\left(\frac{\gamma_d}{d}\right)^\frac d2\frac1{\sqrt{\Delta_k}}$.
All in all we have a tight lower bound in $\frac1{\Delta_k}$ and an upper bound in $\frac{c^d}{\sqrt{\Delta_k}}$ for an small absolute constant $c$. Can we get something finer, or is the upper bound almost tight as well?
