# Translations of rings of integers by algebraic integers

I am curious about a set of related elementary questions in algebraic number theory about translations of rings of integers of number fields by elements algebraic over them. The most basic one is this: suppose that $L / K$ is an extension of number fields, and $\alpha$ is an element of the ring of integers $\mathcal{O}_{L}$ with $\alpha \notin K$. Then for how many elements $t \in \mathcal{O}_{K}$ is $\alpha + t$ a unit in $\mathcal{O}_{L}$? I strongly suspect that there can only be finitely many such $t \in \mathcal{O}_{K}$ (it is easy to show that this is the case when $K = \mathbb{Q}$ or $K$ is an imaginary quadratic field so that it has finitely many units), but I don't see how to prove this.

A more general question would be the following. Given a finite set $S$ of primes of $K$ and $\alpha \in \mathcal{O}_{L} \setminus \mathcal{O}_{K}$ as before, for how many $t \in \mathcal{O}_{K}$ is the ideal $(\alpha + t) \subset \mathcal{O}_{L}$ divisible only by primes lying over $S$?

• If $\alpha \notin K$, then $P(t) = N_{L/K}(\alpha + t) \in \mathcal{O}_K[t]$ is not a perfect power (it has at least two distinct roots). You are looking for values of $P(t)$ with $t \in K$ which lie inside a fixed finitely generated subgroup $\Delta$ of $K^{\times}$, namely, the group $\Delta$ generated by elements of $K$ with non-trivial valuation only in $S$. This is finitely generated because it maps to $\mathbf{Z}^{|S|}$ via the valuation map, and the kernel is the unit group which is finitely generated. It's now a general fact that a polynomial $P(t)$ which is not a perfect power ... – Pound Sterling Mar 10 '17 at 17:35
• ... only takes finitely many values in such a set $\Delta$. Here's a proof which is surely overkill: choose a $N$ large enough such that $P(t) = y^N$ has genus at least two. Choose a set of representatives $\delta$ from $\Delta/\Delta^N$, which is finite because $\Delta$ is finitely generated. Then one is reduced to proving that the finitely many equations $P(t) = \delta y^N$ for each $\delta$ have finitely many rational solutions. But this follows from Faltings' Theorem. – Pound Sterling Mar 10 '17 at 17:37
• Thank you, this looks like it should work! Except that I believe we not only need $P(t)$ to not be a perfect power, but also to be squarefree, otherwise the affine curve $T(t) = y^{N}$ won't be smooth. But this is fine, because $L / K$ is inseparable and so all the Galois conjugates of $\alpha$ are distinct. – Jeff Yelton Mar 10 '17 at 18:21
• Yes it does work, that's why I left the comment. Faltings' theorem applies perfectly well to singular curves providing one is careful about how one defines the genus, because one can replace $X$ by its normalization. And $P(t)$ need not be squarefree if $\alpha$ lies in an intermediate field between $L$ and $K$. – Pound Sterling Mar 10 '17 at 18:27