Let $K$ be a number field. Then is it true that $1+u$ is a non-unit for every $u\in\mathcal{O}_K^*$? I believe this is true. I have worked out a few examples and in each one of them this holds. But I cannot find a general proof anywhere. Any help is appreciated.
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Your title says "ring" but your question says "field". – Q the Platypus Jul 05 '23 at 03:32
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K is a field and $\mathcal{O}_K$ is its ring of integers, where $1+u$ lies. – ShyamalSayak Jul 05 '23 at 03:33
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Also do you mean a finite field. Because the integers don't form a field. – Q the Platypus Jul 05 '23 at 03:34
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2Of course not. A number field is never finite. I think you are misunderstanding the question. – ShyamalSayak Jul 05 '23 at 03:35
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@QthePlatypus https://en.wikipedia.org/wiki/Ring_of_integers – CyclotomicField Jul 05 '23 at 03:36
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3Hint for a counterexample: consider roots of unity. – Greg Martin Jul 05 '23 at 03:37
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@GregMartin Yes, you are right. In $\mathbb{Q}(\zeta_3)$ this is not true. Is there a characterization of such typical ring of integers? This holds for all $K$'s such that $\mathbb{O}_K^*={\pm 1}$ and also when $K=\mathbb{Q}(i)$. – ShyamalSayak Jul 05 '23 at 03:39
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1Generalizing the example with third/sixth roots of unity a little bit: Whenever $u$ satisfies an equation of the form $1+u=\pm u^n$ it follows that $u$ is a unit of the ring of integers of some field. And obviously so is $1+u$. Further generalizations can be cooked up. – Jyrki Lahtonen Jul 05 '23 at 04:02
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The other extreme - the sum of two units is always a unit (or 0) - is discussed here and here. – Bill Dubuque Jul 06 '23 at 11:31