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Is it possible, that $p^t$ (with a prime number $p \in \mathbb N$ and $t \in \mathbb N$) is a unit in an algebraic number field $K$ (e.g. a unit in the ring of integers $\mathcal O_K$) ? And if not, why it is not possible?

Thanks in advance,

David

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What is the minimal polynomial of its inverse in $K$? Can it ever be monic? Equivalently, what is its norm? Can it ever be $\pm 1$? –  Qiaochu Yuan Jun 28 '11 at 19:26

2 Answers 2

If $k \in \mathbb Z$ is a unit in a number ring $\mathcal{O}_K$, then there is $\alpha \in \mathcal{O}_K$ such that $k \alpha=1$. But this implies that $\alpha = 1/k \in \mathbb Q$. Then, $\alpha \in \mathbb Q \cap \mathcal{O}_K = \mathbb Z$ (because $\mathbb Z$ is integrally closed) and $k$ is a unit in $\mathbb Z$.

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This can't happen, and that statement can be proven in a much more general setting. If $A \subset B$ are commutative rings, with $B$ integral over $A$, and $f$ is a non-unit of $A$, then $f$ is a non-unit of $B$.

Proof: Let $(f)$ be the ideal of $A$ generated by $f$. Since $f$ is not a unit, $(f)$ is not $A$, so there is a prime ideal $\mathfrak{p}$ with $(f) \subseteq \mathfrak{p}$. (I may have just used the axiom of choice.) By the going up theorem there is a prime ideal $\mathfrak{q}$ of $B$ with $\mathfrak{q} \cap A = \mathfrak{p}$.

Since $B \cap A = A$, we know that $\mathfrak{q} \neq B$. And since $f \in \mathfrak{p}$, we know that $f \in \mathfrak{q}$. So $f$ is not a unit of $B$.

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+1. Minor comment: isn't ${\frak{q}}\neq B$ contained in the fact that $\frak{q}$ is prime? –  Zev Chonoles Jun 29 '11 at 17:57
    
Thanks, David. :) –  Pete L. Clark Jun 29 '11 at 19:24
    
Welcome! Thanks for spurring me to finally figure out which one is going-up, which is going-down, and why the hypotheses differ. –  David Speyer Jun 29 '11 at 20:05

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