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Something which is not difficult to prove is that if $K$ is a number field generated by an integer $\theta$, then the ring of integers $\mathfrak{O}_K$ is generated over $\mathbb{Z}$ by $\theta$ and by integers of the form $\frac{a_0+a_1\theta+...+a_{n-1}\theta^{n-1}}{m}$, where $m$ is the absolute value of the discriminant of $K$ and $0\leq a_0,a_1,...,a_{n-1}<m$.

What I'd like to prove is this: if for any prime $p$ dividing $m$ we have that there are no nonzero integers of the form $\frac{a_0+a_1\theta+...+a_{n-1}\theta^{n-1}}{p}$ with $0\leq a_0,a_1,...,a_{n-1}<p$, then $\mathbb{Z}[\theta]$ is the ring of integers of $K$. I've been stuck on this question, yet I feel that it should be easy. Any hints?

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  • $\begingroup$ What would really help me is trying a few specific examples, like maybe $\theta = \sqrt{11}$ or $\theta = \sqrt{13}$. $\endgroup$ – Mr. Brooks May 23 '16 at 20:51
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I think you got it slightly wrong. If $\theta\in\frak O$ generates the field extension $K/\mathbf Q$, then any element of $\frak O$ can be written in the form $$ \frac{a_0+a_1\theta+\cdots+a_{n-1}\theta^{n-1}}{m}, $$ for some integers $a_0,\ldots,a_{n-1}$, where $m$ is a natural number such that its square $m^2$ divides the discriminant $$ \Delta(1,\theta,\ldots,\theta^{n-1})=\det(\sigma_i(\theta^j))^2, $$ the $\sigma_i$ being the embeddings of $K$ in $\overline{\mathbf Q}$. Denoting by $G$ the free abelian subgroup of $K$ generated by $1,\theta,\ldots,\theta^{n-1}$, it means that $$ G\subseteq {\frak O}\subseteq \tfrac1m G. $$ To put it otherwise, the quotient ${\frak O}/G$ is a subgroup of the quotient $(\frac1m G)/G$. The latter group is isomorphic to $(\mathbf Z/m)^n$. If ${\frak O}\neq G$ then ${\frak O}/G$ is a nontrivial subgroup of $(\frac1m G)/G$. Then there is a prime number $p$ dividing $m$ such that ${\frak O}/G$ contains an element of order $p$. This means that there is an element in $\frak O$ of the form $$ \frac{a_0+a_1\theta+\cdots+a_{n-1}\theta^{n-1}}{p}, $$ for some integers $a_0,\ldots,a_{n-1}$, satisfying $0\leq a_i<p$, the $a_i$ not being all zero. This is the contrapositive of the statement that you wanted to prove, I think.

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  • $\begingroup$ Do you have a reference for the first result (the one with $m^2$ dividing the discriminant)? $\endgroup$ – Watson Jul 23 '16 at 14:50

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