Let $K$ be an algebraic extension of the rationals of degree $n$ and let $\{ 1, \omega_2, \omega_3, \dots , \omega_n \}$ be an integral basis for the ring of integers $\mathcal{O}_K$ of $K$. Let $\alpha \in \mathcal{O}_K$ be primitive, meaning no rational prime divides $\alpha$, and assume that the ideal $(\alpha )$ is prime to the discriminat of $K$. Does there exist an integer $\beta \in \mathcal{O}_K$ such that $\alpha \beta = 1 + c_2 \omega_2 + c_3 \omega_3 + \dots + c_n \omega_n$ , where the $c_i \in \mathbb{Z}$ ?

The question has been modified after Professor Lozano-Robledo's excellent answer. I added the condition that $(\alpha )$ is prime to the discriminat of $K$.

  • $\begingroup$ Where $c_2,\ldots,c_n$ are arbitrary coefficients in $\mathbb{Q}$, I assume? $\endgroup$ – Alex Becker Jan 14 '12 at 2:09
  • $\begingroup$ Thanks Alex! I edited the question. $\endgroup$ – Samuel Hambleton Jan 14 '12 at 2:17

The answer is no, in general. Here is a counterexample.

Let $K=\mathbb{Q}(\sqrt[3]{2})$. The ring of integers is $\mathcal{O}_K = \mathbb{Z}[\sqrt[3]{2}]$ (for this see for instance these notes by Keith Conrad). Let $\alpha = \sqrt[3]{2}$. Then $\alpha$ is "primitive" (the norm of $\alpha$ is $2$, and $2=\alpha^3$, so no rational prime divides $\alpha$). Now let $\beta\in \mathbb{Z}[\sqrt[3]{2}]$ be arbitrary, i.e., $\beta=a+b\sqrt[3]{2}+c\sqrt[3]{4}$. Then: $$\alpha\cdot \beta = 2c+a\sqrt[3]{2}+b\sqrt[3]{4}.$$ Therefore, $\alpha\cdot\beta$ is never equal to a number of the form $1+d\sqrt[3]{2}+e\sqrt[3]{4}$, because $2c$ is always even.

Edit in response to changes by OP to the original question: What if we further assume that $\alpha$ is prime to the discriminant of the field? The answer is still no, in general.

Let once again $K=\mathbb{Q}(\sqrt[3]{2})$ with ring of integers $\mathcal{O}_K = \mathbb{Z}[\sqrt[3]{2}]$. This time, choose as a $\mathbb{Z}$-basis elements $\{1,w_2,w_3\}$ where $$w_2=3+\sqrt[3]{2},\ w_3=w_2^2 = (3+\sqrt[3]{2})^2.$$ It is easy to see that $\{1,w_2,w_3\}$ forms a $\mathbb{Z}$-basis of $\mathbb{Z}[\sqrt[3]{2}]$ because $\sqrt[3]{2}=w_2-3$ and $\sqrt[3]{4}=w_3-6w_2+9$. Now let $\alpha=w_2=3+\sqrt[3]{2}$. The norm of $\alpha$ is $29$, so $\alpha$ is "primitive" and prime to the discriminant of the field $D_K=2^23^3$. Let $\beta=a+bw_2+cw_2^2$ be an arbitrary element of $\mathbb{Z}[\sqrt[3]{2}]$. Then: $$\alpha\cdot \beta = w_2(a+bw_2+cw_2^2)=aw_2+bw_2^2+cw_2^3.$$ Since $$w_2^3=(3+\sqrt[3]{2})^3 = 29-27w_2+9w_2^3$$ we obtain $$\alpha\cdot \beta = w_2(a+bw_2+cw_2^2)=aw_2+bw_2^2+cw_2^3=29c+(a-27c)w_2+(b+9c)w_2^2.$$ Hence, $\alpha\cdot \beta$ can't be of the form $1+dw_2+ew_2^3$ because $29c$ is always a multiple of $29$.

  • $\begingroup$ Thank you! What if I assume that $\alpha$ is prime to the discriminant of $K$ ? $\endgroup$ – Samuel Hambleton Jan 14 '12 at 2:41
  • $\begingroup$ Sorry I un-accepted your excellent answer! I am requesting more information now so it seems appropriate to un-accept. $\endgroup$ – Samuel Hambleton Jan 14 '12 at 3:11
  • $\begingroup$ I updated my answer accordingly. $\endgroup$ – Álvaro Lozano-Robledo Jan 14 '12 at 3:50
  • $\begingroup$ Thank you very much. I think I'm satisfied now! $\endgroup$ – Samuel Hambleton Jan 14 '12 at 4:12
  • $\begingroup$ @SamuelHambleton : How are you sir, you have answered my previous questions sir, and encouraged me, but after that you never turned to me. Anyway thank you sir for what you did. $\endgroup$ – IDOK Jan 14 '12 at 16:00

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