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$a = 3, b = 1+\sqrt{-26}$ then $(a,b)^3=(a^3,b^3,a^2b,ab^2)$

each generator except $a^3$ has a $b$ factor and $\bar b b=27$, so $"\subseteq"$. Now the question is how to obtain $b$ using these generators. Is there a general formula, which states in which case this is possible or even is there an algorithm. I made it very complicated I think;

$a^3 = 27,\quad$ $b^3 = -77+23\sqrt{-26},\quad$ $a^2b = 9+9\sqrt{-26},\quad$ $ab^2 = -75+6\sqrt{-26}$

$3*a^3+ab^2=81+(-75+6\sqrt{-26})=6+6\sqrt{-26}$

$a^2b-ans = 3+3\sqrt{-26}$

are these correct so far ?

Bill Dubuque
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rsnmrtn
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  • Here it suffices to check $,a,$ is coprime to $\rm\color{#0af}{trace}$ of $,b,,$ indeed $(3,\color{#0af}2)=1,,$ see my answer. – Bill Dubuque Jul 11 '22 at 20:35

3 Answers3

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True by below, $ $ since here: $\,a\!=\!3,\,b+\bar b=\color{#0af}2\,\Rightarrow\, \color{#0a0}{(a,b,\bar b)} = (3,b,\bar b,\color{#0af}2)=1$.

Lemma $\, $ If $\ a^3 = b\bar b\ $ then $\ (a,b)^3 = (b)\iff \color{#0a0}{(a,b,\bar b)}=1$

Proof $\ \ (a,b)^3\! = (b)((\bar b)\!+\!(a,b)^2) = (b)\!\iff\!$ $(\bar b)\!+\!(a,b)^2\! = 1\!\!\!\overset{\color{#c00}{\rm EL}\!\!}\iff\!$ $\color{#0a0}{(\bar b)\!+\!(a,b)}=1$

Bill Dubuque
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I think $b^3$ should be $-77 - 23\sqrt{-26}$ which makes $-(2a^3 + b^3) = 23(1+\sqrt{-26})$ and you have already found $6(1+\sqrt{-26})$ in the ideal. Together they will give you $1+\sqrt{-26}$. I have not come across any general method for these.

Rishi
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    Or $3a^3+\sqrt{-26}a=4b$. As $a^2b=9b$ is in there, the same conclusion follows. Many ways to do this, and yours works. There is a simple algorithmic way of finding generators for this ideal as a free abelian group using Smith normal form type of algorithms. Don't know whether that helps? It should simplify things at the very least. – Jyrki Lahtonen Jul 11 '22 at 18:56
  • I thought there should be algorithms, probably similar to algorithms for solving linear equations. – Rishi Jul 12 '22 at 07:55
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To be explicit, you could use linear algebra:

$$\begin{align} a^3&=27&&\mapsto\begin{bmatrix}27\\0\end{bmatrix}\\ b^3&=-77-23\sqrt{-26}&&\mapsto\begin{bmatrix}-77\\-23\end{bmatrix}\\ a^2b&=9+9\sqrt{-26}&&\mapsto\begin{bmatrix}9\\9\end{bmatrix}\\ ab^2&=-75+6\sqrt{-26}&&\mapsto\begin{bmatrix}-75\\6\end{bmatrix}\\ \end{align}$$

It would suffice to have an integer solution to the equation $$wa^3+xb^3+ya^2b+zab^2=b$$ The corresponding augmented matrix is $$\begin{bmatrix} 27&-77&9&-75&1\\ 0&-23&9&6&1 \end{bmatrix}$$

This reduces to $$\begin{bmatrix} 23&0&-18&-81&-2\\ 0&23&-9&-6&-1 \end{bmatrix}$$

You need choices for $y$ and $z$ such that $18y+81z-2$ and $9y+6z-1$ are each divisible by $23$. In other words, you want to solve $$\begin{bmatrix} 18&81\\ 9&6 \end{bmatrix} \begin{bmatrix} y\\ z \end{bmatrix} \equiv \begin{bmatrix} 2\\ 1 \end{bmatrix}$$

modulo $23$. So we can again use linear algebra (with some division mod $23$) to find

$$ \begin{bmatrix} y\\ z \end{bmatrix} \equiv \begin{bmatrix} 18&81\\ 9&6 \end{bmatrix}^{-1} \begin{bmatrix} 2\\ 1 \end{bmatrix} \equiv \begin{bmatrix} 2\\ 1 \end{bmatrix}$$

We can just go with $y=2$ and $z=1$. This gives $(w,x,y,z)=(5,1,2,1)$. So it seems that $$5a^3+b^3+2a^2b+ab^2=b$$

2'5 9'2
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  • This is good. But numbers like $\sqrt{-26}a$ et cetera are also in the ideal (may or may not be needed to get the desired outcome). – Jyrki Lahtonen Jul 12 '22 at 08:20
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    @JyrkiLahtonen Yes, and maybe it's too subtle but I meant to acknowledge that with the "It would suffice..." line. – 2'5 9'2 Jul 12 '22 at 18:16
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    Sorry, I missed that. Anyway, with this kind of coordinates you always get a basis for the ideal as a free abelian group. And that generalizes to other number fields (once you have identified an integral basis). – Jyrki Lahtonen Jul 12 '22 at 18:57