What other kinds of cubic integer rings are there?












3














Given an integer $n in Bbb{Z}$, we understand $root 3 of n$ to mean the number $x in Bbb{R}$ such that $x^3 - n = 0$. Then $Bbb{Q}(root 3 of n) subset Bbb{R}$, right? The same then goes for the ring of algebraic integers $mathcal{O}_{Bbb{Q}(root 3 of n)}$.



Since $root 3 of {-n} = -root 3 of n$, it follows that $Bbb{Q}(root 3 of n) = Bbb{Q}(root 3 of {-n})$ and likewise $mathcal{O}_{Bbb{Q}(root 3 of n)} = mathcal{O}_{Bbb{Q}(root 3 of {-n})}$. So in order for a ring adjoining a cubic root to $Bbb{Q}$ to have complex numbers we need a cubic root that is imaginary or complex.



Therefore, given $$omega = frac{-1 + sqrt{-3}}{2}$$ and $n > 1$, the ring of integers of $Bbb{Q}(omega root 3 of n)$ should contain complex numbers. What about the ring of $Bbb{Q}(i root 3 of n)$? That is a distinct kind of rings than the other two I've mentioned, right?



Is this all correct? What kinds of cubic rings have I overlooked?






abstract-algebra algebraic-number-theory







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  • 2




    Not every cubic extension of $mathbb{Q}$ is obtained by adjoining the cubic root of an integer (or a rational).
    – Arturo Magidin
    2 days ago






  • 5




    The ring $mathbb{Q}(isqrt[3]{n})$ contains $-sqrt[3]{n^2}$ and $-in$, and hence $-i$. In particular, assuming $n$ is not a perfect cube, this ring is of degree $6$, not $3$, over $mathbb{Q}$, as it contains both a cubic extension and a quadratic one.
    – Arturo Magidin
    2 days ago










  • Related: math.stackexchange.com/q/3060484/328173
    – Kenny Lau
    2 days ago
















3














Given an integer $n in Bbb{Z}$, we understand $root 3 of n$ to mean the number $x in Bbb{R}$ such that $x^3 - n = 0$. Then $Bbb{Q}(root 3 of n) subset Bbb{R}$, right? The same then goes for the ring of algebraic integers $mathcal{O}_{Bbb{Q}(root 3 of n)}$.



Since $root 3 of {-n} = -root 3 of n$, it follows that $Bbb{Q}(root 3 of n) = Bbb{Q}(root 3 of {-n})$ and likewise $mathcal{O}_{Bbb{Q}(root 3 of n)} = mathcal{O}_{Bbb{Q}(root 3 of {-n})}$. So in order for a ring adjoining a cubic root to $Bbb{Q}$ to have complex numbers we need a cubic root that is imaginary or complex.



Therefore, given $$omega = frac{-1 + sqrt{-3}}{2}$$ and $n > 1$, the ring of integers of $Bbb{Q}(omega root 3 of n)$ should contain complex numbers. What about the ring of $Bbb{Q}(i root 3 of n)$? That is a distinct kind of rings than the other two I've mentioned, right?



Is this all correct? What kinds of cubic rings have I overlooked?






abstract-algebra algebraic-number-theory







share|cite|improve this question


















  • 2




    Not every cubic extension of $mathbb{Q}$ is obtained by adjoining the cubic root of an integer (or a rational).
    – Arturo Magidin
    2 days ago






  • 5




    The ring $mathbb{Q}(isqrt[3]{n})$ contains $-sqrt[3]{n^2}$ and $-in$, and hence $-i$. In particular, assuming $n$ is not a perfect cube, this ring is of degree $6$, not $3$, over $mathbb{Q}$, as it contains both a cubic extension and a quadratic one.
    – Arturo Magidin
    2 days ago










  • Related: math.stackexchange.com/q/3060484/328173
    – Kenny Lau
    2 days ago














3












3








3







Given an integer $n in Bbb{Z}$, we understand $root 3 of n$ to mean the number $x in Bbb{R}$ such that $x^3 - n = 0$. Then $Bbb{Q}(root 3 of n) subset Bbb{R}$, right? The same then goes for the ring of algebraic integers $mathcal{O}_{Bbb{Q}(root 3 of n)}$.



Since $root 3 of {-n} = -root 3 of n$, it follows that $Bbb{Q}(root 3 of n) = Bbb{Q}(root 3 of {-n})$ and likewise $mathcal{O}_{Bbb{Q}(root 3 of n)} = mathcal{O}_{Bbb{Q}(root 3 of {-n})}$. So in order for a ring adjoining a cubic root to $Bbb{Q}$ to have complex numbers we need a cubic root that is imaginary or complex.



Therefore, given $$omega = frac{-1 + sqrt{-3}}{2}$$ and $n > 1$, the ring of integers of $Bbb{Q}(omega root 3 of n)$ should contain complex numbers. What about the ring of $Bbb{Q}(i root 3 of n)$? That is a distinct kind of rings than the other two I've mentioned, right?



Is this all correct? What kinds of cubic rings have I overlooked?






abstract-algebra algebraic-number-theory







share|cite|improve this question













Given an integer $n in Bbb{Z}$, we understand $root 3 of n$ to mean the number $x in Bbb{R}$ such that $x^3 - n = 0$. Then $Bbb{Q}(root 3 of n) subset Bbb{R}$, right? The same then goes for the ring of algebraic integers $mathcal{O}_{Bbb{Q}(root 3 of n)}$.



Since $root 3 of {-n} = -root 3 of n$, it follows that $Bbb{Q}(root 3 of n) = Bbb{Q}(root 3 of {-n})$ and likewise $mathcal{O}_{Bbb{Q}(root 3 of n)} = mathcal{O}_{Bbb{Q}(root 3 of {-n})}$. So in order for a ring adjoining a cubic root to $Bbb{Q}$ to have complex numbers we need a cubic root that is imaginary or complex.



Therefore, given $$omega = frac{-1 + sqrt{-3}}{2}$$ and $n > 1$, the ring of integers of $Bbb{Q}(omega root 3 of n)$ should contain complex numbers. What about the ring of $Bbb{Q}(i root 3 of n)$? That is a distinct kind of rings than the other two I've mentioned, right?



Is this all correct? What kinds of cubic rings have I overlooked?






abstract-algebra algebraic-number-theory




abstract-algebra algebraic-number-theory






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked 2 days ago









Bob Happ

2501222




2501222








  • 2




    Not every cubic extension of $mathbb{Q}$ is obtained by adjoining the cubic root of an integer (or a rational).
    – Arturo Magidin
    2 days ago






  • 5




    The ring $mathbb{Q}(isqrt[3]{n})$ contains $-sqrt[3]{n^2}$ and $-in$, and hence $-i$. In particular, assuming $n$ is not a perfect cube, this ring is of degree $6$, not $3$, over $mathbb{Q}$, as it contains both a cubic extension and a quadratic one.
    – Arturo Magidin
    2 days ago










  • Related: math.stackexchange.com/q/3060484/328173
    – Kenny Lau
    2 days ago














  • 2




    Not every cubic extension of $mathbb{Q}$ is obtained by adjoining the cubic root of an integer (or a rational).
    – Arturo Magidin
    2 days ago






  • 5




    The ring $mathbb{Q}(isqrt[3]{n})$ contains $-sqrt[3]{n^2}$ and $-in$, and hence $-i$. In particular, assuming $n$ is not a perfect cube, this ring is of degree $6$, not $3$, over $mathbb{Q}$, as it contains both a cubic extension and a quadratic one.
    – Arturo Magidin
    2 days ago










  • Related: math.stackexchange.com/q/3060484/328173
    – Kenny Lau
    2 days ago








2




2




Not every cubic extension of $mathbb{Q}$ is obtained by adjoining the cubic root of an integer (or a rational).
– Arturo Magidin
2 days ago




Not every cubic extension of $mathbb{Q}$ is obtained by adjoining the cubic root of an integer (or a rational).
– Arturo Magidin
2 days ago




5




5




The ring $mathbb{Q}(isqrt[3]{n})$ contains $-sqrt[3]{n^2}$ and $-in$, and hence $-i$. In particular, assuming $n$ is not a perfect cube, this ring is of degree $6$, not $3$, over $mathbb{Q}$, as it contains both a cubic extension and a quadratic one.
– Arturo Magidin
2 days ago




The ring $mathbb{Q}(isqrt[3]{n})$ contains $-sqrt[3]{n^2}$ and $-in$, and hence $-i$. In particular, assuming $n$ is not a perfect cube, this ring is of degree $6$, not $3$, over $mathbb{Q}$, as it contains both a cubic extension and a quadratic one.
– Arturo Magidin
2 days ago












Related: math.stackexchange.com/q/3060484/328173
– Kenny Lau
2 days ago




Related: math.stackexchange.com/q/3060484/328173
– Kenny Lau
2 days ago










1 Answer
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Things are so easy at algebraic degree $1$, we only have $textbf Z$ to worry about.



At algebraic degree $2$, we have infinitely many real quadratic integer rings and infinitely many imaginary quadratic integer rings, but it still feels manageable because we're only dealing with two kinds of rings, plus $textbf Z$.



Construct a polynomial of the form $x^2 + bx pm 1$, where $b in textbf Z$. The "$pm 1$" bit helps us narrow our focus down to units. By the quadratic equation, if we choose to go with $-1$, we have $$x = frac{-b pm sqrt{b^2 + 4}}{2}$$ (in some cases you might have to rewrite things like $sqrt{200}$ so that you have a squarefree number under the radical symbol).



When we go to algebraic degree $3$, the complexity suddenly explodes. And we find that complexity by doing for cubics something similar to what we did for quadratics just now. Construct a polynomial of the form $x^3 + ax^2 + bx pm 1$, where $a in textbf Z$ also. For example, $1 - root 3 of 2$ is a zero of $x^3 - 3x^2 + 3x + 1$.



But try to solve $x^3 + x^2 - x + 1$, that turns out to be way beyond my meager computational skills. I turn to Wolfram Alpha for help, and it gives three answers all involving $root 3 of {19 - 3 sqrt{33}}$. Whoa.



So yeah, you have overlooked a lot of kinds of rings.






share|cite|improve this answer





















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    Things are so easy at algebraic degree $1$, we only have $textbf Z$ to worry about.



    At algebraic degree $2$, we have infinitely many real quadratic integer rings and infinitely many imaginary quadratic integer rings, but it still feels manageable because we're only dealing with two kinds of rings, plus $textbf Z$.



    Construct a polynomial of the form $x^2 + bx pm 1$, where $b in textbf Z$. The "$pm 1$" bit helps us narrow our focus down to units. By the quadratic equation, if we choose to go with $-1$, we have $$x = frac{-b pm sqrt{b^2 + 4}}{2}$$ (in some cases you might have to rewrite things like $sqrt{200}$ so that you have a squarefree number under the radical symbol).



    When we go to algebraic degree $3$, the complexity suddenly explodes. And we find that complexity by doing for cubics something similar to what we did for quadratics just now. Construct a polynomial of the form $x^3 + ax^2 + bx pm 1$, where $a in textbf Z$ also. For example, $1 - root 3 of 2$ is a zero of $x^3 - 3x^2 + 3x + 1$.



    But try to solve $x^3 + x^2 - x + 1$, that turns out to be way beyond my meager computational skills. I turn to Wolfram Alpha for help, and it gives three answers all involving $root 3 of {19 - 3 sqrt{33}}$. Whoa.



    So yeah, you have overlooked a lot of kinds of rings.






    share|cite|improve this answer


























      1














      Things are so easy at algebraic degree $1$, we only have $textbf Z$ to worry about.



      At algebraic degree $2$, we have infinitely many real quadratic integer rings and infinitely many imaginary quadratic integer rings, but it still feels manageable because we're only dealing with two kinds of rings, plus $textbf Z$.



      Construct a polynomial of the form $x^2 + bx pm 1$, where $b in textbf Z$. The "$pm 1$" bit helps us narrow our focus down to units. By the quadratic equation, if we choose to go with $-1$, we have $$x = frac{-b pm sqrt{b^2 + 4}}{2}$$ (in some cases you might have to rewrite things like $sqrt{200}$ so that you have a squarefree number under the radical symbol).



      When we go to algebraic degree $3$, the complexity suddenly explodes. And we find that complexity by doing for cubics something similar to what we did for quadratics just now. Construct a polynomial of the form $x^3 + ax^2 + bx pm 1$, where $a in textbf Z$ also. For example, $1 - root 3 of 2$ is a zero of $x^3 - 3x^2 + 3x + 1$.



      But try to solve $x^3 + x^2 - x + 1$, that turns out to be way beyond my meager computational skills. I turn to Wolfram Alpha for help, and it gives three answers all involving $root 3 of {19 - 3 sqrt{33}}$. Whoa.



      So yeah, you have overlooked a lot of kinds of rings.






      share|cite|improve this answer
























        1












        1








        1






        Things are so easy at algebraic degree $1$, we only have $textbf Z$ to worry about.



        At algebraic degree $2$, we have infinitely many real quadratic integer rings and infinitely many imaginary quadratic integer rings, but it still feels manageable because we're only dealing with two kinds of rings, plus $textbf Z$.



        Construct a polynomial of the form $x^2 + bx pm 1$, where $b in textbf Z$. The "$pm 1$" bit helps us narrow our focus down to units. By the quadratic equation, if we choose to go with $-1$, we have $$x = frac{-b pm sqrt{b^2 + 4}}{2}$$ (in some cases you might have to rewrite things like $sqrt{200}$ so that you have a squarefree number under the radical symbol).



        When we go to algebraic degree $3$, the complexity suddenly explodes. And we find that complexity by doing for cubics something similar to what we did for quadratics just now. Construct a polynomial of the form $x^3 + ax^2 + bx pm 1$, where $a in textbf Z$ also. For example, $1 - root 3 of 2$ is a zero of $x^3 - 3x^2 + 3x + 1$.



        But try to solve $x^3 + x^2 - x + 1$, that turns out to be way beyond my meager computational skills. I turn to Wolfram Alpha for help, and it gives three answers all involving $root 3 of {19 - 3 sqrt{33}}$. Whoa.



        So yeah, you have overlooked a lot of kinds of rings.






        share|cite|improve this answer












        Things are so easy at algebraic degree $1$, we only have $textbf Z$ to worry about.



        At algebraic degree $2$, we have infinitely many real quadratic integer rings and infinitely many imaginary quadratic integer rings, but it still feels manageable because we're only dealing with two kinds of rings, plus $textbf Z$.



        Construct a polynomial of the form $x^2 + bx pm 1$, where $b in textbf Z$. The "$pm 1$" bit helps us narrow our focus down to units. By the quadratic equation, if we choose to go with $-1$, we have $$x = frac{-b pm sqrt{b^2 + 4}}{2}$$ (in some cases you might have to rewrite things like $sqrt{200}$ so that you have a squarefree number under the radical symbol).



        When we go to algebraic degree $3$, the complexity suddenly explodes. And we find that complexity by doing for cubics something similar to what we did for quadratics just now. Construct a polynomial of the form $x^3 + ax^2 + bx pm 1$, where $a in textbf Z$ also. For example, $1 - root 3 of 2$ is a zero of $x^3 - 3x^2 + 3x + 1$.



        But try to solve $x^3 + x^2 - x + 1$, that turns out to be way beyond my meager computational skills. I turn to Wolfram Alpha for help, and it gives three answers all involving $root 3 of {19 - 3 sqrt{33}}$. Whoa.



        So yeah, you have overlooked a lot of kinds of rings.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered yesterday









        David R.

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