What other kinds of cubic integer rings are there?
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
asked 2 days ago
Bob Happ
2501222
add a comment |
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
asked 2 days ago
Bob Happ
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
add a comment |
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
asked 2 days ago
Bob Happ
2501222
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
asked 2 days ago
Bob Happ
2501222
asked 2 days ago
Bob Happ
2501222
asked 2 days ago
Bob Happ
2501222
asked 2 days ago
Bob Happ
2501222
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
add a comment |
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
add a comment |
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.
answered yesterday
David R.
2511728
add a comment |
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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.
answered yesterday
David R.
2511728
add a comment |
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.
answered yesterday
David R.
2511728
add a comment |
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.
answered yesterday
David R.
2511728
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.
answered yesterday
David R.
2511728
answered yesterday
David R.
2511728
answered yesterday
David R.
2511728
answered yesterday
David R.
2511728
2511728
add a comment |
add a comment |
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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