Mixing
What other kinds of cubic integer rings are there?
$begingroup$
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 Jan 3 at 22:32
Bob HappBob Happ
2731222
$endgroup$
add a comment |
$begingroup$
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 Jan 3 at 22:32
Bob HappBob Happ
2731222
$endgroup$
2
$begingroup$
Not every cubic extension of $mathbb{Q}$ is obtained by adjoining the cubic root of an integer (or a rational).
$endgroup$
– Arturo Magidin
Jan 3 at 22:34
8
$begingroup$
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.
$endgroup$
– Arturo Magidin
Jan 3 at 22:36
1
$begingroup$
Related: math.stackexchange.com/q/3060484/328173
$endgroup$
– Kenny Lau
Jan 4 at 7:28
add a comment |
$begingroup$
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 Jan 3 at 22:32
Bob HappBob Happ
2731222
$endgroup$
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 Jan 3 at 22:32
Bob HappBob Happ
2731222
asked Jan 3 at 22:32
Bob HappBob Happ
2731222
asked Jan 3 at 22:32
Bob HappBob Happ
2731222
asked Jan 3 at 22:32
Bob HappBob Happ
2731222
asked Jan 3 at 22:32
Bob HappBob Happ
2731222
2731222
2
$begingroup$
Not every cubic extension of $mathbb{Q}$ is obtained by adjoining the cubic root of an integer (or a rational).
$endgroup$
– Arturo Magidin
Jan 3 at 22:34
8
$begingroup$
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.
$endgroup$
– Arturo Magidin
Jan 3 at 22:36
1
$begingroup$
Related: math.stackexchange.com/q/3060484/328173
$endgroup$
– Kenny Lau
Jan 4 at 7:28
add a comment |
2
$begingroup$
Not every cubic extension of $mathbb{Q}$ is obtained by adjoining the cubic root of an integer (or a rational).
$endgroup$
– Arturo Magidin
Jan 3 at 22:34
8
$begingroup$
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.
$endgroup$
– Arturo Magidin
Jan 3 at 22:36
1
$begingroup$
Related: math.stackexchange.com/q/3060484/328173
$endgroup$
– Kenny Lau
Jan 4 at 7:28
2
2
$begingroup$
Not every cubic extension of $mathbb{Q}$ is obtained by adjoining the cubic root of an integer (or a rational).
$endgroup$
– Arturo Magidin
Jan 3 at 22:34
$begingroup$
Not every cubic extension of $mathbb{Q}$ is obtained by adjoining the cubic root of an integer (or a rational).
$endgroup$
– Arturo Magidin
Jan 3 at 22:34
8
8
$begingroup$
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.
$endgroup$
– Arturo Magidin
Jan 3 at 22:36
$begingroup$
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.
$endgroup$
– Arturo Magidin
Jan 3 at 22:36
1
1
$begingroup$
Related: math.stackexchange.com/q/3060484/328173
$endgroup$
– Kenny Lau
Jan 4 at 7:28
$begingroup$
Related: math.stackexchange.com/q/3060484/328173
$endgroup$
– Kenny Lau
Jan 4 at 7:28
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
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 Jan 4 at 22:27
David R.David R.
3391728
$endgroup$
add a comment |
$begingroup$
I am copying the information below from Daniel Harrer's diploma thesis, section 2.3; I believe it is originally due to Delone and Fadeev.
For any integers $a$, $b$, $c$, $d$, there is a commutative associative ring with basis $1$, $alpha$, $beta$ and multiplication table
$$begin{array}{rccc}
alpha^2 &=& b alpha &- a beta &-ac \
beta^2 &=& d alpha &- c beta &-bd \
alpha beta &=& & & -ad \
end{array}$$
All cubic integral domains, and more generally all nilpotent-free cubic rings, are of this form.
These formulas define an integral domain if and only if the cubic polynomial $a z^3+b z^2 y + c z + d$ is irreducible. Indeed, if $omega$ is a root of the irreducible $a z^3+b z^2 y + c z + d$, then we can embed this ring in $mathbb{C}$ by $alpha = - a omega$ and $beta = d/omega$. (This paragraph isn't in Harrer.)
Two such cubic rings are isomorphic if and only if the cubics $ax^3+bx^2y+cxy^2+dy^3$ and $a'x^3+b'x^2y+c'xy^2+d'y^3$ are related by $GL_2(mathbb{Z})$ change of variables in $(x,y)$. (Here $GL_2(mathbb{Z})$ is $2 times 2$ integer matrices with determinant $pm 1$.) See Harrer for details.
I haven't seen anyone do this, but it seems natural to me: Let $omega_1$, $omega_2$, $omega_3$ be the roots of the cubic $az^3+bz^2+cz+d$ in $mathbb{C}$. Then $GL_2(mathbb{Z})$ acts on these by Mobius transformations. If we restrict ourself to cubic fields with one real embedding, then one of the $omega$'s will be in the upper half plane. We can use the $GL_2(mathbb{Z})$ action to move it into the fundamental domain $|mathrm{Re}(omega)|<1/2$, $|omega|>1$. So it seems to me that cubic rings with one real embedding have a natural canonical form where we require $az^3+bz^2+cz+d$ to have a root in the fundamental domain.
answered Jan 11 at 15:40
David E SpeyerDavid E Speyer
45.4k4125205
$endgroup$
add a comment |
$begingroup$
It is a bit unclear to me exactly what kind of an answer you are looking for. Listing the following points that occurred to me while reading the question.
- You don't need to use $omega$ to get a cubic extension field containing non-real numbers. Any cubic polynomial $f(x)$, irreducible over $Bbb{Q}$, such that $f$ has only a single real zero will yield such fields. For example, $f(x)=x^3-2x+2$, irreducible by Eisenstein, has a single real zero $in(-2,-1)$. If $alpha$ is one of the complex zeros of $f$, then $K_1=Bbb{Q}(alpha)$ is a cubic field, and $Bbb{Z}[alpha]$ is a ring that as a an abelian group is free of rank three. The field $K_1$ cannot be gotten by adjoining a cube root (real or complex) of an integer to $Bbb{Q}$. A way of seeing that is to use the discriminants. The discriminant of $f(x)$ is $d(f)=-76$, implying that the discriminant of $mathcal{O}_{K_1}$ is either $-76$ or $-19$. The discriminant of $g(x)=x^3-n$ is $d(g)=-27n^2$. The discriminants of orders of the same field can only differ by a factor that is a square, making it impossible for $mathcal{O}_{K_1}$ to contain $Bbb{Z}[omega^jroot3of n]$ for any alternative $j=0,1,2$ (so any root of $g(x)$).
- Some cubic extension fields of $Bbb{Q}$ are Galois extensions, implying that the minimal polynomials of all the elements have real zeros (all in that same field). An example appearing possibly the most frequently in our website is $K=Bbb{Q}(2cos(2pi/9))$. The zeros of the irreducible polynomial $p(x)=x^3-3x+1$ are $2cos(2^kpi/9)$ with $k=1,2,3$, cyclically permuted by the mapping $rmapsto r^2-2$. The ring of integers of $K$ is $Bbb{Z}[2cos(2pi/9)]$. Here the discriminant is $d(p)=81$, the square of a rational number. This happens always with cubic Galois extensions.
- The third primitive root of unity $omega$ does play the following important role (look up Kummer theory for details and generalizations). If, instead of $Bbb{Q}$, we use $F=Bbb{Q}(omega)$ as our base field, then we have the following characterization of cubic Galois extensions of $F$.
Assume that $L$ is a cubic Galois extension of $F$, then there exists an element $zin L$ such that $alpha=z^3in F$. In other words $L=F(root3ofalpha)$.
Of course, in the last bullet, $F$ has other types of cubic extensions, but none of those are Galois over $F$.
answered Jan 6 at 14:35
Jyrki LahtonenJyrki Lahtonen
108k12166369
$endgroup$
$begingroup$
Very helpful, thank you. What I still don't know is a lot of the terminology: like that ring you mentioned in point 1, from the polynomial $x^3 - 2x + 2$, what do you call that one? For a cubefree integer $n$, what do you call $mathbb Q(root 3 of n)$? Adjoining a "trigonometric" number like in point 2, that never gives a quadratic ring, right?
$endgroup$
– Bob Happ
Jan 9 at 21:55
$begingroup$
One more thing: from the link Brooks gave, I found lmfdb.org/NumberField/3.1.140.1 Is that the same as or different from the ring you gave in point 1?
$endgroup$
– Bob Happ
Jan 9 at 21:56
1
$begingroup$
$Bbb{Q}(cos(pi/4))$ and $Bbb{Q}(2cos(2pi/5))$ are quadratic. The former is $Bbb{Q}(sqrt2)$ and the latter $Bbb{Q}(sqrt5)$. I don't think that cubic in item 1. has a special name.
$endgroup$
– Jyrki Lahtonen
Jan 9 at 23:24
3
$begingroup$
No, The example in item 1. is this cubic in LMFDB. The splitting fields of $f(x)$ and $f(-x)$ are always the same.
$endgroup$
– Jyrki Lahtonen
Jan 9 at 23:31
add a comment |
$begingroup$
Hmmm... our own Franz Lemmermeyer really needs to weigh in on this one. Seven years ago he wrote a paper on the "simplest cubic fields" in Manuscripta Mathematica.
There are
- Simply real cubic rings
- Totally real cubic rings
I'm not sure what the distinction is. Then of course there are imaginary cubic rings, whether or not they're obtained by adjoining the cubic root of an integer or by some other adjunction.
$endgroup$
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3061094%2fwhat-other-kinds-of-cubic-integer-rings-are-there%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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 Jan 4 at 22:27
David R.David R.
3391728
$endgroup$
add a comment |
$begingroup$
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 Jan 4 at 22:27
David R.David R.
3391728
$endgroup$
add a comment |
$begingroup$
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 Jan 4 at 22:27
David R.David R.
3391728
$endgroup$
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 Jan 4 at 22:27
David R.David R.
3391728
answered Jan 4 at 22:27
David R.David R.
3391728
answered Jan 4 at 22:27
David R.David R.
3391728
answered Jan 4 at 22:27
David R.David R.
3391728
3391728
add a comment |
add a comment |
$begingroup$
I am copying the information below from Daniel Harrer's diploma thesis, section 2.3; I believe it is originally due to Delone and Fadeev.
For any integers $a$, $b$, $c$, $d$, there is a commutative associative ring with basis $1$, $alpha$, $beta$ and multiplication table
$$begin{array}{rccc}
alpha^2 &=& b alpha &- a beta &-ac \
beta^2 &=& d alpha &- c beta &-bd \
alpha beta &=& & & -ad \
end{array}$$
All cubic integral domains, and more generally all nilpotent-free cubic rings, are of this form.
These formulas define an integral domain if and only if the cubic polynomial $a z^3+b z^2 y + c z + d$ is irreducible. Indeed, if $omega$ is a root of the irreducible $a z^3+b z^2 y + c z + d$, then we can embed this ring in $mathbb{C}$ by $alpha = - a omega$ and $beta = d/omega$. (This paragraph isn't in Harrer.)
Two such cubic rings are isomorphic if and only if the cubics $ax^3+bx^2y+cxy^2+dy^3$ and $a'x^3+b'x^2y+c'xy^2+d'y^3$ are related by $GL_2(mathbb{Z})$ change of variables in $(x,y)$. (Here $GL_2(mathbb{Z})$ is $2 times 2$ integer matrices with determinant $pm 1$.) See Harrer for details.
I haven't seen anyone do this, but it seems natural to me: Let $omega_1$, $omega_2$, $omega_3$ be the roots of the cubic $az^3+bz^2+cz+d$ in $mathbb{C}$. Then $GL_2(mathbb{Z})$ acts on these by Mobius transformations. If we restrict ourself to cubic fields with one real embedding, then one of the $omega$'s will be in the upper half plane. We can use the $GL_2(mathbb{Z})$ action to move it into the fundamental domain $|mathrm{Re}(omega)|<1/2$, $|omega|>1$. So it seems to me that cubic rings with one real embedding have a natural canonical form where we require $az^3+bz^2+cz+d$ to have a root in the fundamental domain.
answered Jan 11 at 15:40
David E SpeyerDavid E Speyer
45.4k4125205
$endgroup$
add a comment |
$begingroup$
I am copying the information below from Daniel Harrer's diploma thesis, section 2.3; I believe it is originally due to Delone and Fadeev.
For any integers $a$, $b$, $c$, $d$, there is a commutative associative ring with basis $1$, $alpha$, $beta$ and multiplication table
$$begin{array}{rccc}
alpha^2 &=& b alpha &- a beta &-ac \
beta^2 &=& d alpha &- c beta &-bd \
alpha beta &=& & & -ad \
end{array}$$
All cubic integral domains, and more generally all nilpotent-free cubic rings, are of this form.
These formulas define an integral domain if and only if the cubic polynomial $a z^3+b z^2 y + c z + d$ is irreducible. Indeed, if $omega$ is a root of the irreducible $a z^3+b z^2 y + c z + d$, then we can embed this ring in $mathbb{C}$ by $alpha = - a omega$ and $beta = d/omega$. (This paragraph isn't in Harrer.)
Two such cubic rings are isomorphic if and only if the cubics $ax^3+bx^2y+cxy^2+dy^3$ and $a'x^3+b'x^2y+c'xy^2+d'y^3$ are related by $GL_2(mathbb{Z})$ change of variables in $(x,y)$. (Here $GL_2(mathbb{Z})$ is $2 times 2$ integer matrices with determinant $pm 1$.) See Harrer for details.
I haven't seen anyone do this, but it seems natural to me: Let $omega_1$, $omega_2$, $omega_3$ be the roots of the cubic $az^3+bz^2+cz+d$ in $mathbb{C}$. Then $GL_2(mathbb{Z})$ acts on these by Mobius transformations. If we restrict ourself to cubic fields with one real embedding, then one of the $omega$'s will be in the upper half plane. We can use the $GL_2(mathbb{Z})$ action to move it into the fundamental domain $|mathrm{Re}(omega)|<1/2$, $|omega|>1$. So it seems to me that cubic rings with one real embedding have a natural canonical form where we require $az^3+bz^2+cz+d$ to have a root in the fundamental domain.
answered Jan 11 at 15:40
David E SpeyerDavid E Speyer
45.4k4125205
$endgroup$
add a comment |
$begingroup$
I am copying the information below from Daniel Harrer's diploma thesis, section 2.3; I believe it is originally due to Delone and Fadeev.
For any integers $a$, $b$, $c$, $d$, there is a commutative associative ring with basis $1$, $alpha$, $beta$ and multiplication table
$$begin{array}{rccc}
alpha^2 &=& b alpha &- a beta &-ac \
beta^2 &=& d alpha &- c beta &-bd \
alpha beta &=& & & -ad \
end{array}$$
All cubic integral domains, and more generally all nilpotent-free cubic rings, are of this form.
These formulas define an integral domain if and only if the cubic polynomial $a z^3+b z^2 y + c z + d$ is irreducible. Indeed, if $omega$ is a root of the irreducible $a z^3+b z^2 y + c z + d$, then we can embed this ring in $mathbb{C}$ by $alpha = - a omega$ and $beta = d/omega$. (This paragraph isn't in Harrer.)
Two such cubic rings are isomorphic if and only if the cubics $ax^3+bx^2y+cxy^2+dy^3$ and $a'x^3+b'x^2y+c'xy^2+d'y^3$ are related by $GL_2(mathbb{Z})$ change of variables in $(x,y)$. (Here $GL_2(mathbb{Z})$ is $2 times 2$ integer matrices with determinant $pm 1$.) See Harrer for details.
I haven't seen anyone do this, but it seems natural to me: Let $omega_1$, $omega_2$, $omega_3$ be the roots of the cubic $az^3+bz^2+cz+d$ in $mathbb{C}$. Then $GL_2(mathbb{Z})$ acts on these by Mobius transformations. If we restrict ourself to cubic fields with one real embedding, then one of the $omega$'s will be in the upper half plane. We can use the $GL_2(mathbb{Z})$ action to move it into the fundamental domain $|mathrm{Re}(omega)|<1/2$, $|omega|>1$. So it seems to me that cubic rings with one real embedding have a natural canonical form where we require $az^3+bz^2+cz+d$ to have a root in the fundamental domain.
answered Jan 11 at 15:40
David E SpeyerDavid E Speyer
45.4k4125205
$endgroup$
I am copying the information below from Daniel Harrer's diploma thesis, section 2.3; I believe it is originally due to Delone and Fadeev.
For any integers $a$, $b$, $c$, $d$, there is a commutative associative ring with basis $1$, $alpha$, $beta$ and multiplication table
$$begin{array}{rccc}
alpha^2 &=& b alpha &- a beta &-ac \
beta^2 &=& d alpha &- c beta &-bd \
alpha beta &=& & & -ad \
end{array}$$
All cubic integral domains, and more generally all nilpotent-free cubic rings, are of this form.
These formulas define an integral domain if and only if the cubic polynomial $a z^3+b z^2 y + c z + d$ is irreducible. Indeed, if $omega$ is a root of the irreducible $a z^3+b z^2 y + c z + d$, then we can embed this ring in $mathbb{C}$ by $alpha = - a omega$ and $beta = d/omega$. (This paragraph isn't in Harrer.)
Two such cubic rings are isomorphic if and only if the cubics $ax^3+bx^2y+cxy^2+dy^3$ and $a'x^3+b'x^2y+c'xy^2+d'y^3$ are related by $GL_2(mathbb{Z})$ change of variables in $(x,y)$. (Here $GL_2(mathbb{Z})$ is $2 times 2$ integer matrices with determinant $pm 1$.) See Harrer for details.
I haven't seen anyone do this, but it seems natural to me: Let $omega_1$, $omega_2$, $omega_3$ be the roots of the cubic $az^3+bz^2+cz+d$ in $mathbb{C}$. Then $GL_2(mathbb{Z})$ acts on these by Mobius transformations. If we restrict ourself to cubic fields with one real embedding, then one of the $omega$'s will be in the upper half plane. We can use the $GL_2(mathbb{Z})$ action to move it into the fundamental domain $|mathrm{Re}(omega)|<1/2$, $|omega|>1$. So it seems to me that cubic rings with one real embedding have a natural canonical form where we require $az^3+bz^2+cz+d$ to have a root in the fundamental domain.
answered Jan 11 at 15:40
David E SpeyerDavid E Speyer
45.4k4125205
edited Jan 12 at 14:21
answered Jan 11 at 15:40
David E SpeyerDavid E Speyer
45.4k4125205
answered Jan 11 at 15:40
David E SpeyerDavid E Speyer
45.4k4125205
answered Jan 11 at 15:40
David E SpeyerDavid E Speyer
45.4k4125205
45.4k4125205
add a comment |
add a comment |
$begingroup$
It is a bit unclear to me exactly what kind of an answer you are looking for. Listing the following points that occurred to me while reading the question.
- You don't need to use $omega$ to get a cubic extension field containing non-real numbers. Any cubic polynomial $f(x)$, irreducible over $Bbb{Q}$, such that $f$ has only a single real zero will yield such fields. For example, $f(x)=x^3-2x+2$, irreducible by Eisenstein, has a single real zero $in(-2,-1)$. If $alpha$ is one of the complex zeros of $f$, then $K_1=Bbb{Q}(alpha)$ is a cubic field, and $Bbb{Z}[alpha]$ is a ring that as a an abelian group is free of rank three. The field $K_1$ cannot be gotten by adjoining a cube root (real or complex) of an integer to $Bbb{Q}$. A way of seeing that is to use the discriminants. The discriminant of $f(x)$ is $d(f)=-76$, implying that the discriminant of $mathcal{O}_{K_1}$ is either $-76$ or $-19$. The discriminant of $g(x)=x^3-n$ is $d(g)=-27n^2$. The discriminants of orders of the same field can only differ by a factor that is a square, making it impossible for $mathcal{O}_{K_1}$ to contain $Bbb{Z}[omega^jroot3of n]$ for any alternative $j=0,1,2$ (so any root of $g(x)$).
- Some cubic extension fields of $Bbb{Q}$ are Galois extensions, implying that the minimal polynomials of all the elements have real zeros (all in that same field). An example appearing possibly the most frequently in our website is $K=Bbb{Q}(2cos(2pi/9))$. The zeros of the irreducible polynomial $p(x)=x^3-3x+1$ are $2cos(2^kpi/9)$ with $k=1,2,3$, cyclically permuted by the mapping $rmapsto r^2-2$. The ring of integers of $K$ is $Bbb{Z}[2cos(2pi/9)]$. Here the discriminant is $d(p)=81$, the square of a rational number. This happens always with cubic Galois extensions.
- The third primitive root of unity $omega$ does play the following important role (look up Kummer theory for details and generalizations). If, instead of $Bbb{Q}$, we use $F=Bbb{Q}(omega)$ as our base field, then we have the following characterization of cubic Galois extensions of $F$.
Assume that $L$ is a cubic Galois extension of $F$, then there exists an element $zin L$ such that $alpha=z^3in F$. In other words $L=F(root3ofalpha)$.
Of course, in the last bullet, $F$ has other types of cubic extensions, but none of those are Galois over $F$.
answered Jan 6 at 14:35
Jyrki LahtonenJyrki Lahtonen
108k12166369
$endgroup$
$begingroup$
Very helpful, thank you. What I still don't know is a lot of the terminology: like that ring you mentioned in point 1, from the polynomial $x^3 - 2x + 2$, what do you call that one? For a cubefree integer $n$, what do you call $mathbb Q(root 3 of n)$? Adjoining a "trigonometric" number like in point 2, that never gives a quadratic ring, right?
$endgroup$
– Bob Happ
Jan 9 at 21:55
$begingroup$
One more thing: from the link Brooks gave, I found lmfdb.org/NumberField/3.1.140.1 Is that the same as or different from the ring you gave in point 1?
$endgroup$
– Bob Happ
Jan 9 at 21:56
1
$begingroup$
$Bbb{Q}(cos(pi/4))$ and $Bbb{Q}(2cos(2pi/5))$ are quadratic. The former is $Bbb{Q}(sqrt2)$ and the latter $Bbb{Q}(sqrt5)$. I don't think that cubic in item 1. has a special name.
$endgroup$
– Jyrki Lahtonen
Jan 9 at 23:24
3
$begingroup$
No, The example in item 1. is this cubic in LMFDB. The splitting fields of $f(x)$ and $f(-x)$ are always the same.
$endgroup$
– Jyrki Lahtonen
Jan 9 at 23:31
add a comment |
$begingroup$
It is a bit unclear to me exactly what kind of an answer you are looking for. Listing the following points that occurred to me while reading the question.
- You don't need to use $omega$ to get a cubic extension field containing non-real numbers. Any cubic polynomial $f(x)$, irreducible over $Bbb{Q}$, such that $f$ has only a single real zero will yield such fields. For example, $f(x)=x^3-2x+2$, irreducible by Eisenstein, has a single real zero $in(-2,-1)$. If $alpha$ is one of the complex zeros of $f$, then $K_1=Bbb{Q}(alpha)$ is a cubic field, and $Bbb{Z}[alpha]$ is a ring that as a an abelian group is free of rank three. The field $K_1$ cannot be gotten by adjoining a cube root (real or complex) of an integer to $Bbb{Q}$. A way of seeing that is to use the discriminants. The discriminant of $f(x)$ is $d(f)=-76$, implying that the discriminant of $mathcal{O}_{K_1}$ is either $-76$ or $-19$. The discriminant of $g(x)=x^3-n$ is $d(g)=-27n^2$. The discriminants of orders of the same field can only differ by a factor that is a square, making it impossible for $mathcal{O}_{K_1}$ to contain $Bbb{Z}[omega^jroot3of n]$ for any alternative $j=0,1,2$ (so any root of $g(x)$).
- Some cubic extension fields of $Bbb{Q}$ are Galois extensions, implying that the minimal polynomials of all the elements have real zeros (all in that same field). An example appearing possibly the most frequently in our website is $K=Bbb{Q}(2cos(2pi/9))$. The zeros of the irreducible polynomial $p(x)=x^3-3x+1$ are $2cos(2^kpi/9)$ with $k=1,2,3$, cyclically permuted by the mapping $rmapsto r^2-2$. The ring of integers of $K$ is $Bbb{Z}[2cos(2pi/9)]$. Here the discriminant is $d(p)=81$, the square of a rational number. This happens always with cubic Galois extensions.
- The third primitive root of unity $omega$ does play the following important role (look up Kummer theory for details and generalizations). If, instead of $Bbb{Q}$, we use $F=Bbb{Q}(omega)$ as our base field, then we have the following characterization of cubic Galois extensions of $F$.
Assume that $L$ is a cubic Galois extension of $F$, then there exists an element $zin L$ such that $alpha=z^3in F$. In other words $L=F(root3ofalpha)$.
Of course, in the last bullet, $F$ has other types of cubic extensions, but none of those are Galois over $F$.
answered Jan 6 at 14:35
Jyrki LahtonenJyrki Lahtonen
108k12166369
$endgroup$
$begingroup$
Very helpful, thank you. What I still don't know is a lot of the terminology: like that ring you mentioned in point 1, from the polynomial $x^3 - 2x + 2$, what do you call that one? For a cubefree integer $n$, what do you call $mathbb Q(root 3 of n)$? Adjoining a "trigonometric" number like in point 2, that never gives a quadratic ring, right?
$endgroup$
– Bob Happ
Jan 9 at 21:55
$begingroup$
One more thing: from the link Brooks gave, I found lmfdb.org/NumberField/3.1.140.1 Is that the same as or different from the ring you gave in point 1?
$endgroup$
– Bob Happ
Jan 9 at 21:56
1
$begingroup$
$Bbb{Q}(cos(pi/4))$ and $Bbb{Q}(2cos(2pi/5))$ are quadratic. The former is $Bbb{Q}(sqrt2)$ and the latter $Bbb{Q}(sqrt5)$. I don't think that cubic in item 1. has a special name.
$endgroup$
– Jyrki Lahtonen
Jan 9 at 23:24
3
$begingroup$
No, The example in item 1. is this cubic in LMFDB. The splitting fields of $f(x)$ and $f(-x)$ are always the same.
$endgroup$
– Jyrki Lahtonen
Jan 9 at 23:31
add a comment |
$begingroup$
It is a bit unclear to me exactly what kind of an answer you are looking for. Listing the following points that occurred to me while reading the question.
- You don't need to use $omega$ to get a cubic extension field containing non-real numbers. Any cubic polynomial $f(x)$, irreducible over $Bbb{Q}$, such that $f$ has only a single real zero will yield such fields. For example, $f(x)=x^3-2x+2$, irreducible by Eisenstein, has a single real zero $in(-2,-1)$. If $alpha$ is one of the complex zeros of $f$, then $K_1=Bbb{Q}(alpha)$ is a cubic field, and $Bbb{Z}[alpha]$ is a ring that as a an abelian group is free of rank three. The field $K_1$ cannot be gotten by adjoining a cube root (real or complex) of an integer to $Bbb{Q}$. A way of seeing that is to use the discriminants. The discriminant of $f(x)$ is $d(f)=-76$, implying that the discriminant of $mathcal{O}_{K_1}$ is either $-76$ or $-19$. The discriminant of $g(x)=x^3-n$ is $d(g)=-27n^2$. The discriminants of orders of the same field can only differ by a factor that is a square, making it impossible for $mathcal{O}_{K_1}$ to contain $Bbb{Z}[omega^jroot3of n]$ for any alternative $j=0,1,2$ (so any root of $g(x)$).
- Some cubic extension fields of $Bbb{Q}$ are Galois extensions, implying that the minimal polynomials of all the elements have real zeros (all in that same field). An example appearing possibly the most frequently in our website is $K=Bbb{Q}(2cos(2pi/9))$. The zeros of the irreducible polynomial $p(x)=x^3-3x+1$ are $2cos(2^kpi/9)$ with $k=1,2,3$, cyclically permuted by the mapping $rmapsto r^2-2$. The ring of integers of $K$ is $Bbb{Z}[2cos(2pi/9)]$. Here the discriminant is $d(p)=81$, the square of a rational number. This happens always with cubic Galois extensions.
- The third primitive root of unity $omega$ does play the following important role (look up Kummer theory for details and generalizations). If, instead of $Bbb{Q}$, we use $F=Bbb{Q}(omega)$ as our base field, then we have the following characterization of cubic Galois extensions of $F$.
Assume that $L$ is a cubic Galois extension of $F$, then there exists an element $zin L$ such that $alpha=z^3in F$. In other words $L=F(root3ofalpha)$.
Of course, in the last bullet, $F$ has other types of cubic extensions, but none of those are Galois over $F$.
answered Jan 6 at 14:35
Jyrki LahtonenJyrki Lahtonen
108k12166369
$endgroup$
It is a bit unclear to me exactly what kind of an answer you are looking for. Listing the following points that occurred to me while reading the question.
- You don't need to use $omega$ to get a cubic extension field containing non-real numbers. Any cubic polynomial $f(x)$, irreducible over $Bbb{Q}$, such that $f$ has only a single real zero will yield such fields. For example, $f(x)=x^3-2x+2$, irreducible by Eisenstein, has a single real zero $in(-2,-1)$. If $alpha$ is one of the complex zeros of $f$, then $K_1=Bbb{Q}(alpha)$ is a cubic field, and $Bbb{Z}[alpha]$ is a ring that as a an abelian group is free of rank three. The field $K_1$ cannot be gotten by adjoining a cube root (real or complex) of an integer to $Bbb{Q}$. A way of seeing that is to use the discriminants. The discriminant of $f(x)$ is $d(f)=-76$, implying that the discriminant of $mathcal{O}_{K_1}$ is either $-76$ or $-19$. The discriminant of $g(x)=x^3-n$ is $d(g)=-27n^2$. The discriminants of orders of the same field can only differ by a factor that is a square, making it impossible for $mathcal{O}_{K_1}$ to contain $Bbb{Z}[omega^jroot3of n]$ for any alternative $j=0,1,2$ (so any root of $g(x)$).
- Some cubic extension fields of $Bbb{Q}$ are Galois extensions, implying that the minimal polynomials of all the elements have real zeros (all in that same field). An example appearing possibly the most frequently in our website is $K=Bbb{Q}(2cos(2pi/9))$. The zeros of the irreducible polynomial $p(x)=x^3-3x+1$ are $2cos(2^kpi/9)$ with $k=1,2,3$, cyclically permuted by the mapping $rmapsto r^2-2$. The ring of integers of $K$ is $Bbb{Z}[2cos(2pi/9)]$. Here the discriminant is $d(p)=81$, the square of a rational number. This happens always with cubic Galois extensions.
- The third primitive root of unity $omega$ does play the following important role (look up Kummer theory for details and generalizations). If, instead of $Bbb{Q}$, we use $F=Bbb{Q}(omega)$ as our base field, then we have the following characterization of cubic Galois extensions of $F$.
Assume that $L$ is a cubic Galois extension of $F$, then there exists an element $zin L$ such that $alpha=z^3in F$. In other words $L=F(root3ofalpha)$.
Of course, in the last bullet, $F$ has other types of cubic extensions, but none of those are Galois over $F$.
answered Jan 6 at 14:35
Jyrki LahtonenJyrki Lahtonen
108k12166369
edited Jan 8 at 4:33
answered Jan 6 at 14:35
Jyrki LahtonenJyrki Lahtonen
108k12166369
answered Jan 6 at 14:35
Jyrki LahtonenJyrki Lahtonen
108k12166369
answered Jan 6 at 14:35
Jyrki LahtonenJyrki Lahtonen
108k12166369
108k12166369
$begingroup$
Very helpful, thank you. What I still don't know is a lot of the terminology: like that ring you mentioned in point 1, from the polynomial $x^3 - 2x + 2$, what do you call that one? For a cubefree integer $n$, what do you call $mathbb Q(root 3 of n)$? Adjoining a "trigonometric" number like in point 2, that never gives a quadratic ring, right?
$endgroup$
– Bob Happ
Jan 9 at 21:55
$begingroup$
One more thing: from the link Brooks gave, I found lmfdb.org/NumberField/3.1.140.1 Is that the same as or different from the ring you gave in point 1?
$endgroup$
– Bob Happ
Jan 9 at 21:56
1
$begingroup$
$Bbb{Q}(cos(pi/4))$ and $Bbb{Q}(2cos(2pi/5))$ are quadratic. The former is $Bbb{Q}(sqrt2)$ and the latter $Bbb{Q}(sqrt5)$. I don't think that cubic in item 1. has a special name.
$endgroup$
– Jyrki Lahtonen
Jan 9 at 23:24
3
$begingroup$
No, The example in item 1. is this cubic in LMFDB. The splitting fields of $f(x)$ and $f(-x)$ are always the same.
$endgroup$
– Jyrki Lahtonen
Jan 9 at 23:31
add a comment |
$begingroup$
Very helpful, thank you. What I still don't know is a lot of the terminology: like that ring you mentioned in point 1, from the polynomial $x^3 - 2x + 2$, what do you call that one? For a cubefree integer $n$, what do you call $mathbb Q(root 3 of n)$? Adjoining a "trigonometric" number like in point 2, that never gives a quadratic ring, right?
$endgroup$
– Bob Happ
Jan 9 at 21:55
$begingroup$
One more thing: from the link Brooks gave, I found lmfdb.org/NumberField/3.1.140.1 Is that the same as or different from the ring you gave in point 1?
$endgroup$
– Bob Happ
Jan 9 at 21:56
1
$begingroup$
$Bbb{Q}(cos(pi/4))$ and $Bbb{Q}(2cos(2pi/5))$ are quadratic. The former is $Bbb{Q}(sqrt2)$ and the latter $Bbb{Q}(sqrt5)$. I don't think that cubic in item 1. has a special name.
$endgroup$
– Jyrki Lahtonen
Jan 9 at 23:24
3
$begingroup$
No, The example in item 1. is this cubic in LMFDB. The splitting fields of $f(x)$ and $f(-x)$ are always the same.
$endgroup$
– Jyrki Lahtonen
Jan 9 at 23:31
$begingroup$
Very helpful, thank you. What I still don't know is a lot of the terminology: like that ring you mentioned in point 1, from the polynomial $x^3 - 2x + 2$, what do you call that one? For a cubefree integer $n$, what do you call $mathbb Q(root 3 of n)$? Adjoining a "trigonometric" number like in point 2, that never gives a quadratic ring, right?
$endgroup$
– Bob Happ
Jan 9 at 21:55
$begingroup$
Very helpful, thank you. What I still don't know is a lot of the terminology: like that ring you mentioned in point 1, from the polynomial $x^3 - 2x + 2$, what do you call that one? For a cubefree integer $n$, what do you call $mathbb Q(root 3 of n)$? Adjoining a "trigonometric" number like in point 2, that never gives a quadratic ring, right?
$endgroup$
– Bob Happ
Jan 9 at 21:55
$begingroup$
One more thing: from the link Brooks gave, I found lmfdb.org/NumberField/3.1.140.1 Is that the same as or different from the ring you gave in point 1?
$endgroup$
– Bob Happ
Jan 9 at 21:56
$begingroup$
One more thing: from the link Brooks gave, I found lmfdb.org/NumberField/3.1.140.1 Is that the same as or different from the ring you gave in point 1?
$endgroup$
– Bob Happ
Jan 9 at 21:56
1
1
$begingroup$
$Bbb{Q}(cos(pi/4))$ and $Bbb{Q}(2cos(2pi/5))$ are quadratic. The former is $Bbb{Q}(sqrt2)$ and the latter $Bbb{Q}(sqrt5)$. I don't think that cubic in item 1. has a special name.
$endgroup$
– Jyrki Lahtonen
Jan 9 at 23:24
$begingroup$
$Bbb{Q}(cos(pi/4))$ and $Bbb{Q}(2cos(2pi/5))$ are quadratic. The former is $Bbb{Q}(sqrt2)$ and the latter $Bbb{Q}(sqrt5)$. I don't think that cubic in item 1. has a special name.
$endgroup$
– Jyrki Lahtonen
Jan 9 at 23:24
3
3
$begingroup$
No, The example in item 1. is this cubic in LMFDB. The splitting fields of $f(x)$ and $f(-x)$ are always the same.
$endgroup$
– Jyrki Lahtonen
Jan 9 at 23:31
$begingroup$
No, The example in item 1. is this cubic in LMFDB. The splitting fields of $f(x)$ and $f(-x)$ are always the same.
$endgroup$
– Jyrki Lahtonen
Jan 9 at 23:31
add a comment |
$begingroup$
Hmmm... our own Franz Lemmermeyer really needs to weigh in on this one. Seven years ago he wrote a paper on the "simplest cubic fields" in Manuscripta Mathematica.
There are
- Simply real cubic rings
- Totally real cubic rings
I'm not sure what the distinction is. Then of course there are imaginary cubic rings, whether or not they're obtained by adjoining the cubic root of an integer or by some other adjunction.
$endgroup$
add a comment |
$begingroup$
Hmmm... our own Franz Lemmermeyer really needs to weigh in on this one. Seven years ago he wrote a paper on the "simplest cubic fields" in Manuscripta Mathematica.
There are
- Simply real cubic rings
- Totally real cubic rings
I'm not sure what the distinction is. Then of course there are imaginary cubic rings, whether or not they're obtained by adjoining the cubic root of an integer or by some other adjunction.
$endgroup$
add a comment |
$begingroup$
Hmmm... our own Franz Lemmermeyer really needs to weigh in on this one. Seven years ago he wrote a paper on the "simplest cubic fields" in Manuscripta Mathematica.
There are
- Simply real cubic rings
- Totally real cubic rings
I'm not sure what the distinction is. Then of course there are imaginary cubic rings, whether or not they're obtained by adjoining the cubic root of an integer or by some other adjunction.
$endgroup$
Hmmm... our own Franz Lemmermeyer really needs to weigh in on this one. Seven years ago he wrote a paper on the "simplest cubic fields" in Manuscripta Mathematica.
There are
- Simply real cubic rings
- Totally real cubic rings
I'm not sure what the distinction is. Then of course there are imaginary cubic rings, whether or not they're obtained by adjoining the cubic root of an integer or by some other adjunction.
answered Jan 10 at 22:43
community wiki
Lisa
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3061094%2fwhat-other-kinds-of-cubic-integer-rings-are-there%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
2
$begingroup$
Not every cubic extension of $mathbb{Q}$ is obtained by adjoining the cubic root of an integer (or a rational).
$endgroup$
– Arturo Magidin
Jan 3 at 22:34
8
$begingroup$
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.
$endgroup$
– Arturo Magidin
Jan 3 at 22:36
1
$begingroup$
Related: math.stackexchange.com/q/3060484/328173
$endgroup$
– Kenny Lau
Jan 4 at 7:28