Units of the set of all Eisenstein Integers

[Math Amateur]

In Chapter 1: "Integral Domains", of Saban Alaca and Kenneth S. Williams' (A&W) book "Introductory Algebraic Number Theory", the set of all Eisenstein integers, $$\mathbb{Z} + \mathbb{Z} \omega$$ is defined as follows:https://www.physicsforums.com/attachments/3392Then, Exercise 2 on page 23 of A&W reads as follows:

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Prove that $$U( \mathbb{Z} + \mathbb{Z} \omega) = \{ \pm 1, \pm \omega, \pm \omega^2 \}$$

where

U(D) is the set of units of D.

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Can someone please help me with this exercise.

I have tried to get a start with considering the equation

$$(a_1 + b_1 \omega ) (a_2 + b_2 \omega ) = (a_2 + b_2 \omega ) (a_1 + b_1 \omega ) = 1 $$

but the approach led me nowhere ...

Some help would be appreciated ...

Peter

 

[mathbalarka]

Everything-by-definition approach will only work out 10% of the time when you are working with integral closure of algebraic number fields.

You need to use the norm-approach again. Let $N : \Bbb Z +\Bbb Z \omega \to \Bbb Z$ be defined as $N(a + b\omega) = (a + b\omega)(a + b\omega^2) = a^2 - ab + b^2$. Show that the norm is multiplicative, and that it is positive. Thus show that for any unit $\alpha$, $N(\alpha) = 1$. Solve the corresponding diophantine equation.

This is actually easier than the Pell equation stuffs we encountered previously. Have fun.

 

[Math Amateur]

Everything-by-definition approach will only work out 10% of the time when you are working with integral closure of algebraic number fields.

You need to use the norm-approach again. Let $N : \Bbb Z +\Bbb Z \omega \to \Bbb Z$ be defined as $N(a + b\omega) = (a + b\omega)(a + b\omega^2) = a^2 - ab + b^2$. Show that the norm is multiplicative, and that it is positive. Thus show that for any unit $\alpha$, $N(\alpha) = 1$. Solve the corresponding diophantine equation.

This is actually easier than the Pell equation stuffs we encountered previously. Have fun.

Thanks for the help and advice, Mathbalarka ...

Just a question, though ... you write:

" ... ... You need to use the norm-approach again. Let $N : \Bbb Z +\Bbb Z \omega \to \Bbb Z$ be defined as $N(a + b\omega) = (a + b\omega)(a + b\omega^2) = a^2 - ab + b^2$. ... ... "... ... how did you know to define the norm as you did?


Peter

 

[Math Amateur]

Thanks for the help and advice, Mathbalarka ...

Just a question, though ... you write:

" ... ... You need to use the norm-approach again. Let $N : \Bbb Z +\Bbb Z \omega \to \Bbb Z$ be defined as $N(a + b\omega) = (a + b\omega)(a + b\omega^2) = a^2 - ab + b^2$. ... ... "... ... how did you know to define the norm as you did?


Peter

Euge, Mathbalarka ... thanks for your help on this post and other posts as well on algebraic number theory ...

In this exercise on the finding the units of the Eisenstein Domain, I checked that the norm you gave me (not sure how you arrived at it, mind!) was multiplicative (very tedious exercise indeed! ... BUT ... maybe I did it in a very inefficient way ... ) but ... ... then you suggest showing that for a unit $$\alpha$$ that $$N( \alpha ) = 1 $$ ... I am having trouble showing this ... can you help?

Peter

 

[Euge]

Euge, Mathbalarka ... thanks for your help on this post and other posts as well on algebraic number theory ...

In this exercise on the finding the units of the Eisenstein Domain, I checked that the norm you gave me (not sure how you arrived at it, mind!) was multiplicative (very tedious exercise indeed! ... BUT ... maybe I did it in a very inefficient way ... ) but ... ... then you suggest showing that for a unit $$\alpha$$ that $$N( \alpha ) = 1 $$ ... I am having trouble showing this ... can you help?

Peter

To find $N(a + b\omega)$, compute the product of the conjugates of $a + b\omega$. Let $x = a + b\omega$. Since $\omega$ is a cube root of unity, $(x - a)^3 = (b\omega)^3 = b^3$. So $a + b\omega$ satisfies the polynomial $(x - a)^3 - b^3$. By Demoivre's theorem, the roots of this polynomial are $a + b$, $a + b\omega$, and $a + b\omega^2$. Since

$$(x - a)^3 - b^3 = (x - a - b)[(x - a)^2 + (x - a)b + b^2] = [x - (a + b)][x^2 + (b - 2a)x + (a^2 - ab + b^2)]$$

and $a + b\omega$ does not satisfy $x - (a + b)$, the minimal polynomial of $a + b\omega$ is the second factor of the last equality above. Its roots are $a + b\omega$ and $a + b\omega^2$, and the product its roots is the constant term $a^2 - ab + b^2$, i.e., $(a + b\omega)(a + b\omega^2) = a^2 - ab + b^2$. So $N(a + b\omega) = a^2 - ab + b^2$.

If $\alpha$ is a unit of $\Bbb Z + \Bbb Z\omega$, then $\alpha \beta = 1$ for some $\beta \in \Bbb Z + \Bbb Z\omega$. Hence $N(\alpha \beta) = N(1)$, and since $N$ is multiplicative, $N(\alpha)N(\beta) = N(\alpha\beta) = N(1)$. Now $N(1) = N(1 + 0\omega) = 1^2 - 1(0) + 0^2 = 1$. Thus $N(\alpha) N(\beta) = 1$. Since $N$ is integer-valued, we must have $N(\alpha) = \pm 1$. However, $N(\alpha)$ is never negative. To see this, let's return the quadratic expression $a^2 - ab + b^2$, $a, b\in \Bbb Z$. If $a$ and $b$ have opposite signs, $-ab > 0$, and thus $a^2 - ab + b^2 > 0$. If $a$ and $b$ have the same sign, then $ab > 0$ and thus $a^2 - ab + b^2 = (a - b)^2 + ab \ge ab > 0$. If either $a$ or $b$ is zero, then $a^2 - ab + b^2$ is either $a^2$ or $b^2$, both of which can never be negative. Thus, in all cases, $a^2 - ab + b^2 \ge 0$. So we rule out $N(\alpha) = -1$ and deduce $N(\alpha) = 1$.

 

[mathbalarka]

...how did you know to define the norm as you did?

Very good. I like these kind of questions. Reading a proof line-by-line mechanically is not mathematics. *Understanding* the idea behind is.

If you are familiar with a bit of Galois theory, you'll recall that given a Galois extension $L/K$, there is a canonical homomorphism $L \to K$ defined by "multiplying through" by the Galois conjugates. Explicitly,

$$\alpha \mapsto \prod_{g \in \mathbf{Gal}(L/K)} g(\alpha)$$

In particular, this is the constant coefficient of the minimal polynomial of $\alpha$ in $K$, as the roots of the minimal poly consists of the Galois conjugates of $\alpha$.

Now what we are working with is $\mathbb{Z}[\omega]$. Note that this is the integral closure of the field $\Bbb{Q}[\omega]$ so we can quite similarly define the norm for $\Bbb{Z}[\omega]$ since Galois conjugate of integers of a field $L/K$ are integers in $L$ for Galois extensions $L/K$.

Just as a note, while most of the fun properties are lost when coming down from $L \supseteq K$ to the integral closure of both, $\mathcal{O}_L \supseteq \mathcal{O}_K$, some Galoisness still remains in a sense. It is in fact true that $\mathcal{O}_L$ is a $\mathbf{Gal}(L/K)$-module over $\mathcal{O}_K$ and this property whips up some interesting algebraic number theory later on.

 

[Math Amateur]

Very good. I like these kind of questions. Reading a proof line-by-line mechanically is not mathematics. *Understanding* the idea behind is.

If you are familiar with a bit of Galois theory, you'll recall that given a Galois extension $L/K$, there is a canonical homomorphism $L \to K$ defined by "multiplying through" by the Galois conjugates. Explicitly,

$$\alpha \mapsto \prod_{g \in \mathbf{Gal}(L/K)} g(\alpha)$$

In particular, this is the constant coefficient of the minimal polynomial of $\alpha$ in $K$, as the roots of the minimal poly consists of the Galois conjugates of $\alpha$.

Now what we are working with is $\mathbb{Z}[\omega]$. Note that this is the integral closure of the field $\Bbb{Q}[\omega]$ so we can quite similarly define the norm for $\Bbb{Z}[\omega]$ since Galois conjugate of integers of a field $L/K$ are integers in $L$ for Galois extensions $L/K$.

Just as a note, while most of the fun properties are lost when coming down from $L \supseteq K$ to the integral closure of both, $\mathcal{O}_L \supseteq \mathcal{O}_K$, some Galoisness still remains in a sense. It is in fact true that $\mathcal{O}_L$ is a $\mathbf{Gal}(L/K)$-module over $\mathcal{O}_K$ and this property whips up some interesting algebraic number theory later on.

You are convincing me that I need to learn Galois Theory and, perhaps also revise Field Theory ... particularly field extensions ...

Peter