# Help with Algebraic Number Theory

• Jun 11th 2008, 12:45 PM
jamix
Help with Algebraic Number Theory
Hello

I'm currently working through a text titled 'Introductory Algebraic Number Theory' by Saban Alaca and Kenneth S. Williams, and I'm having trouble with some of the practise exercises throughout the text.

The first problem I've been trying to figure out is the following:

Consider the integral domain A = Z + Z((1 + sqrt(m))/2) where m = 1(mod4) and is less than -3

Prove that the set of units U(Z +Z((1 + sqrt(m)/2)= (+1, -1).

Can anyone help me out with this?
• Jun 11th 2008, 02:53 PM
ThePerfectHacker
Quote:

Originally Posted by jamix
Consider the integral domain A = Z + Z((1 + sqrt(m))/2) where m = 1(mod4) and is less than -3

Prove that the set of units U(Z +Z((1 + sqrt(m)/2)= (+1, -1).

I did not work on this problem, but did you try doing it by definition? Meaning $\alpha$ is a unit iff there is $\beta$ such that $\alpha \beta = 1$. Now if $\alpha \in A$ then $\alpha = a + b\cdot \tfrac{1+\sqrt{m}}{2}$. Thus, we are saying we can find $\beta = c + d \cdot \tfrac{1+\sqrt{n}}{2}$ so that $\left( a + b\cdot \tfrac{1+\sqrt{m}}{2} \right) \cdot \left( c + d\cdot \tfrac{1+\sqrt{n}}{2} \right) = 1$. Now multiply out compare coefficients with RHS and argue that $b=d=0$ and $ac=1$. Thus, $a=c=1$ or $a=c=-1$. Which tells you the units at $\pm 1$.
• Jun 11th 2008, 05:54 PM
NonCommAlg
Quote:

Originally Posted by jamix
Hello

I'm currently working through a text titled 'Introductory Algebraic Number Theory' by Saban Alaca and Kenneth S. Williams, and I'm having trouble with some of the practise exercises throughout the text.

The first problem I've been trying to figure out is the following:

Consider the integral domain A = Z + Z((1 + sqrt(m))/2) where m = 1(mod4) and is less than -3

Prove that the set of units U(Z +Z((1 + sqrt(m)/2)= (+1, -1).

Can anyone help me out with this?

let $x=a+\frac{b}{2}(1 + \sqrt{m}) \in A.$ then: $N(x)=x \bar{x}=\left(a + \frac{b}{2} + \frac{b\sqrt{m}}{2} \right)\left(a + \frac{b}{2} - \frac{b\sqrt{m}}{2}\right)=\left(a + \frac{b}{2}\right)^2 - \frac{b^2m}{4}.$

now $x$ is a unit iff $N(x)=1,$ i.e. iff: $\left(a + \frac{b}{2}\right)^2-\frac{b^2m}{4}=1.$ call this (1). so $1+\frac{b^2m}{4} = \left(a+\frac{b}{2} \right)^2 \geq 0.$ hence

$b^2m \geq -4.$ call this (2). now we're given that $m < -3.$ suppose that $b \neq 0.$ then $b^2 \geq 1,$ because $b$ is an integer.

thus $b^2 m < - 3.$ so by (2) we'll have: $-4 \leq b^2m < -3,$ which is possible only when $m=-4, \ b^2 = 1.$ but since

$m \equiv 1 \mod 4,$ we cannot have $m=-4.$ so our assumption that $b \neq 0$ is false and hence $b = 0.$ thus by (1) we

will have $a^2=1$, which gives us: $x=a=\pm 1. \ \ \ \square$
• Jun 11th 2008, 08:35 PM
jamix
Thanks NonCommAlg

I had to do the proof of the theorem you presented that N(x)=1 iff x is a unit. Is this theorem true for all integral domains? More specifically, are there integral domains out there for which N(x)= (+1, -1) iff x is a unit?
• Jun 11th 2008, 08:39 PM
ThePerfectHacker
Quote:

Originally Posted by jamix
Thanks NonCommAlg

I had to do the proof of the theorem you presented that N(x)=1 iff x is a unit. Is this theorem true for all integral domains? More specifically, are there integral domains out there for which N(x)= (+1, -1) iff x is a unit?

If $N$ is such a function so that $N(ab) = N(a)N(b)$ then $N(x) = 1$ if $x$ is a unit. To prove this we first show $N(1) = 1$. Since $1\cdot 1 = 1$ it means $N(1)N(1) = N(1)$ and so $N(1) = 1$ (since it is non-zero). Now if $a$ is a unit it means $ab=1$ for some $b$. Thus, $N(a)N(b) = 1\implies N(a) = 1$. I am not sure about the converse direction, I would imagine no. However, I cannot think of a good counterexample because all the known integral domains I worked with (integers, Gaussian, Eisenstein) have this property.
• Jun 11th 2008, 09:28 PM
jamix
ThePerfectHacker

You said that N(a)N(b) = 1 implies that N(a) = 1. What if N(a) = N(b) = -1?

With regards to the converse (ie that x is a unit if N(x) = 1), this should follow immediately since the element y such that xy=1 is just the conjugate of x.
• Jun 12th 2008, 06:33 AM
ThePerfectHacker
Quote:

Originally Posted by jamix
ThePerfectHacker

You said that N(a)N(b) = 1 implies that N(a) = 1. What if N(a) = N(b) = -1?

With regards to the converse (ie that x is a unit if N(x) = 1), this should follow immediately since the element y such that xy=1 is just the conjugate of x.

It cannot happen that N(a) = -1, because norms are non-negative.

I think I have an answer to your question about converses. It is necessary for us to define what an Euclidean domain is, since you are studying algebraic number theory you will definitely run into these. A Euclidean domain is an integral domain $D$ so that there exists a function $N:D^{\times} \mapsto \mathbb{N}$ (non-negative) which satisfies two conditions. The first condition is the "division algorithm", that is, given $a,b\in D$ with $b\not = 0$ we can divide $a$ by $b$ 'with a remainder' i.e. write $a = qb + r$, where $r=0$ or otherwise $N(r) < N(b)$. The second condition is for all $a,b\in \mathbb{D}^{\times}$ we have $N(a) \leq N(ab)$. The function $N$ is called an Euclidean norm.

Just a warning an Euclidean domain does not need to satisfy $N(ab) = N(a)N(b)$. It turns out that in the classic examples I will give below this Euclidean norm is multiplicative.

The first example are the ordinary integers. Let $D = \mathbb{Z}$, this is an integral domain. We define $N(x) = |x|$, and it follows that this is an Euclidean domain, also, $N(xy) = N(x)N(y)$. The second example are the Gaussian integers, $D = \mathbb{Z}[i] = \{a+bi|a,b\in \mathbb{Z} \}$. We define $N(\alpha) = |\alpha| = \sqrt{a^2+b^2}$ if $\alpha = a+bi$. Furthermore, it is multiplicative. It takes work showing that it satisfies the division algorithm, the way we show this is let $\beta \not = 0$ and $\alpha$ be arbitrary, divide $\alpha / \beta = q+ir$ in $\mathbb{Q}$ and pick $n,m$ so that $|n-q|,|r-m|\leq \tfrac{1}{2}$; now argue that this is the quotient and remainder you are looking form. Another example are the Eisenstein integers, $\mathbb{Z}[\omega] = \{a+b\omega |a,b\in \mathbb{Z}\}$ where $\omega = e^{2\pi i/3}$. Here we define $N(\alpha) = \alpha \bar \alpha = a^2 - ab+b^2$. And again $N(\alpha \beta) = N(\alpha)N(\beta)$. It satisfies the division algorithm by using the similar argument as above. Some useful facts here, if you are planning to do some computations with them, is that $\bar \omega = \omega^2 = -1-\omega$. And finally $D = F[x]$, the polynomials over a field, are an Euclidean domain. Define $N(f(x)) = \deg f(x)$ for non-zero polynomials. And this is the standard division algorithm you are used to. And again $N$ is multiplicative.

There are two really useful theorems about Euclidean domains. The first one is that Euclidean domains are principal ideal domains, meaning if $I$ is an ideal then it is principal i.e. $I = \left< a \right>$ for some $a\in D$. The proof is really simple, assume that $I$ is a non-trivial ideal and considier $\{ N(x) | x\in I^{\times} \}$, this set is a set of non-negative integers so there is a minimal element. Argue that if $\alpha \in I$ is what gives the minimal norm then $I = \left< \alpha \right>$ (note we never use condition #2 of integral domains!). The second useful theorem is what you asked: $N(x) = N(1)$ iff $x$ is a unit. The proof is simple also. Note, $N(1) \leq N(1a) = N(a)$ for all $a\in D^{\times}$, thus $N(1)$ is a lower bound for all the norms. While if $a$ is unit we can write $N(a) \leq N(aa^{-1}) = N(1)$ and so it means $N(a) = N(1)$. For the converse let $a\in D^{\times}$ be such that $N(a) = N(1)$ then write $1 = qa+r$, it cannot be that $N(r) < N(a)=N(1)$ for $N(1)$ is minimal by above, so $r=0$ which means $1=qa$ and therefore it is a unit.

Note, this is not exactly what you asked, you rather asked $N(x) = 1$ iff $x$ is unit. But as I said in almost all examples you will see that will be the case. In fact, my number theory book defines Euclidean domain differently how I posted above, so I was not even familar with the most general definition.
• Jun 12th 2008, 12:08 PM
jamix
Hi PerfectHacker

Thanks for the useful info. Euclidean domains are part of chapter 2, but hopefully I'll get there soon. For now I just want to make sure I can get through all the exercises in this first chapter.

Thanks for help (I'll post some more questions on this same thread again if I get stuck again).