Problem in ordering of the $\mathbb{C}$

Question

The question given in Rudin is that ,

prove that there doesn't exist an order such that $\mathbb{C} $turns into an ordered field.

An order on S is a relation with the following properties (i) If x,y $\in S$ then any one of the following should hold true $x<y$ ,$x=y$ or $y<x$ (ii) If $x<y$,$y<z$ then $x<z$

Since I have read some portions of euclidean domain, my question is what is wrong if I define a Norm in $\mathbb{C}$ as $a^2 + b^2$ and then try to order the complex numbers as $a+ib < c+id$ if and only if the norm is less .What is wrong in this ?

Answer 1

The main thing that is wrong with this is that trichotomy fails.

We need exactly one of the following to happen: i) $w < z$ or ii) $z < w$ or iii) $z = w$.

Consider if $w=a + bi$ and $z = c + di$ and $a^2 + b^2 = c^2 + d^2$ but $a\ne c$ and $b\ne d$. (Any two points on a circle centered at $0$ will have this property. Ex. $w = i$ and $z = 1$.)

Then you have none of $w < z$ or $z < w$ or $z =w$ true.

.......

But consider the dictionary order:

$a+ bi < c + di$ if either $a < c$ or if $a=c$ and $b < d$.

That is a valid ordet that doesn satisfy trichotomy.

If $w =a +bi$ and $z= c+di$ then we have the follow 5 cases:

Case 1: $a < c$. Then $w < z$; it is not true that $z < w$; and $w\ne z$.

Case 2: $a > c$ then $z < w$; it is not true that $w < z$ and $w\ne z$.

Case 3: $a =c$ and $b < d$. Then $w < z$; it is not true that $z < w$; and $w\ne z$.

Case 4: $a=c$ and $b > d$. then $z < w$; it is not true that $w < z$ and $w\ne z$.

Case 5: $a=c$ and $b = d$. Then $w=z$ and it is not true that $w < z$ and it is not true if $z< w$.

That is a perfectly okay order.... but it is not an ordered field.

.......

There are many ways to order $\mathbb C$ but none of these allow $\mathbb C$ to be an ordered field.

You left out the THIRD requirement.

An ordered field is a field with an order relation where the followin are true: i) for all $a, b,c$ with $b < c$ then $a + b < a+c$ and ii) for all $a > 0$ and $b,c$ with $b< c$ then $ab < ac$.

This will lead to a requirements that if $x >0$ then $-x < 0$ (because $x + (-x) > 0 + (-x)$) and that if $x \ne 0$ then $ x^2 > 0$. (if $x > 0$ then $x*x > 0*x$ and if $x < 0$ then $(-x) > 0$ and $(-x)(-x) > 0 (-x)$)

So $\mathbb C$ can never be an ordered field becuase $i^2 = -1$ and $1^2 = 1$ and we can't have both $1 > 0$ and $-1> 0$.

.......

Note the dictionary order fails to make an ordered field. We have $i > 0$ (because $i =0 + i > 0 + 0i=0$ because $0 = 0$ but $1 > 0$) and yet $i^2 = -1 < 0$ (because $-1 < 0$ so $-1 + 0i < 0 + 0i=0$) . But we if $i>0$ we need to have $i^2 = i*i > i*0=0$. And we don't.

So the dictionary order fails to be an ordered field.

Answer 2

In your definition there is no way to distinguish $1$, $-1$, $i$ and $-i$ because $N(\pm i)=N(\pm 1)=1$.

Answer 3

In any ordered field, every square must be positive. Because if $a > 0$ then $a^2 > 0$ and if $a < 0$ then $a^2 > 0$ (multiplying both sides by $a$ preserves the inequality if $a > 0$ and flips it if $a < 0$). But this is a problem because $1 = 1^2 > 0$ means that $1 - 1 > 0 - 1$ means that $-1 < 0$. Since $-1 < 0$ it cannot be a square because as we've established, every square must be positive.