2

I am currently reading this paper on the axioms of complex numbers. The set of complex numbers is characterized (and defined up to a natural isomorphism) by the following axioms:

  • $\mathbf{C}$ is a commutative field.
  • $\mathbf{C}\ni z\mapsto\overline{z}\in\mathbf{C}$ is a ring homomorphism.
  • $\overline{\overline{z}}=z$ for all $z\in\mathbf{C}$.
  • There exists some $z\in\mathbf{C}\setminus\{0\}$ with $\overline{z}=-z$ (equivalently, there exists some $z\in\mathbf{C}$ with $\overline{z}\neq z$).
  • $\mathbf{R}:=\{z\in\mathbf{C}:\overline{z}=z\}$ is a complete ordered field.

In the proof of theorem 3, the authors make the following assumption:

If $\overline{z}=-z$ and $z\neq 0$, then $z\cdot z<0$.

It is clear that $z\cdot z$ is a real, nonzero number, since $\overline{z\cdot z}=\overline{z}\cdot \overline{z}=z\cdot z$. Thus, it remains to be proven that $z\cdot z$ is negative. The paper suggests a proof by contradiction:

Suppose $z\cdot z>0$, then we consider the real numbers $X:=\sqrt{z\cdot z}$ and $Y:=-\sqrt{z\cdot z}$. Since either $z=X$ or $z=Y$, $z$ is a real number - contradiction.

However, I don't see how the claim that either $z=X$ or $z=Y$ can be proven based on the axioms listed above.

Filippo
  • 3,536
  • 2
    That's definitely not a characterization of $\Bbb C$. $\Bbb R$ with the ring homomorphism $z \mapsto z$ satisfies those axioms. – jjagmath Aug 27 '21 at 11:37
  • @jjagmath You are right, I forgot to mention one additional axiom mentioned in the paper, see the edit. I left it out on purpose, because it wasn't relevant for answering the question, but I understand that this wasn't a good idea. – Filippo Aug 27 '21 at 11:53
  • After the edit of your axioms, they are still different from the ones from the article. For example, you omitted $\bar{\bar{z}}=z$. Why? Are you sure that's after omitting that, those axioms still characterize $\Bbb C$? – jjagmath Aug 27 '21 at 11:56
  • Thanks for edits. Now your question doesn't begin with a false statement :) – jjagmath Aug 27 '21 at 12:09
  • @jjagmath I am sorry, I forgot this point although it's very important. I should have checked the paper before posting the question. – Filippo Aug 27 '21 at 12:09
  • @jjagmath Thank YOU! :) – Filippo Aug 27 '21 at 12:10
  • Very good point, re the definition of a Complex Number. I am clearly out of my depth here, so I will delete my answer. – user2661923 Aug 27 '21 at 15:28
  • @user2661923 Nevertheless, thank you very much for your time! If we find some other way to prove that $z\cdot\overline{z}>0$, this would really be an alternative approach. – Filippo Aug 27 '21 at 15:34
  • @user2661923 I have thought about it. As you already said, it follows from theorem 4 in the paper that $z\cdot\overline{z}>0$ for all $z\in\mathbf{C}\setminus{0}$. (Theorem 4 says that if $i$ is one of the two complex numbers satisfying $i\cdot i=-1$, then each $z$ can be written uniquely in the form $x+iy$, where $x$ and $y$ are real). However you do not need this theorem to show that $z\cdot\overline{z}>0$: Consider the following set: $$\mathbf{I}:={z\in\mathbf{C}:\overline{z}=-z}$$ It can be shown that $\mathbf{C}=\mathbf{R}\oplus\mathbf{I}$ (I can elaborate if you want), – Filippo Aug 27 '21 at 17:29
  • i.e. each $z$ can be written uniquely in the form $x+y$ with $\overline{x}=x$ and $\overline{y}=-y$. Now consider some $z=x+y$, then $$z\cdot\overline{z}=x\cdot x-y\cdot y$$ and it follows from the title of my question that $z\cdot\overline{z}>0$. – Filippo Aug 27 '21 at 17:30

1 Answers1

3

On any field, you have$$a^2=b^2\iff a=b\text{ or }a=-b,$$since $a^2=b^2\iff(a-b)(a+b)=0$. So,$$z^2=X^2\iff z=X\text{ or }z=-X=Y.$$

José Carlos Santos
  • 427,504
  • Thank you very much. I just wanted to let you know that I initially forgot to mention a few very important points needed for a complete description of the complex numbers. Fortunately, they do not affect the question or your answer. Sorry for that! – Filippo Aug 27 '21 at 12:29