0
$\begingroup$

Most Complex Analysis books introduce the complex numbers by defining a multiplication operation on $\mathbb{R}^2$ in the usual way and shows that this defines a field. Multiplying $(0,1)(0,1)$ gives $(-1, 0)$, by definition, thus $z^2 + 1 = 0$ has a solution in this system.

In Nevanlinna's Introduction to Complex Analysis, he gives a different approach. He starts by giving an axiomatic definition of a two-dimensional vector space over the reals, which he calls $R$. He then extends this 2D vector space to a vector algebra, by saying that for any $x, y \in R$, there is a 'product'

$$ z = xy \in R $$

that satisfies the following four axioms:

(1) The product is commutative: $xy = yx$

(2) The product is bilinear, that is, linear in each factor.

(3) The product is associative: $x(yz) = (xy)z$

(4) The product $xy$ vanishes, $xy = 0$, if and only if at least one factor vanishes.

He then shows that for any fixed $y \neq 0$, the linear map

$$ x \rightarrow xy $$

is onto, thus our algebra is a field and there exists a unit vector $e$ such that $ex = xe = x$ for all $x\in R$. He shows $e$ is unique in the usual way.

He then considers the equation $x^2 = a$ and leaves it as an exercise to show using the four axioms listed above that $x^2 + e = 0$ has a solution in $R$. Only then does he show that the product must have the form that is the usual starting point of most complex analysis texts.

I am stuck on how to show that $x^2 + e = 0$ has a solution only relying on the algebraic axioms listed above. I have searched quite a bit online and have not found any help.

  • 1
    Reminds me of a funny post that tries to solve $x^2+1=0$ over the naturals, and by assuming it exists, it gets pretty far, though seemingly nowhere, non-trivial results end up being produced. It gave me quite the laugh. – 2017-01-02
  • 1
    Take $f \in R\setminus \mathbb{R}e$. Then $\{e,f\}$ is a basis of $R$, and we can write $f^2 = \alpha e + \beta f$. Look at $(u e + vf)^2$. – 2017-01-02
  • 0
    @SimpleArt : can you find this post again, please? – 2017-01-02
  • 1
    @Alphonse I'll try :D – 2017-01-02
  • 0
    @DanielFischer which gives showing that there exists $u,v$ such that $(u^2 + v^2\alpha)e + (2uv + \beta v^2)f = -e$. Is this what you had in mind? – 2017-01-02
  • 1
    Would the downvoters please justify what their issue with my question is? – 2017-01-02
  • 0
    Yes. It's not particularly elegant, but it works. – 2017-01-02

0 Answers 0