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I'm reading Section I.11 The Complex Numbers from textbook Analysis I by Amann/Escher. I am trying to construct $\mathbb C$ from authors' hints.

There is a smallest extension field $\mathbb{C}$ of $\mathbb{R}$, the field of complex numbers, in which the equation $z^{2}=-1$ is solvable. $\mathbb{C}$ is unique up to isomorphism.

Could you please verify if my attempt contains logical gaps/errors? Any suggestion is greatly appreciated.


My attempt:

1.

Let $K$ be any extension field of $\mathbb R$ in which the equation $z^{2}=-1$ is solvable. Then there exists $i \in K$ such that $i^2 = -1$. Let $C_K := \{x + iy \mid x,y \in \mathbb R\}$. Then $\mathbb R \subseteq C_K$.

For $z=x+i y \in C_K$ and $w=a+i b \in C_K$, we have

$$\begin{aligned} z+w &=x+a+i(y+b) \in C_K \\-z &=-x+i(-y) \in C_K \\ z w &=x a+i x b+i y a+i^{2} y b=x a-y b+i(x b+y a) \in C_K,\quad i^2 = -1 \end{aligned}$$

If $z=x+i y \neq 0$ then $x \in \mathbb{R}^{ \times}$ or $y \in \mathbb{R}^{ \times}$. Thus we have

$$\frac{1}{z}=\frac{1}{x+i y}=\frac{x-i y}{(x+i y)(x-i y)}=\frac{x}{x^{2}+y^{2}}+i \frac{-y}{x^{2}+y^{2}} \in C_K$$

It follows that $C_K$ is a subfield of $K$ and extension field of $\mathbb R$ in which the equation $z^{2}=-1$ is solvable. Let $K'$ be another extension field of $\mathbb R$ in which the equation $z^{2}=-1$ is solvable and $i' \in K'$ such that ${i'}^2 = -1$. Define $f: C_K \to C_{K'}$ by $$f(x+i y) = x+i' y, \quad x,y \in \mathbb R$$

It is easy to verify that $f$ is an isomorphism between $C_K$ and $C_{K'}$.

As a result, every extension field $K$ of $\mathbb R$, in which the equation $z^{2}=-1$ is solvable, contains an extension field $C_K$ of $\mathbb R$, in which the equation $z^{2}=-1$ is solvable.

It follows that $C_K$ is the smallest extension field of $\mathbb R$ in which the equation $z^{2}=-1$ is solvable, if such an extension field exists. Moreover, $C_K$ is unique up to isomorphism.

2.

We show the existence of such fields $K, C_K$. We define addition and multiplication on $\mathbb R^2$ by

$$\begin{aligned}+: &\mathbb{R}^{2} \times \mathbb{R}^{2} \rightarrow \mathbb{R}^{2}, \quad((x, y),(a, b)) \mapsto (x+a, y+b)\\ \cdot: &\mathbb{R}^{2} \times \mathbb{R}^{2} \rightarrow \mathbb{R}^{2}, \quad((x, y),(a, b)) \mapsto (x a-y b, x b+y a)\end{aligned}$$

and set $K :=\left(\mathbb{R}^{2},+, \cdot\right)$. One can easily check that $K$ is a field with additive identity $(0,0)$, unity $(1,0)$, additive inverse $-(x, y)=(-x,-y)$, and multiplicative inverse $(x, y)^{-1}=\left(x /\left(x^{2}+y^{2}\right),-y /\left(x^{2}+y^{2}\right)\right)$ if $(x, y) \neq(0,0)$.

It is easy to verify that $$\mathbb{R} \rightarrow K, \quad x \mapsto(x, 0)$$ is an injective homomorphism. Consequently we can identify $\mathbb{R}$ with its image in $K$ and so consider $\mathbb{R}$ to be a subfield of $K$. The equation $(0,1)^{2}=(0,1)(0,1)=(-1,0)=-(1,0)$ implies that $(0,1) \in K$ is a solution of $z^{2}=-1_{K}$. Let $i:= (0,1)$.

For $(x,y) \in K$, $(x,y) = (x,0) + (0,1)(y,0) = (x,0)+i(y,0)$ in which $(x,0),(y,0) \in \mathbb{R}$. Then $C_K = K$. From now on, let $\mathbb{C} := C_K$ and call it the field of complex numbers.

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    $\begingroup$ The proof is very clear to me! However there is an alternate approach by using polynomial ring. $\endgroup$ – Sujit Bhattacharyya Jun 28 at 3:18
  • $\begingroup$ Thank you so much @SujitBhattacharyya. I am reading on your reference ^^ $\endgroup$ – MadnessFor MATH Jun 28 at 3:48

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