# There is a smallest and unique up to isomorphism extension field $\mathbb{C}$ of $\mathbb{R}$, where the equation $z^{2}=-1$ is solvable

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.

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