Exercise 3.2.5

Proof.

(a)

Since $-1$ is not a square in $F$ by Exercise 4, the polynomial $x^2+1$ has no root in $F$, and it has degree 2, thus it is irreducible over $F$.

Therefore $F(i)=F[x]∕⟨x^2+1⟩$ is a field, where $i=x+⟨x^2+1⟩$, by Proposition 3.1.1.

The division of any polynomial $f$ by $x^2+1$ gives

$$f=q(x^2+1)+\mathit{bx}+a,$$

so every $y\in F(i)$ is of the form $y=a+\mathit{ib}$.

(b)

Let $a{x}^2+\mathit{bx}+c,a,b,c\in F(i),a\ne 0$, any quadratic polynomial.

$$\begin{array}{llll}\hfill a{x}^2+\mathit{bx}+c& =a\left(x^2+\frac{b}{a}x+\frac{c}{a}\right)& \hfill & \\ \hfill & =a\left[{\left(x+\frac{b}{2a}\right)}^2-\frac{b^2}{4a^2}+\frac{c}{a}\right]& \hfill & \\ \hfill & =a\left[{\left(x+\frac{b}{2a}\right)}^2-\frac{\mathrm{\Delta }}{4a^2}\right],\mathrm{\Delta }=b^2-4\mathit{ac}& \hfill & \\ \hfill & & \hfill \end{array}$$

By definition of a real closed field, every positive element of $F$ has a square root in $F$. With the same proof as in Ex 3.2.2, we can prove that every $z\in F(i)$ has a square root. One square root of $z=x+\mathit{iy},x,y\in F,$ is given by

$$z_0=\sqrt{\frac{1}{2}\left(\sqrt{x^2+y^2}+x\right)}+i\mathrm{sgn}(y)\sqrt{\frac{1}{2}\left(\sqrt{x^2+y^2}-x\right)}.$$

We will write $\sqrt{\mathrm{\Delta }}$ one of the square roots of $\mathrm{\Delta }$. Then

$$\begin{array}{llll}\hfill a{x}^2+\mathit{bx}+c& =a\left[{\left(x+\frac{b}{2a}\right)}^2-{\left(\frac{\sqrt{\mathrm{\Delta }}}{2a}\right)}^2\right]& \hfill & \\ \hfill & =a(x-x_1)(x-x_2),x_1=\frac{-b-\sqrt{\mathrm{\Delta }}}{2}\in F(i),x_2=\frac{-b+\sqrt{\mathrm{\Delta }}}{2}\in F(i)& \hfill & \end{array}$$

splits completely over $F(i)$.

(c)

By definition of a real closed field, and by (b),

$\text{∙}$ every polynomial of odd degree in $F[x]$ has a root in $F$,

$\text{∙}$ every element $a\in F(i)$ has a square root in $F(i)$,

$\text{∙}$ every quadratic polynomial $f\in F(i)[x]$ splits completely over $F(i)$.

The Proposition 3.2.2 and the Lemme 3.2.3 are so satisfied if we replace $ℝ$ by $F$ and $ℂ$ by $F(i)$.

Theorem 3.2.4 for $F(i)$ follows, with the same proof: $F(i)$ is an algebraically closed field.