Exercise 4.4.8

Proof.

(a)

Let $ℝ\subset L$ be an algebraic extension.

If $x^2+1$ has a root $\alpha$ in $L$, we can take $K=L$. Otherwise $x^2+1$, being of degree 2, is irreducible over $L$, thus $K=L[x]∕⟨x^2+1⟩$ is an extension of $L$ containing $\alpha =\bar{x}=x+⟨x^2+1⟩$, root of $x^2+1$ in $K$.

In the two cases, there exists an extension $L\subset K$ such that $x^2+1$ has a root $\alpha$ in $K$ (and $[L[\alpha ]:L]\le \mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->(x^2+1)=2$).

(b)

Let $\beta \in L(\alpha )$. As $L[\alpha ]$ is algebraic over $L$ (since $[L(\alpha ):L]\le 2)$, and as $L$ is algebraic over $ℝ$, the Theorem 4.4.7 shows that $\beta$ is algebraic over $ℝ$. As the coefficients of the minimal polynomial of $\beta$ over $ℝ$ are real, these coefficients are a fortiori in $ℝ(\alpha )$, thus $L(\alpha )$ is algebraic over $ℝ(\alpha )$.

As $\alpha$ is a root of $x^2+1$, irreducible over $ℝ$, $ℝ(\alpha )=ℝ[\alpha ]\simeq ℝ[x]∕⟨x^2+1⟩\simeq ℂ$.

(c)

As $ℝ(\alpha )$ is isomorphic to $ℂ$, $ℝ(\alpha )$ is an algebraically closed field. Moreover $L(\alpha )$ is algebraic over $ℝ(\alpha )$. By Exercise 4.4.7, $L(\alpha )=ℝ(\alpha )$.

Since

$$\begin{array}{lll}\hfill 2=[ℝ(\alpha ):ℝ]=[L(\alpha ):ℝ]=[L(\alpha ):L]\times [L:ℝ],& & \hfill \text{(1)}\end{array}$$

$[L:ℝ]$ divides 2, thus $[L:ℝ]=1$ or $2$.

Conclusion: Every algebraic extension of $ℝ$ is finite of degree at most 2.

By (1),

$$\begin{array}{llll}\hfill [L:ℝ]=2& ⟺[L(\alpha ):L]=1& \hfill & \\ \hfill & ⟺L(\alpha )=L& \hfill & \\ \hfill & \Rightarrow ℂ\simeq L& \hfill & \end{array}$$

Conversely, if $ℂ\simeq L$, then $L(\alpha )\simeq L$. Let $\phi :L(\alpha )\to L$ be an isomorphism. Then $\beta =\phi (\alpha )\in L$ satisfies ${\beta }^2+1=0$, thus $\beta \notin ℝ$. Consequently $ℝ⊊L$, $1<[L:ℝ]\le 2$, thus $[L:ℝ]=2$.

$$[L:ℝ]=2⟺L\simeq ℂ.$$