To show that \(M_{\infty }\) is a subfield of \(\mathbb {C}\) we have to show that \(0,1\in M_{\infty }\) and \(M_{\infty }\) is closed under addition, multiplication, subtraction and division.

• This follows from \(0,1 \in M\) and Lemma 1.22.

• For \(z_1,z_2 \in M_{\infty }\) we can construct \(z_1 + z_2 \in M_{\infty }\). 1.31

• For \(z_1,z_2 \in M_{\infty }\) we can construct \(z_1 \cdot z_2 \in M_{\infty }\). 1.42

• For \(z \in M_{\infty }\) we can construct \(-z \in M_{\infty }\). 1.30

• For \(z \in M_{\infty }\) with \(z \ne 0\) we can construct \(z^{-1} \in M_{\infty }\). 1.44

This lemma is a direct consequence of section 1.4.

• We can apply construction 1.36

• We can apply construction 1.39 and 1.36.

• We can apply construction 1.48 and 1.45.

It is established that \(M_{\infty }\) is a field (see remark 2.3). Furthermore, the corollary 1.51 provides a root \(z^{\frac{1}{2}}\) of \(z \in M_{\infty }\).

Therefore \(M_{\infty }\) is quadratic closed.

For \(z \in Conj(M\cup N)\) there is a \(w \in M\cup N\) such that \(\overline{w} = z\), therefore \( z = \overline{w} \in Conj(M) \cup Conj(N)\). The other direction is analog.

We can apply construction 1.34 and the fact that \(\overline{\overline z} = z\) for all \(z \in \mathbb {C}\).

We can apply Lemma 2.9 and 2.10.

For every \(r \in \mathbb {Q}\) we have \(\overline{r} = r\).

For \(z \in Conj(M)\) there is a \(w \in M\) such that \(\overline{w} = z\) and since \(M \subseteq N\) we have \(w \in N\) and therefore \(z \in Conj(N)\).

We have to show that \(0,1 \in Conj(F)\) and \(Conj(F)\) is closed under addition, multiplication, negation and inversion.

• Since \(0,1 \in F\) and \(\overline{0} = 0, \overline{1} = 1\) we have \(0,1 \in Conj(F)\).

• For \(z_1,z_2 \in F\) we have \(\overline{z_1 + z_2} = \Re (z_1 + z_2) - \imath\cdot \Im (z_1 + z_2) = \Re (z_1) + \Re (z_2) - \imath\cdot (\Im (z_1) + \Im (z_2)) = \overline{z_1} + \overline{z_2}\). Therefore \(Conj(F)\) is closed under addition.

• For \(z \in F\) we have \(\overline{-z} = \Re (-z) - \imath\cdot \Im (-z) = -(\Re (z) - \imath\cdot (\Im (z))) = -\overline{z}\). Therefore \(Conj(F)\) is closed under negation.

• Since \(\overline{z_1 \cdot z_2} = \overline{z_1} \cdot \overline{z_2}\) we have \(Conj(F)\) is closed under multiplication.

• Since \(\overline{z^{-1}} = \overline{z}^{-1}\) we have \(Conj(F)\) is closed under inversion.

Let \(z = x + \imath y \in L\). Since \(L\) is conjugate closed we know that \(\overline{z}=x-\imath y \in L\). This implies

and therefore also \(\imath y = z - x \in L\).

For \(z_1 = x_1 + \imath y_1\) and \(z_2 = x_2 + \imath y_2\) we have

and therefore

After applying Lemma 2.17 we get \(r^2 \in L\).

The proof is divided into two parts. Initially, it is demonstrated that \(z\) belongs to the intersection of \(G_1\) and \(G_2\), if and only if there exist real numbers \(\lambda , \mu \in \mathbb {R}\), such that the equations \(1\), \(2\) and \(3\) are satisfied. Subsequently, it follows that \(z\) is an element of \(L\).
Equations \(1\) and \(2\) are equivalent to the following:

This is the definition of \(z \in G_1 \cap G_2\), expressed in terms of its real and imaginary parts.
The third equation is equivalent to \(z = \lambda z_1 + (1 - \lambda )z_2\). This allows us to conclude that \(z\) belongs to \(G_1\) at the point where \(G_1\) and \(G_2\) intersect. Consequently, we can assume that \(z\) belongs to the intersection of \(G_1\) and \(G_2\).

Now we can show that \(z\) is an element of \(L\).
Since we know that z is equal to \(\lambda z_1 + (1-\lambda )z_2\) and \(z_1, z_2 \in L\) we only have to show that \(\lambda \in L\). Here for we use the equations from the first part of the proof.

Now we solve 2 for \(\mu \)

Since we divided by \(\imath(y_4-y_3)\) we need to assume that \(y_4 \ne y_3\), so we need to first handle the case \(y_4 = y_3\).
If \(y_4 = y_3\) we have \(\lambda (\imath y_1 - \imath y_2) = \imath y_4-\imath y_2\) and since \(y_4 = y_3\) \(y_1 \ne y_2\), because otherwise both Lines would be parallel to the real line and therefore \(G_1 = G_2\) or \(G_1 \cap G_2 = \varnothing \). Therefore \(\lambda = \frac{\imath y_4-\imath y_2}{\imath y_1 - \imath y_2}\). Using the fact that real part and the imaginary part times \(\)

First we have to show \(z \in G \cap C\) iff and only iff there exists a \(\lambda \in \mathbb {R}\) with \(\lambda ^2 a+ \lambda b + c = 0\) and \(z = \lambda z_1 + (1-\lambda )z_2\).

"\(\Rightarrow :\)" If z belongs to the intersection of G and C, then z satisfies the equations of C and G. Consequently

"\(\Leftarrow :\)" Since \(z = \lambda z_1 + (1-\lambda )z_2\) we get \(z\) in \(G\) and can use the equations from the first part of the proof to show that \(z\) is in \(C\).

Now we can show that there exists a \(w \in L\) such that \(z \in L(\sqrt{w})\).
Since we know that \(z = \lambda z_1 + (1-\lambda )z_2\) and \(z_1, z_2 \in L\) we only have to show that \(\lambda \in L(\sqrt{w})\). To do this, we use the equations from the first part of the proof. Since \(\lambda \) is a solution of a quadratic equation we now get that \(\lambda \) is equal to \(\frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). Since \(a,b,c \in L\) we get \(w = b^2 - 4ac \in L\) so \(\lambda \in L(\sqrt{w})\). Therefore \(z = \lambda z_1 + (1-\lambda )z_2\) is in \(L(\sqrt{w})\).

The initial step is to demonstrate that \( z\in C_1 \cap C_2 \Leftrightarrow \exists G, z\in G \cap C_1 \land z \in G \cap C_2\).
"\(\Rightarrow \)": If \(z\) is to be in both \(G \cap C_1\) and \(G \cap C_2\), it must be the case that \(z\) is in both \(C_1\) and \(C_2\). Consequently, it is also in \(C_1 \cap C_2\).
"\(\Leftarrow \)": We begin by establishing that \(z=x + iy\in C_1\cap C_2\) is equivalent to:

The remaining task is to identify two elements in \(L\) that satisfy the given equation. This can be achieved by considering three cases: In the initial case, where \(x_1 = x_2\), it can be demonstrated that \(y_1 \ne y_2\), as otherwise it would follow that \(C_1=C_2\). Here, we can choose

In the second case we have \(y_1 = y_2\) and \(x_1 \ne x_2\). Here we can choose ¹

For \( x_1 \ne x_2\) and \(y_1 \ne y_2\) chose

Since the points \(z_1\) and \(z_2\) lie in \(L\), we can conclude that the line \(G\) lies in the set of lines of \(L\). This allows us to apply the results stated in lemma 2.20 to obtain \(w\), with the result that \(z\) is contained within the set \(L(w)\).

In reference 2.16, it was demonstrated that for a field F, the field of complex numbers, \(Conj(F)\) is a field. It can thus be concluded that \(Conj(K(M))\) is also a field. As both \(K\) and \(M\) are subsets of \(K(M)\), it can be inferred from lemma 2.15 that \(Conj(K) \overset {ConjClosed}{=} K\) and \(Conj(M) \overset {ConjClosed}{=} M\) are subsets of \(Conj(K(M))\). As \(K(M)\) is the smallest subfield of \(\mathbb {C}\) that includes \(K\) and \(M\), it can be concluded that

Furthermore, if we apply \(Conj\) to both sides and again infer 2.15, we obtain

which leads to the conclusion that \(Conj(K(M)) = K(M)\).

By employing the preceding lemma, it is sufficient to demonstrate that \(\mathbb {Q}\) and \(M \cup Conj(M)\) are conjugate closed, which can be inferred from 2.14 and 2.13

From the definition of \(K_0 := \mathbb {Q}(M\cup Conj(M))\), it can be seen that this is the smallest subfield of \(\mathbb {C}\) containing both \(Q\) and \(M\cup Conj(M)\). Consequently, it is sufficient to demonstrate that both \(\mathbb {Q}\) and \(M\cup Conj(M)\) are contained within \(M_{\infty }\). Since \(\mathbb {Q}\) is contained in every subfield of \(\mathbb {C}\), it is therefore also contained in \(M_{\infty }\). Furthermore, since \(M\) is contained in \(M_{\infty }\) (see 1.22) and \(M\) is conjugate closed (2.12), we can conclude that \(M\cup Conj(M) \subseteq \mathbb {Q}\).

\((ii)\implies (i)\): Let \(w\) be as in \((ii)\).Then \(\sqrt{w}\) is a root of \(X^2 - w \in K[X]\). Since \(\sqrt{w} \notin K\) this polynomial is irreducible in \(K[X]\). Therefore \([L:K] = 2\).
\((i)\implies (ii)\): Let \([L:K] = 2\) and \(\alpha \in L \setminus K\). Then \(K(\alpha ) = L\) and

This implies

Now let \(w := \frac{b^2}{4} - c \in K\) then we get \(L = K(\alpha ) = K(\sqrt{w})\).

"\(\Rightarrow :\)" Let \(n\) be the smallest natural number such that there exists a chain of subfields \(L_1 \subset L_2 \subset \ldots \subset L_n\) with \(z \in L_n\) and \([L_{i+1}:L_i] = 2\) for \(i = 0, \ldots , n-1\). Now lemma 2.26 gives us that there exists a \(w \in L_{n-1}\) such that \(L_n\) arises from \(L_{n-1}\) by adjoining \(\sqrt{w}\).
"\(\Leftarrow :\)" Let \(L\) be an intermediate field of \(\mathbb {C}/K_0\) with \(z \in L\) and \(L\) arises from \(K_0\) by a finte sequence of adjunctions of square roots. Then there exists a chain of subfields \(L_1 \subset L_2 \subset \ldots \subset L_n\) with \(z \in L_n\) and \(L_{i+1}:= L_i(\sqrt{w_i})\) for \(i = 0, \ldots , n-1\).

Induction over \(n\)

• Base case \(n=1\):
  \(K_0 \le K_0\) and \(K_0\) is conjugation closed 2.24.

• induction hypothesis:
  Assume that for \(n\) there is a chain of conjugation closed intermediate fields \(K_0 \le K_1 \le \ldots \le K_n\) such that \(M_i\subset K_i\) and \(K_i:= K_{i-1}(X_i)\) for a set of square roots \(X_i\) of elements of \(K_{i-1}\).

• Inductive step \(n \to n+1\):
  For \(z\in M_{n+1}\) there are four cases:

  • By the induction hypothesis \(z \in K_n\) and \(K_n\) is conjugation closed and arises from \(K_0\) by a sequence of adjunctinos of square roots.

  • By the induction hypothesis \(z \in ILL (K_n)\) and using 2.19 we get that \(z \in K_{n}\).

  • By the induction hypothesis \(z \in ILC (K_n)\) and using 2.20 there is a \(w \in K_n\) such that \(z \in K_n(\sqrt{w})\) insert \(\sqrt{w}, \overline{\sqrt{w}}\) to \(X_n\).

  • By the induction hypothesis \(z \in ICC (K_n)\) and using 2.21 there is a \(w \in K_n\) such that \(z \in K_n(\sqrt{w})\) insert \(\sqrt{w}, \overline{\sqrt{w}}\) to \(X_n\).

"\(\Leftarrow :\)" It can be shown by induction that \(L_n\) is contained within \(M_{\infty }\). Therefore, it can be inferred that \(z\) is also contained within \(M_{\infty }\).

• Base case \(n=1\):
  \(L_1 = K_0 \subseteq M_{\infty }\) 2.25.

• induction hypothesis:
  Assume that for \(n\): \(\forall i {\lt} n: L_i \le L_{i+1} \land [L_{i+1}:L_i]=2\) implies \(L_n \subseteq M_{\infty }\).

• Inductive step \(n \to n+1\):
  Given that \([L_{n+1}:L_n] = 2\), it follows from the conclusions of Lemma 2.26 that there exists a \(w \in L_n\) with the property that \( \sqrt{w} \notin L_n\) and \(L_{n+1} = L_n(w)\). By the induction hypothesis, it can be inferred that \(L_n \subseteq M_{\infty }\). Since \(w \in L_n \subseteq M_{\infty }\) and \( M_{\infty }\) is quadratic closed (2.7) \(L_n(w) = L_{n+1} \subseteq M_{\infty }\).

"\(\Rightarrow :\)" There exists a \(n\) such that \(z\in M_n\), and we know that there exists a \(K_n\) with \(M_n \subseteq K_n\) which is derived from \(K_0\) by successive adjoing square roots 2.28. We can conclude that there is a \(K\), which is derived from \(K_0\) by successive adjoining square roots, and that \(z\in K\). Since \(M_i\) is finite, we get that we adjoin finitely many square roots and so we evoke 2.27.

By Theorem 2.29, it can be inferred that there exists a chain of subfields, \(K_0 = L_1 \subset L_2 \subset \dots \subset L_n \subset \mathbb {C}\), with \(z \in L_n\) and \([L_i : L_{i+1}] = 2\) for \(i = 0, 1, \dots , n-1\). Moreover, the multiplicativity formula for degrees indicates that the degree of the extension \([L_n : K_0]\) is equal to the product of the degrees of the extensions \([L_n : L_{n-1}] \cdot [L_{n-1} : L_{n-2}] \cdot \dots [L_2 : L_1]\). Thus, we have that \([L_n : K_0] = 2^n\). The fact that \(z\in L_n\) implies that \(K_0(z)\subseteq L_n\). It thus follows that the index of the field extension \([L_n : K_0] = [L_n : K_0(z)]\cdot [K_0(z) : K_0]\), which implies that \([K_0(z) : K_0]\) is a divisor of \(2^n\).

Contraposition of Lemma 2.30.

A combination of the multiplicativity formula for degrees and corollary 2.31.