Definition 17.20. Intermediate Field.
Given a field extension \(F \subseteq L\text{,}\) an intermediate field is a subfield \(E\) of \(L\) that contains \(F\text{,}\) so that \(F \subseteq E\subseteq L\text{.}\)
Section 17.2 The Fundamental Theorem of Galois Theory
Subsection The Fundamental Theorem: Statement and Uses
Proposition 17.21. Intermediate Extensions are Galois.
If \(F \subseteq L\) is a (finite) Galois extension, then so is \(E \subseteq L\) for any intermediate field \(E\text{.}\)
Proof.
If \(L\) is the splitting field over \(F\) of a separable polynomial \(f(x) \in F[x]\text{,}\) then \(L\) is also the splitting field over \(E\) of the same polynomial, now regarded as belonging to \(E[x]\text{.}\)
Warning 17.22.
Example 17.23.
\(L = \Q(\sqrt[3]{2}, e^{2 \pi i/3})\) is Galois over \(F = \Q\) but \(E = \Q(\sqrt[3]{2})\) is not Galois over \(\Q\text{.}\)
Theorem 17.24. Fundamental Theorem of Galois Theory.
Suppose \(F \subseteq L\) is a (finite) Galois field extension. Then the functions
\begin{equation*}
\Psi:\left\{\text{intermediate fields $E$, with $F\subseteq E \subseteq L$}\right\} \to\left\{\text{subgroups $H$ of $\Gal(L/F)$}\right\}
\end{equation*}
given by
\begin{equation*}
\Psi(E)=\Gal(L/E)
\end{equation*}
is a bijection whose inverse
\begin{equation*}
\Psi^{-1}:
\left\{\text{subgroups $H$ of $\Gal(L/F)$}\right\} \to
\left\{\text{intermediate fields $E$, with $F \subseteq E \subseteq L$}\right\}
\end{equation*}
is given by
\begin{equation*}
\Psi^{-1}(H)=L^H.
\end{equation*}
Moreover, this bijective correspondence enjoys the following properties:
- \(\Psi\) and \(\Psi^{-1}\) each reverse the order of inclusions.
- \(\Psi\) and \(\Psi^{-1}\) convert degrees of extensions to indices of subgroups:
- \([\Gal(L/F) : H] = [L^H:F]\) or, equivalently,
- \([\Gal(L/F) : \Gal(L/E)] = [E:F]\text{.}\)
- Normal subgroups correspond to intermediate fields that are Galois over \(F\text{:}\)
- If \(N \nsg G\) then \(L^N/F\) is Galois.
- If \(E/F\) is Galois, then \(\Gal(L/E) \nsg \Gal(L/F)\text{.}\)
- If \(E = L^N\) for a normal subgroup \(N \nsg \Gal(L/F)\text{,}\) then \(\Gal(E/F) \cong \Gal(L/F)/N\text{.}\)
- If \(H_1, H_2\) are subgroups of \(G\) with corresponding fixed subfields \(E_1=L^{H_1}\) and \(E_2=L^{H_2}\text{,}\) then
- \(E_1\cap E_2=L^{<H_1,H_2>}\) and \(\Gal(L/E_1\cap E_2)=\langle H_1,H_2\rangle\)
- \(E_1 E_2=L^{H_1\cap H_2}\) and \(\Gal(L/E_1E_2)=H_1\cap H_2\text{.}\)
Proof.
First, we need to check that both functions are well-defined. For each intermediary field \(E\text{,}\) we know from Corollary 6.31 that \(L / E\) is also Galois, and hence it makes sense to write \(\operatorname{Gal}(L / E)\text{;}\) moreover, any \(\sigma \in \operatorname{Gal}(L / E)\) is an automorphism of \(L\) that fixes \(E\text{,}\) and thus \(F \subseteq E\text{,}\) so \(\sigma \in \operatorname{Gal}(L / F)\text{.}\) This shows that \(\Psi\) is well-defined. Conversely, given a subgroup \(H\) of \(\operatorname{Gal}(L / F), L^{H}\) is a subfield of \(L\) by Exercise 18.
Next, we need to check that \(\Psi\) and \(\Psi^{-1}\) are indeed inverse functions. Given a subgroup \(H\) of \(\operatorname{Gal}(L / F)\text{,}\) we have \(\operatorname{Gal}\left(L / L^{H}\right)=H\) by Artin’s Theorem. Thus
\begin{equation*}
\Psi \circ \Psi^{-1}(H)=\Psi\left(L^{H}\right)=\operatorname{Gal}\left(L / L^{H}\right)=H .
\end{equation*}
Conversely, given an intermediate field \(E, L / E\) is Galois by Corollary 6.31, and hence \(L^{\operatorname{Gal}(L / E)}=E\) by Corollary 6.24. Thus
\begin{equation*}
\Psi^{-1} \circ \Psi(E)=\Psi(\operatorname{Gal}(L / E))=L^{\operatorname{Gal}(L / E)}=E .
\end{equation*}
This establishes the fact that \(\Psi\) is indeed a bijective correspondence.
Now we check that \(\Psi\) satisfies the given list of properties. For brevity, set \(G:=\operatorname{Gal}(L / F)\text{.}\)
- The fact that the correspondence is order reversing follows from the definitions. Given intermediate fields \(E_{1} \subseteq E_{2}\text{,}\) any automorphism of \(L\) that preserves \(E_{2}\) must also preserve \(E_{1}\text{,}\) thus \(\operatorname{Gal}\left(L / E_{2}\right) \supseteq \operatorname{Gal}\left(L / E_{1}\right)\text{.}\) Conversely, if \(H_{1} \leq H_{2} \leq \operatorname{Gal}(L / E)\text{,}\) then every \(x \in L\) that is fixed by every \(\sigma \in H_{2}\) must also be fixed in particular by every element of \(H_{1}\text{,}\) so \(L^{H_{2}} \supseteq L^{H_{1}}\text{.}\)
- By definition of Galois extension, \([L: F]=|G|\text{.}\) By Artin’s Theorem, for any subgroup \(H \leq G\) the extension \(L^{H} \subseteq L\) is also Galois, and thus by definition \(\left[L: L^{H}\right]=|H|\text{.}\) Using the Degree Formula, we have\begin{equation*} \left[L^{H}: F\right]=\frac{[L: F]}{\left[L: L^{H}\right]}=\frac{|G|}{|H|}=[G: H] . \end{equation*}So if \(H=\Psi(E)=\operatorname{Gal}(E / F)\text{,}\) then \(L^{H}=E\) and the formula above can be rewritten as\begin{equation*} [\operatorname{Gal}(L / F): \operatorname{Gal}(L / E)]=[E: F] . \end{equation*}
-
Suppose \(E\) is an intermediate field that is Galois over \(F\text{.}\) Fix \(\sigma \in G\) and \(\alpha \in E\text{.}\) Since \(E / F\) is Galois, by Corollary 6.26 the polynomial \(m_{\alpha, F}\) is separable and all of its roots are in \(E\text{.}\) By Lemma \(6.10, \sigma(\alpha)\) is also a root of \(m_{\alpha, F}\text{,}\) and thus \(\sigma(\alpha) \in E\text{.}\)Suppose \(\tau \in \operatorname{Gal}(L / E)\text{.}\) For any \(\alpha \in E\) we have \(\sigma(\alpha) \in E\text{,}\) so \(\tau(\sigma(\alpha))=\sigma(\alpha)\text{.}\) Thus\begin{equation*} \sigma^{-1}\left(\tau(\sigma(\alpha))=\sigma^{-1}(\sigma(\alpha))=\alpha\right). \end{equation*}This proves that \(\sigma^{-1} \tau \sigma \in \operatorname{Gal}(L / E)\) and hence that \(\operatorname{Gal}(L / E) \unlhd G\text{.}\) We have shown that if \(E\) is Galois over \(F\text{,}\) then the corresponding subgroup \(\operatorname{Gal}(L / E)\) of \(G\) is normal.For the converse, consider a normal subgroup \(N \unlhd G\) and the corresponding intermediate field \(E=L^{N}\text{,}\) so that \(N=\operatorname{Gal}(L / E)\text{.}\) We will show that \(E\) is the splitting field over \(F\) of a separable polynomial in \(F[x]\text{,}\) and hence is Galois over \(F\) by Corollary 6.18.Pick any \(\alpha \in E\) and set \(m:=m_{\alpha, F}\text{.}\) By Corollary 6.26, \(m\) is separable and all of its roots belong to \(L\text{.}\) We claim that all the roots must in fact belong to \(E\text{.}\) Since \(m\) is irreducible and \(L / F\) is Galois, by Corollary \(6.26 G\) acts transitively on the set of roots of \(m\text{.}\) Thus, given be any other root \(\beta \in L\) of \(m\text{,}\) there is a \(\sigma \in G\) with \(\sigma(\alpha)=\beta\text{.}\) Since \(N\) is normal, for any \(\tau \in N\) we have \(\sigma \tau^{\prime}=\tau \sigma\) for some \(\tau^{\prime} \in N\text{.}\) But \(\tau^{\prime} \in N\) fixes \(E\text{,}\) so \(\tau^{\prime}(\alpha)=\alpha\text{.}\) Therefore,\begin{equation*} \beta=\sigma(\alpha)=\sigma \tau^{\prime}(\alpha)=\tau \sigma(\alpha)=\tau(\beta) \end{equation*}which shows that \(\beta\) is also fixed by \(N\text{.}\) But then \(\beta \in L^{N}=E\text{.}\) Therefore, \(E\) contains all the roots of \(m_{\alpha, F}\text{,}\) and thus \(E\) must contain the splitting field of \(m_{\alpha, F}\text{.}\)We have \(E=F\left(\alpha_{1}, \ldots, \alpha_{l}\right)\) for some \(\alpha_{1}, \ldots, \alpha_{l} \in E\text{.}\) If \(m_{1}, \ldots, m_{n}\) are the distinct polynomials among \(m_{\alpha_{1}, F}, \ldots, m_{\alpha_{l}, F}\text{,}\) then \(E\) is the splitting field of the separable polynomial \(m_{1} \cdots m_{n}\text{.}\) By Corollary 6.28, \(E\) is Galois over \(F\text{.}\)If \(E=L^{N}\) for a normal subgroup \(N \unlhd \operatorname{Gal}(L / F)\text{,}\) then \(\operatorname{Gal}(E / F) \cong \operatorname{Gal}(L / F) / N\text{.}\)
-
Let \(E=L^{N}\) for a normal subgroup \(N\) of \(G\text{.}\) We want to show that \(\operatorname{Gal}(E / F)\) is isomorphic to \(G / N\text{.}\)For each \(\sigma \in G\text{,}\) we claim that \(\sigma(E) \subseteq E\text{.}\) By Lemma 6.10, for all \(\alpha \in E\) the element \(\sigma(\alpha)\) is also a root of \(m_{\alpha, F}\text{.}\) But since \(E / F\) is Galois, it must contain all of the roots of \(m_{\alpha, F}\text{,}\) by Corollary 6.26, so \(\sigma(\alpha) \in E\text{.}\) Thus \(\sigma(E) \subseteq E\text{,}\) so the restriction of \(\sigma\) to \(E\) determines an injective field homomorphism \(\left.\sigma\right|_{E}: E \rightarrow E\text{.}\) Since \(\left.\sigma\right|_{F}=\mathrm{id}_{F}\text{,}\) this map is also a linear transformation of vector spaces over \(F\text{.}\) But \(E\) is a finite vector space over \(F\text{,}\) and any injective linear transformation \(E \rightarrow E\) must be bijective. We conclude that \(\left.\sigma\right|_{E}\) is an automorphism of \(E\text{.}\) We thus have a well-defined function\begin{equation*} \begin{gathered} \phi: G \longrightarrow \operatorname{Gal}(E / F) \\ \sigma \longmapsto \phi(\sigma)=\left.\sigma\right|_{E} . \end{gathered} \end{equation*}Moreover, this map is a group homomorphism by construction. The kernel is the subgroup of \(G\) of automorphisms that restrict to the identity on \(E\text{,}\) which is precisely \(N\text{.}\) Hence we have an induced injective group homomorphism\begin{equation*} \bar{\phi}: G / N \rightarrow \operatorname{Gal}(E / F) . \end{equation*}But \(|N|=|\operatorname{Gal}(E / F)|<\infty\text{,}\) so this map \(\bar{\phi}\) must be an isomorphism.
Corollary 17.25. Galois Correspondence and Lattices.
The Galois correspondence induces a lattice isomorphism between the lattice of intermediate fields of a Galois extension \(F \subseteq L\) and the “dual’’ of the lattice of subgroups of \(\Gal(L/F)\text{.}\)
Example 17.26. FTGT and \((x^4-2)\).
Let \(L\) be the splitting field of \(x^4 - 2\) over \(\Q\text{.}\) Let’s use the fundamental theorem to list all intermediate fields for \(\Q \subseteq L\) and to determine which are Galois over \(\Q\text{.}\) Notice that without the theorem, it isn’t even obvious that there are only a finite number of such intermediate fields.
Solution.
We know \(G := \Gal(L/\Q)\) corresponds to the \(8\) element subgroup of \(S_4\) generated by \(\s = (2 \, 4)\) and \(\tau = (1 \, 2 \,3 \, 4)\) where we number the roots as \(\a_1 = \sqrt[4]{2}, \a_2 = i \a_1, \a_3 = -\a_1, \a_4 =-i\a_1\text{.}\)
This group is isomorphic to \(D_8\) and we can make this isomorphism explicit by labeling the four corners of a square by \(\a_1, \dots,\a_4\text{,}\) counter-clockwise. So, \(\tau\) is rotation by \(90\) degrees and \(\s\) is reflection about the line joining vertices \(1\) and \(3\text{.}\)
The subgroup lattice and intermediate field lattice are represented below, with normal subgroups and Galois extensions highlighted (boxed).
\begin{equation*}
\begin{array}{ll}
\{e\} & G=\langle(24),(1234)\rangle \\
H_{1}=\langle(24)\rangle & H_{5}=\langle(13)(24)\rangle \\
H_{2}=\langle(13)\rangle & H_{6}=\langle(1234)\rangle \\
H_{3}=\langle(12)(34)\rangle & H_{7}=\langle(13),(24)\rangle \\
H_{4}=\langle(14)(23)\rangle & H_{8}=\langle(12)(34),(14)(23)\rangle
\end{array}
\end{equation*}
and the lattices are of subgroups of \(G\) and intermediate fields of \(\mathbb{Q} \subseteq \mathbb{Q}\left(\alpha_{1}, i\right)\) are
The intermediate fields are the fixed subfields of \(L\) associated to each of these subgroups. In some sense, this answers the question, but let’s find explicit generators for at least some of these.
\(G\) corresponds to \(\Q\) and \(e\) corresponds to \(L = \Q(\a_1, i)\text{.}\)
Set \(E_i = L^{H_i}\text{.}\) \(E_1\) has degree \(4 = [G:H_1]\) over \(\Q\text{.}\) It is clear \(\a_1\) (and \(\a_3\)) belongs to \(E_1\) and since \([\Q(\a_1):\Q]=4\text{,}\) we must have \(E_1 = \Q(\a_1)\text{.}\)
Likewise \(E_2 = \Q(\a_2)\text{.}\)
\(E_3\) also has degree four over \(\Q\text{.}\) Let \(\beta = \a_1 + \a_2 = (1+i)\sqrt[4]{2}\) and note \(\beta \in E_3\text{.}\) If \([\Q(\beta):\Q] = 2\text{,}\) then \(\beta\) would be fixed by a subgroup of index \(2\) that contains \((1 \,2)(3 \, 4)\text{,}\) and the only possibility is \(H_8\text{.}\) But \((1 \, 4)(2 \, 3)\) sends \(\beta\) to \(\a_4 + \a_3 = - \beta \ne \beta\text{.}\) So we must have \([\Q(\beta):\Q] = 4\) and hence \(E_3 = \Q(\beta)\text{.}\)
I’ll skip the details on \(E_4\) and \(E_5\text{,}\) but they are \(E_4 =\Q((1-i)\a_1)\) and \(E_5 = \Q(\sqrt{2},i)\text{.}\)
\(E_6\) has degree equal to \([G:H_6] = 2\) over \(\Q\) and so we merely need to find a single, non-rational element of \(L\) fixed by \(\tau\text{.}\) Since \(\tau(i) = i\) (which can be seen by looking back at how we built \(\tau\) originally or by noting that \(\tau(i) = \tau(\a_2/\a_1) = \a_3/\a_2= i\)), we get \(E_6 = \Q(i)\text{.}\)
I’ll skip the details on \(E_7\text{,}\) but it is \(E_7 = \Q(\sqrt{2})\text{.}\)
\(E_8\) also has degree two over \(\Q\) and so we just need to find a single non-rational element fixed by the two generators of \(H_8\text{.}\) Note that \(\a_1\a_2 = \a_3\a_4 = i \sqrt{2}\) and so \(i \sqrt{2}\) is fixed by both \((1 \, 2)(3 \, 4)\) and \((1 \,4)(2 \, 3)\text{.}\) Thus \(E_8 = \Q(i \sqrt{2})\text{.}\)
Finally, we note that \(G, \{e\}, H_5, H_6, H_7, H_8\) are normal subgroups of \(D_8\text{,}\) since \(H_5\) is the center of \(D_8\) and each of \(H_6, H_7, H_8\) has index two. Some messy checking reveals these to be the only normal subgroups. It follows from the Fundamental Theorem of Galois Theory that \(\Q, L, E_5, E_6, E_7, E_8\) are the only intermediate fields that are Galois over \(\Q\text{.}\) As an example, to see directly that \(E_3\) is not Galois over \(\Q\text{,}\) note that \((1+i)\sqrt[4]{2}\) is a root of \(x^4 + 4\text{,}\) which is irreducible. But \((1-i)\sqrt[4]{2}\) is also a root of this polynomial and it is not in \(E_3\text{.}\) We conclude from Proposition
[provisional cross-reference: cite] that \(E_3\) is not a normal extension of \(\Q\text{.}\)
Example 17.27. Cyclotomic Extensions Revisited.
Let \(F\) be a field, let \(n\) be a positive integer such that \(\char(F)\) does not divide \(n\text{,}\) and let \(\ov{F}\) be the algebraic closure of \(F\text{.}\) If \(\zeta\ov{F}\) is a primitive \(n\)-th root of \(1\) over \(F\text{,}\) then \(F(\zeta)/F\) is a finite Galois extension, and \(\Gal(F(\zeta)/F)\) is a cyclic group that is isomorphic to a subgroup of \((\Z/n,+)\text{.}\)
Remark 17.28.
If \(F\) is a field of prime characteristic \(p\) and \(n\) is an integer that is divisible by \(p\text{,}\) then \(x^n - 1\) is not separable.
Example 17.29.
Let \(f\) be an irreducible, fifth degree polynomial in \(\Q[x]\) with exactly three real roots and let \(L\) be its splitting field over \(\Q\text{.}\) Let the real roots be \(\a_1, \a_2, \a_3\) and the imaginary ones be \(\a_4\) and \(\a_5\text{.}\) Then \(G = \Gal(L/\Q)\) is isomorphic to \(S_5\text{.}\) You will prove this on one of the final exam problems (for a specific case), but let me get you started with a couple {}: Show that \(G\) is isomorphic to a subgroup \(H\) of \(S_5\) with \(5 \mid |H|\text{.}\) You may use (without justifying it) that if \(\tau\) is a \(2\)-cycle in \(S_5\) and \(\s\) is any five cycle in \(S_5\text{,}\) then the subgroup of \(S_5\) they generate is all of \(S_5\text{.}\)
Example 17.30. Finding Unique Intermediate Field.
If \(F \subseteq L\) is a finite Galois extension of degree \(21\text{,}\) then I claim there is a unique intermediate field \(E\) with \([E:F] = 3\) and that \(E\) must be a Galois extension over \(F\text{.}\)
To see this, set \(G= \Gal(L/F)\text{.}\) Then \(|G| = 21 = 3 \cdot 7\) and by the Sylow’s Theorems, there is a unique Sylow \(7\)-subgroup, call it \(H\text{,}\) and hence \(H\) is normal in \(G\text{.}\) It follows from the Fundamental Theorem of Galois Theory that \(E := L^H\) is an intermediate field that
- is Galois over \(F\) and
- satisfies \([E: F] = [G :H] = 3\text{.}\) Moreover, it is unique since \(G\) has just one subgroup of index \(3\text{.}\)
In fact, there are exactly two groups of order \(21\) up to isomorphism, the cyclic one and one that is a (non-trivial) semi-direct product of \(\Z/7\) by \(\Z/3\text{.}\) So, there are just two possible lattices of intermediate fields for such a field extension.
