General Galois Extensions
Fall 2021 #4 #algebra/qual/work
Recall that for a given positive integer \(n\), the cyclotomic field \(\mathbb{Q}\left(\zeta_{n}\right)\) is generated by a primitive \(n\)th root of unity \(\zeta_{n}\).

What is the degree of \(Q\left(\zeta_{n}\right)\) over \(Q\) ?

Define what it means for a finite field extension \(L / K\) to be Galois, and prove that the cyclotomic field \(Q\left(\zeta_{n}\right)\) is Galois over \(\mathbb{Q}\).

What is the Galois group of \(\mathbb{Q}\left(\zeta_{n}\right)\) over \(\mathbb{Q}\) ?

How many subfields of \(\mathbb{Q}\left(\zeta_{2021}\right)\) have degree 2 over Q? Note that \(2021=43 \cdot 47\)
Fall 2020 #4 #algebra/qual/completed
Let \(K\) be a Galois extension of \(F\), and let \(F \subset E \subset K\) be inclusions of fields. Let \(G \coloneqq{ \mathsf{Gal}}(K/F)\) and \(H \coloneqq{ \mathsf{Gal}}(K/E)\), and suppose \(H\) contains \(N_G(P)\), where \(P\) is a Sylow \(p\)subgroup of \(G\) for \(p\) a prime. Prove that \([E: F] \equiv 1 \operatorname{mod}p\).
The correspondence:
Normalizers: \begin{align*} N_G(P) = \left\{{g\in G {~\mathrel{\Big\vert}~}gPg^{1}= P}\right\} .\end{align*}

Reduce to a group theory problem: \([E:F] = [G:H]\), despite the fact that \(E/F\) is not necessarily Galois. This is because we can count in towers: \begin{align*} [K:F] = [K:E][E:F] &\implies [G:1] = [K:E][H:1] \\ &\implies {\sharp}G = [K:E] {\sharp}H \\ &\implies [G:H] = {{\sharp}G \over {\sharp}H} = [K:E] .\end{align*}

Essential fact: if \(P \in {\operatorname{Syl}}_p(G)\), we can use that \(P \subseteq N_G(P) \subset H\) and so \(P\in {\operatorname{Syl}}_p(H)\) as well.

Now use that \(N_G(P) \subseteq H\), and do Sylow theory for \(P\) in both \(G\) and \(H\):
 Sylow 3 on \(G\) yields \(n_p(G) = [G: N_G(P)] \equiv 1 \operatorname{mod}p\).
 Sylow 3 on \(H\) yields \(n_p(H) = [G: N_H(P)] \equiv 1 \operatorname{mod}p\).

Claim: \(N_H(P) = N_G(P)\).
 We have \(N_H(P) \subseteq N_G(P)\) since \(H \subseteq G\), so \(hPh^{1}= P\) remains true regarding either \(h\in H\) or \(h\in G\).
 For \(N_G(P) \subseteq N_H(P)\), use that \(N_G(P) \subseteq H\) and so \(gPg^{1}= P\) implies \(g\in H\), so \(g\in N_H(P)\).

Now morally one might want to apply an isomorphism theorem: \begin{align*} {G/ N_G(P) \over H/N_H(P)}= {G/ N_H(P) \over H/N_H(P)}\cong {G\over H} ,\end{align*} but we don’t have normality. However, we can still get away with the corresponding counting argument if everything is finite: \begin{align*} {[G: N_G(P)] \over [H:N_H(P)] }= {[G: N_H(P)] \over [H:N_H(P)] }= {{\sharp}G / {\sharp}N_H(P) \over {\sharp}H / {\sharp}N_H(P)} = {{\sharp}G \over {\sharp}H} = [G: H] .\end{align*}

We have an equation of the form \(n_p(G)/n_p(H) = m\), and we want to show \(m\equiv 1 \operatorname{mod}p\). So write \begin{align*} {n_p(G) \over n_p(H) } = m \implies m n_p(H) &= n_p(G) \\ \implies m n_p(H) &\equiv n_p(G) \operatorname{mod}p \\ \implies m\cdot 1 &\equiv 1 \operatorname{mod}p \\ \implies m &\equiv 1 \operatorname{mod}p .\end{align*}
Fall 2019 Midterm #9 #algebra/qual/completed
Let \(n\geq 3\) and \(\zeta_n\) be a primitive \(n\)th root of unity. Show that \([{\mathbf{Q}}(\zeta_n + \zeta_n^{1}): {\mathbf{Q}}] = \phi(n)/2\) for \(\phi\) the totient function.

Some notation: let \(\alpha_k \coloneqq\zeta_n^k + \zeta_n^{k}\).

Let \(m(x)\) be the minimal polynomial of \(\alpha_1 \coloneqq\zeta_n + \zeta_n^{1}\). Note that \(\alpha_1 \in {\mathbf{Q}}(\zeta_n)\).

Use that \({ \mathsf{Gal}}({\mathbf{Q}}(\zeta_n)/{\mathbf{Q}}) \cong C_n^{\times}\), consisting of maps \(\sigma_k: \zeta \mapsto \zeta^k\) for \(\gcd(k, n) = 1\), of which there are \(\phi(n)\) many.

Galois transitively permutes the roots of irreducible polynomials, so the roots of \(m\) are precisely the Galois conjugates of \(\alpha\), i.e. the Galois orbit of \(\alpha\), so we can just compute it. For illustrative purposes, suppose \(n\) is prime, then \begin{align*} \sigma_1(\zeta_n + \zeta_n^{1}) &= \zeta_n + \zeta_n^{1}=\alpha_1 \\ \sigma_2(\zeta_n + \zeta_n^{1}) &= \zeta_n^2 + \zeta_n^{2} = \alpha_2 \\ \sigma_3(\zeta_n + \zeta_n^{1}) &= \zeta_n^3 + \zeta_n^{3} = \alpha_3 \\ \vdots&\\ \sigma_{n1}(\zeta_n + \zeta_n^{1}) &= \zeta_n^{n1} + \zeta_n^{(n1)} = \zeta_n^{1} + \zeta_n^{1} = \alpha_1 \\ \sigma_{n2}(\zeta_n + \zeta_n^{1}) &= \zeta_n^{n2} + \zeta_n^{(n2)} = \zeta_n^{2} + \zeta_n^{2} = \alpha_2 \\ \sigma_{n3}(\zeta_n + \zeta_n^{1}) &= \zeta_n^{n3} + \zeta_n^{(n3)} = \zeta_n^{3} + \zeta_n^{3} = \alpha_3 ,\end{align*} where we’ve used that \(\zeta^{k} = \zeta^{k\operatorname{mod}n}\). From this, we see that \(\sigma_{k}(\alpha_1)=\sigma_{nk}(\alpha_1)\) and we pick up \((n1)/2\) distinct conjugates.

For \(n\) not prime, the exact same argument runs through \(\phi(n)\) values of \(k\) for \(\sigma_k\), and again yields \(\sigma_{k}(\alpha_1) = \sigma_{\phi(n)  k}(\alpha_1)\). Matching them up appropriately yields \(\phi(n)/2\) distinct roots.
Fall 2019 Midterm #10 #algebra/qual/completed
Let \(L/K\) be a finite normal extension.

Show that if \(L/K\) is cyclic and \(E/K\) is normal with \(L/E/K\) then \(L/E\) and \(E/K\) are cyclic.

Show that if \(L/K\) is cyclic then there exists exactly one extension \(E/K\) of degree \(n\) with \(L/E/K\) for each divisor \(n\) of \([L:K]\).
The setup:
Part 1:
 \(L/K\) is cyclic means \(L/K\) is Galois and \(G\coloneqq{ \mathsf{Gal}}(L/K) = C_n\) for some \(n\).
 By the FTGT, setting \(H \coloneqq{ \mathsf{Gal}}(L/E)\), we get \(H {~\trianglelefteq~}G\) precisely because \(E/K\) is normal, and \({ \mathsf{Gal}}(L/E) = G/H\).
 But then if \(G\) is cyclic, \(H \leq G\) must be cyclic, and \(G/H\) is cyclic as well since writing \(G = C_n = \left\langle{x}\right\rangle\), we have \(G/H = \left\langle{xH}\right\rangle\).
Part 2:

Letting \(G\coloneqq{ \mathsf{Gal}}(L/K) = C_n\), by elementary group theory we have subgroups \(H\coloneqq C_d \leq C_n\) for every \(d\) dividing \(n\).
 A observation we’ll need: every subgroup is normal here since \(G\) is abelian.
 By the fundamental theorem, taking the fixed field of \(H \leq { \mathsf{Gal}}(L/K)\), we obtain some intermediate extension \(E\coloneqq K^H\) fitting into a tower \(L/E/K\).
 By the fundamental theorem, \([E: K] = [G:H] = n/d\), where we’ve used that \(H{~\trianglelefteq~}G\).
 Letting \(d\) range through divisors lets \(n/d\) range through divisors, so we get extensions of every degree \(d\) dividing \(n\).
Fall 2019 Midterm #8 #algebra/qual/work
Let \(k\) be a field of characteristic \(p\neq 0\) and \(f\in k[x]\) irreducible. Show that \(f(x) = g(x^{p^d})\) where \(g(x) \in k[x]\) is irreducible and separable.
Conclude that every root of \(f\) has the same multiplicity \(p^d\) in the splitting field of \(f\) over \(k\).
Fall 2019 Midterm #7 #algebra/qual/work
Show that a field \(k\) of characteristic \(p\neq 0\) is perfect \(\iff\) for every \(x\in k\) there exists a \(y\in k\) such that \(y^p=x\).
Spring 2012 #4 #algebra/qual/work
Let \(f(x) = x^7  3\in {\mathbf{Q}}[x]\) and \(E/{\mathbf{Q}}\) be a splitting field of \(f\) with \(\alpha \in E\) a root of \(f\).

Show that \(E\) contains a primitive 7th root of unity.

Show that \(E\neq {\mathbf{Q}}(\alpha)\).
Fall 2013 #5 #algebra/qual/completed
Let \(L/K\) be a finite extension of fields.

Define what it means for \(L/K\) to be separable.

Show that if \(K\) is a finite field, then \(L/K\) is always separable.
 Give an example of a finite extension \(L/K\) that is not separable.

\(L/k\) is separable iff every element \(\alpha\) is separable, i.e. the minimal polynomial \(m(x)\) of \(\alpha\) is a separable polynomial, i.e. \(m(x)\) has no repeated roots in (say) the algebraic closure of \(L\) (or just any splitting field of \(m\)).

If \(\operatorname{ch}k = p\), suppose toward a contradiction that \(L/k\) is not separable. Then there is some \(\alpha\) with an inseparable (and irreducible) minimal polynomial \(f(x)\in k[x]\).

Claim: since \(f\) is inseparable and irreducible, \(f(x) = g(x^p)\) for some \(g\in k[x]\).
 Note: write \(g(x) \coloneqq\sum a_k x^k\), so that \(f(x) = \sum a_k (x^p)^k = \sum a_k x^{pk}\).

This is a contradiction, since it makes \(f\) reducible by using the “Freshman’s dream”: \begin{align*} f(x) = \sum a_k x^{pk} = \qty{ \sum a_k^{1\over p} x^k}^p \coloneqq(h(x))^p .\end{align*}

Proof of claim: in \(\operatorname{ch}k = p, f\) inseparable \(\implies f(x) = g(x^p)\).
 Use that \(f\) is inseparable iff \(\gcd(f, f') \neq 1\), and since \(f\) is irreducible this forces \(f' \equiv 0\), so \(ka_k = 0\) for all \(k\).
 Then \(a_k\neq 0\) forces \(p\divides k\), so \(f(x) = a_0 + a_px^p + a_{2p}x^{2p} + \cdots\) and one takes \(g(x) \coloneqq\sum a_{kp}x^{kp}\).

A finite inseparable extension:
 It’s a theorem that finite extensions of perfect fields are separable, so one needs a nonperfect field.
 Take \(L/k \coloneqq{ \mathbf{F} }_p(t^{1\over p}) / { \mathbf{F} }_p(t)\), which is a degree \(p\) extension (although both fields are infinite are characteristic \(p\)).
 Then the minimal polynomial of \(t\) is \(f(x) \coloneqq x^p  t \in { \mathbf{F} }_p(t)[x]\), where \(f'(x) = px^p \equiv 0\) Alternatively, just note that \(f\) factors as \(f(x) = (xt^{1\over p})^p\) in \(L[x]\), which has multiple roots.
Fall 2012 #4 #algebra/qual/work
Let \(f(x) \in {\mathbf{Q}}[x]\) be a polynomial and \(K\) be a splitting field of \(f\) over \({\mathbf{Q}}\). Assume that \([K:{\mathbf{Q}}] = 1225\) and show that \(f(x)\) is solvable by radicals.
Galois Groups: Concrete Computations
Exercise: \(G(x^22)\)
Compute the Galois group of \(x^22\).
\({\mathbf{Z}}/2{\mathbf{Z}}\)?
Exercise: \(G(x^p2)\)
Let \(p \in \mathbb{Z}\) be a prime number. Then describe the elements of the Galois group of the polynomial \(x^{p}2\).
\({\mathbf{Q}}(2^{1\over p}, \zeta_p)\), which has degree \(p(p1)\) and is generated by the maps \begin{align*} \sqrt[p]{2} & \mapsto \sqrt[p]{2} \zeta^{a} \\ \zeta & \mapsto \zeta^{b} .\end{align*}
Fall 2020 #3 #algebra/qual/work

Define what it means for a finite extension of fields \(E\) over \(F\) to be a Galois extension.

Determine the Galois group of \(f(x) = x^3  7\) over \({\mathbf{Q}}\), and justify your answer carefully.
 Find all subfields of the splitting field of \(f(x)\) over \({\mathbf{Q}}\).
Part a:

A finite extension \(E/F\) is Galois if it is normal and separable:
 Normal: every \(f\in F[x]\) either has no roots in \(E\) or all roots in \(E\).
 Separable: every element \(e\in E\) has a separable minimal polynomial \(m(x)\), i.e. \(m\) has no repeated roots.
Part b:

Note \(f\) is irreducible by Eisenstein with \(p=7\), and since \({\mathbf{Q}}\) is perfect, irreducible implies separable.

Writing \(L \coloneqq\operatorname{SF}(f)/{\mathbf{Q}}\), this is a Galois extension:
 \(L\) is separable: it is a finite extension of a perfect field, which is automatically separable.
 \(L\) is normal: \(L\) is the splitting field of a separable polynomial, and thus normal.

Since \(f\) is degree 3, we have \(G\coloneqq{ \mathsf{Gal}}(L/k) \leq S_3\), and since \(G\) is a transitive subgroup the only possibilities are \begin{align*} G = S_3 \cong D_3, A_3 \cong C_3 .\end{align*}

Factor \(x^3  7 = (x\omega)(x\zeta_3\omega)(x\zeta_3^2\omega)\) where \(\omega \coloneqq 7^{1\over 3}\) and \(\zeta_3\) is a primitive 3rd root of unity. Then \(L = {\mathbf{Q}}(\zeta_3, \omega)\).
 Aside: label the roots in this order, so \(r_1 = \omega, r_2 = \zeta_3\omega, r_3 = \zeta_3^2\omega\).

Write \(\min_{\omega, {\mathbf{Q}}}(x) = x^3  7\) and let \(L_0/{\mathbf{Q}}\coloneqq{\mathbf{Q}}(\omega)/{\mathbf{Q}}\) yields \([L_0: {\mathbf{Q}}] = 3\).

Write \(\min_{\zeta_3, {\mathbf{Q}}}(x) = (x^31)/(x1) = x^2 + x + 1\), and note that this is still the minimal polynomial over \(L_0\) since \(L_0 \subseteq {\mathbf{R}}\) and \(\zeta_3 \in {\mathbf{C}}\setminus{\mathbf{R}}\). So \([L:L_0] = 2\).

Counting in towers, \begin{align*} [L:{\mathbf{Q}}] = [L:L_0][L_0: {\mathbf{Q}}] = (2)(3) = 6 .\end{align*}

But \({\sharp}S_3 = 6\) and \({\sharp}A_3 = 3\), so \(G = S_3\).

Explicitly, since we can write \(\operatorname{SF}(f) = {\mathbf{Q}}(\omega, \zeta_3)\), we can find explicit generators: \begin{align*} \sigma: &\begin{cases} \omega &\mapsto \omega \\ \zeta_3 &\mapsto \zeta_3\cdot \zeta_3. \end{cases} && \implies \sigma \sim (1,2,3) \\ \tau: &\begin{cases} \omega &\mapsto \omega \\ \zeta_3 &\mapsto \overline{\zeta_3}. \end{cases} && \implies \tau \sim (2, 3) .\end{align*} So \(G = \left\langle{\sigma, \tau {~\mathrel{\Big\vert}~}\sigma^3, \tau^2}\right\rangle\).
Part c:
 Note that the subgroup lattice for \(S_3\) looks like the following:

Note that we can identify
 \(\tau = (2,3)\) which fixes \(r_1\)
 \(\sigma \tau = (1,2)\) which fixes \(r_3\)
 \(\sigma^2\tau = (1, 3)\) which fixes \(r_2\)
 \(\sigma = (1,2,3)\), for which we need to calculate the fixed field. Using that \(\sigma(\omega) =\zeta\omega\) and \(\sigma(\zeta)=\zeta\), supposing \(\sigma(\alpha) = \alpha\) we have \begin{align*} \sigma(\alpha) &\coloneqq\sigma(a + b\zeta_3 + c\zeta_3^2 + d\omega + e\zeta_3\omega + f\zeta_3^2\omega) \\ &= a + b\zeta_3 + c\zeta_3^2 + d\zeta_3\omega + e\zeta_3^2\omega + f\omega \\ \implies \alpha &= a + b\zeta_3 + c\zeta_3^2 + t_1(\omega + \zeta_3\omega + \zeta_3^2\omega) \\ \implies \alpha &= a + b\zeta_3 + c\zeta_3^2 + t_1\omega (1 + \zeta_3+ \zeta_3^2) \\ \implies \alpha &= a + b\zeta_3 + c\zeta_3^2 ,\end{align*} using the general fact that \(\sum_{k=0}^{n1}\zeta_n^k = 0\). So the fixed field is \({\mathbf{Q}}(1, \zeta, \zeta^2) = {\mathbf{Q}}(\zeta)\).
 We thus get the following lattice correspondence:
Spring 2021 #4 #algebra/qual/work
Define \begin{align*} f(x) \coloneqq x^4 + 4x^2 + 64 \in {\mathbf{Q}}[x] .\end{align*}

Find the splitting field \(K\) of \(f\) over \({\mathbf{Q}}\).

Find the Galois group \(G\) of \(f\).
 Exhibit explicitly the correspondence between subgroups of \(G\) and intermediate fields between \({\mathbf{Q}}\) and \(K\).
 Useful trick: given \(a + \sqrt{b}\), try to rewrite this as \((\sqrt{c} + \sqrt{d})^2\) for some \(c, d\) to get a better basis for \(\operatorname{SF}(f)\).

First consider \(g(z) \coloneqq z^2 + 4z + 64\). Applying the quadratic formula yields \begin{align*} z = {4 \pm \sqrt{16  64} \over 2} = 2 \pm {1\over 2}\sqrt{ 15 \cdot 16 } = 2 \pm 2i \sqrt{15} .\end{align*}

Substituting \(z=x^2\) yields the splitting field of \(f\) as \(L\coloneqq{\mathbf{Q}}(\pm \sqrt{ 2 \pm 2i\sqrt{15}})\).
 Note that this factorization shows that \(f\) is irreducible over \({\mathbf{Q}}\), since the two quadratic factors have irrational coefficients and none of the roots are real.
 Irreducible implies separable over a perfect field, so \(L/{\mathbf{Q}}\) is a separable extension.
 \(L\) is the splitting field of a separable polynomial and thus normal, making \(L\) Galois.

In this form, it’s not clear what the degree \([L:{\mathbf{Q}}]\) is, so we can find a better basis by rewriting the roots of \(g\): \begin{align*} z = 2 \pm 2i\sqrt{15} = \qty{\sqrt{5}}^2  \qty{\sqrt 3}^2 \pm 2i\sqrt{5}\sqrt{3} = (\sqrt 5 \pm i\sqrt{3})^2 ,\end{align*} and so the roots of \(f\) are \(x = \pm \sqrt{5} \pm i\sqrt{3}\) and \(L = {\mathbf{Q}}(\sqrt 5, i\sqrt 3)\).

Counting in towers, \begin{align*} [L:{\mathbf{Q}}] = [{\mathbf{Q}}(\sqrt 5, i \sqrt{3} ) : {\mathbf{Q}}\sqrt{5} ][{\mathbf{Q}}\sqrt{5} : {\mathbf{Q}}] = (2)(2) = 4 ,\end{align*} where we’ve used that \(\min_{\sqrt 5, {\mathbf{Q}}}(x) = x^25\) and \(\min_{i\sqrt 3, {\mathbf{Q}}}(x) = x^2 + 3\), which remains the minimal polynomial over \({\mathbf{Q}}(\sqrt 5) \subseteq {\mathbf{R}}\) since both roots are not real.

So \(G\coloneqq{ \mathsf{Gal}}(L/{\mathbf{Q}}) \leq S_4\) is a transitive subgroup of size 4, making it either \(C_4\) or \(C_2^2\).

Label the roots: \begin{align*} r_1 &= \sqrt 5 + i\sqrt 3 \\ r_2 &= \sqrt{5}  i \sqrt{3} \\ r_3 &=  \sqrt 5 + i\sqrt 3 = r_2 \\ r_4 &= \sqrt{5}  i\sqrt{3} = r_1 .\end{align*}

We can start writing down automorphisms: \begin{align*} \sigma_1: \begin{cases} \sqrt 5 &\mapsto \sqrt 5 \\ i\sqrt 3 &\mapsto i\sqrt 3 . \end{cases} && \sigma_1 \sim (1,3)(2,4) \\ \sigma_2 \begin{cases} \sqrt 5 &\mapsto \sqrt 5 \\ i\sqrt 3 &\mapsto i\sqrt 3 . \end{cases} && \sigma_2 \sim (1, 2)(3, 4) .\end{align*} Note that these define automorphisms because we’ve specified what happens to a basis and they send roots to other roots.

Checking that \(\sigma_1^2 = \sigma_2^2 = \operatorname{id}\), this produces two distinct order 2 elements, forcing \(G \cong C_2^2\) since \(C_4\) only has one order 2 element. Explicitly, we have \begin{align*} C_2^2 \cong G = \left\langle{\tau_1, \tau_2}\right\rangle = \left\{{\operatorname{id}, \tau_1, \tau_2, \tau_1 \tau_2}\right\} = \left\{{\operatorname{id}, (1,3)(2,4), (1,2)(3,4), (1,4)(2,3) }\right\} ,\end{align*} and the generic subgroup lattice looks like:

Computing some fixed fields. Write \(i \sqrt{3} = x, \sqrt{5} = y\), then elements in the splitting field are of the form \(\alpha = 1 + ax + by + cxy\).

For \(\sigma_1\), we have \(x\mapsto x\), so \begin{align*} \sigma_1(\alpha) = 1  ax + by  cxy = \alpha \implies a=a=0, c=c=0 ,\end{align*} so this preserves \(1+by\), making the fixed field \({\mathbf{Q}}(1, y) = {\mathbf{Q}}(i \sqrt{3})\).

For \(\sigma_2\), we have \(y\mapsto y\), so \begin{align*} \sigma_2(\alpha) = 1 +ax by cxy = \alpha \implies b=b=0,c=c=0 ,\end{align*} preserving \(1 + ax\) and making the fixed field \({\mathbf{Q}}(1, x) = {\mathbf{Q}}(\sqrt 5)\).

For \(\sigma_1 \sigma_2\), we have \(x\mapsto x\) and \(y\mapsto y\), so \begin{align*} \sigma_1\sigma_2(\alpha) = 1 ax by +cxy = \alpha \implies a=a=, b=b=0 ,\end{align*} preserving \(1 + cxy\) and yielding \({\mathbf{Q}}(xy) = {\mathbf{Q}}(i\sqrt 3 \sqrt 5)\).


So the lattice correspondence we get here is
Fall 2019 Midterm #6 #algebra/qual/work
Compute the Galois group of \(f(x) = x^33x 3\in {\mathbf{Q}}[x]/{\mathbf{Q}}\).
Spring 2018 #2 #algebra/qual/completed
Let \(f(x) = x^4  4x^2 + 2 \in {\mathbf{Q}}[x]\).

Find the splitting field \(K\) of \(f\), and compute \([K: {\mathbf{Q}}]\).

Find the Galois group \(G\) of \(f\), both as an explicit group of automorphisms, and as a familiar abstract group to which it is isomorphic.
 Exhibit explicitly the correspondence between subgroups of \(G\) and intermediate fields between \({\mathbf{Q}}\) and \(k\).
\todo[inline]{Not the nicest proof! Would be better to replace the adhoc computations at the end.}
Note that \(g(x) = x^2  4x + 2\) has roots \(\beta = 2 \pm \sqrt{2}\), and so \(f\) has roots \begin{align*} \alpha_1 &= \sqrt{2 + \sqrt 2} \\ \alpha_2 &= \sqrt{2  \sqrt 2} \\ \alpha_3 &= \alpha_1 \\ \alpha_4 &= \alpha_2 .\end{align*}
and splitting field \(K = {\mathbf{Q}}(\left\{{\alpha_i}\right\})\).
\(K\) is the splitting field of a separable polynomial and thus Galois over \({\mathbf{Q}}\). Moreover, Since \(f\) is irreducible by Eisenstein with \(p=2\), the Galois group is a transitive subgroup of \(S^4\), so the possibilities are:
 \(S_4\)
 \(A_4\)
 \(D_4\)
 \({\mathbf{Z}}/(2) \times{\mathbf{Z}}/(2)\)
 \({\mathbf{Z}}/(4)\)
We can note that \(g\) splits over \(L \coloneqq{\mathbf{Q}}(\sqrt 2)\), an extension of degree 2.
We can now note that \(\min(\alpha, L)\) is given by \(p(x) = x^2  (2 + \sqrt 2)\), and so \([K: L] = 2\).
We then have \begin{align*} [K: {\mathbf{Q}}] = [K: L] [L : {\mathbf{Q}}] = (2)(2) = 4 .\end{align*}
This \({\left\lvert {{ \mathsf{Gal}}(K/{\mathbf{Q}})} \right\rvert} = 4\), which leaves only two possibilities:
 \({\mathbf{Z}}/(2) \times{\mathbf{Z}}/(2)\)
 \({\mathbf{Z}}/(4)\)
We can next check orders of elements. Take \begin{align*} \sigma &\in { \mathsf{Gal}}(K/{\mathbf{Q}}) \\ \alpha_1 &\mapsto \alpha_2 .\end{align*}
Computations show that
 \(\alpha_1^2 \alpha_2^2 = 2\), so \(\alpha_1 \alpha_2 = \sqrt 2\)
 \(\alpha_1^2 = 2 + \sqrt 2 \implies \sqrt 2 = \alpha_1^2  2\)
and thus \begin{align*} \sigma^2(\alpha_1) &= \sigma(\alpha_2) \\ &= \sigma\left(\frac{\sqrt 2}{\alpha_1}\right) \\ &= \frac{\sigma(\sqrt 2)}{\sigma(\alpha_1)} \\ &= \frac{\sigma(\alpha_1^2  2)}{\alpha_2} \\ &= \frac{\alpha_2^2  2}{\alpha_2} \\ &= \alpha_2 2\alpha_2^{1}\\ &= \alpha_2  \frac{2\alpha_1}{\sqrt 2} \\ &= \alpha_2 \alpha_1 \sqrt 2 \\ &\neq \alpha_1 ,\end{align*}
and so the order of \(\sigma\) is strictly greater than 2, and thus 4, and thus \({ \mathsf{Gal}}(K/{\mathbf{Q}}) = \left\{{\sigma^k {~\mathrel{\Big\vert}~}1\leq k \leq 4}\right\} \cong {\mathbf{Z}}/(4)\).
?? The subgroup of index 2 \(\left\langle{\sigma^2}\right\rangle\) corresponds to the field extension \(Q(\sqrt 2) / {\mathbf{Q}}\).
\todo[inline]{Finish (c)}
Spring 2020 #4 #algebra/qual/work
Let \(f(x) = x^42 \in {\mathbf{Q}}[x]\).

Define what it means for a finite extension field \(E\) of a field \(F\) to be a Galois extension.

Determine the Galois group \({ \operatorname{Gal}}(E/{\mathbf{Q}})\) for the polynomial \(f(x)\), and justify your answer carefully.
 Exhibit a subfield \(K\) in \((b)\) such that \({\mathbf{Q}}\leq K \leq E\) with \(K\) not a Galois extension over \({\mathbf{Q}}\). Explain.
Spring 2017 #8 #algebra/qual/work

Let \(K\) denote the splitting field of \(x^5  2\) over \({\mathbf{Q}}\). Show that the Galois group of \(K/{\mathbf{Q}}\) is isomorphic to the group of invertible matrices \begin{align*} \left(\begin{array}{ll} a & b \\ 0 & 1 \end{array}\right) {\quad \operatorname{where} \quad} a\in { \mathbf{F} }_5^{\times}\text{ and } b\in { \mathbf{F} }_5 .\end{align*}

Determine all intermediate fields between \(K\) and \({\mathbf{Q}}\) which are Galois over \({\mathbf{Q}}\).
Fall 2016 #4 #algebra/qual/work
Set \(f(x) = x^3  5 \in {\mathbf{Q}}[x]\).

Find the splitting field \(K\) of \(f(x)\) over \({\mathbf{Q}}\).

Find the Galois group \(G\) of \(K\) over \({\mathbf{Q}}\).
 Exhibit explicitly the correspondence between subgroups of \(G\) and intermediate fields between \({\mathbf{Q}}\) and \(K\).
Spring 2016 #2 #algebra/qual/work
Let \(K = {\mathbf{Q}}[\sqrt 2 + \sqrt 5]\).

Find \([K: {\mathbf{Q}}]\).

Show that \(K/{\mathbf{Q}}\) is Galois, and find the Galois group \(G\) of \(K/{\mathbf{Q}}\).
 Exhibit explicitly the correspondence between subgroups of \(G\) and intermediate fields between \({\mathbf{Q}}\) and \(K\).
Fall 2015 #5 #algebra/qual/work
Let \(u = \sqrt{2 + \sqrt{2}}\), \(v = \sqrt{2  \sqrt{2}}\), and \(E = {\mathbf{Q}}(u)\).

Find (with justification) the minimal polynomial \(f(x)\) of \(u\) over \({\mathbf{Q}}\).

Show \(v\in E\), and show that \(E\) is a splitting field of \(f(x)\) over \({\mathbf{Q}}\).
 Determine the Galois group of \(E\) over \({\mathbf{Q}}\) and determine all of the intermediate fields \(F\) such that \({\mathbf{Q}}\subset F \subset E\).
Spring 2015 #5 #algebra/qual/work
Let \(f(x) = x^4  5 \in {\mathbf{Q}}[x]\).

Compute the Galois group of \(f\) over \({\mathbf{Q}}\).

Compute the Galois group of \(f\) over \({\mathbf{Q}}(\sqrt{5})\).
Fall 2014 #3 #algebra/qual/work
Consider the polynomial \(f(x) = x^4  7 \in {\mathbf{Q}}[x]\) and let \(E/{\mathbf{Q}}\) be the splitting field of \(f\).

What is the structure of the Galois group of \(E/{\mathbf{Q}}\)?

Give an explicit description of all of the intermediate subfields \({\mathbf{Q}}\subset K \subset E\) in the form \(K = {\mathbf{Q}}(\alpha), {\mathbf{Q}}(\alpha, \beta), \cdots\) where \(\alpha, \beta\), etc are complex numbers. Describe the corresponding subgroups of the Galois group.
Fall 2013 #6 #algebra/qual/work
Let \(K\) be the splitting field of \(x^42\) over \({\mathbf{Q}}\) and set \(G = { \operatorname{Gal}}(K/{\mathbf{Q}})\).

Show that \(K/{\mathbf{Q}}\) contains both \({\mathbf{Q}}(i)\) and \({\mathbf{Q}}(\sqrt[4]{2})\) and has degree 8 over \({\mathbf{Q}}\)/

Let \(N = { \operatorname{Gal}}(K/{\mathbf{Q}}(i))\) and \(H = { \operatorname{Gal}}(K/{\mathbf{Q}}(\sqrt[4]{2}))\). Show that \(N\) is normal in \(G\) and \(NH = G\).
Hint: what field is fixed by \(NH\)?

Show that \({ \operatorname{Gal}}(K/{\mathbf{Q}})\) is generated by elements \(\sigma, \tau\), of orders 4 and 2 respectively, with \(\tau \sigma\tau^{1}= \sigma^{1}\).
Equivalently, show it is the dihedral group of order 8.

How many distinct quartic subfields of \(K\) are there? Justify your answer.
Spring 2014 #4 #algebra/qual/work
Let \(E\subset {\mathbf{C}}\) denote the splitting field over \({\mathbf{Q}}\) of the polynomial \(x^3  11\).

Prove that if \(n\) is a squarefree positive integer, then \(\sqrt{n}\not\in E\).
Hint: you can describe all quadratic extensions of \({\mathbf{Q}}\) contained in \(E\).

Find the Galois group of \((x^3  11)(x^2  2)\) over \({\mathbf{Q}}\).
 Prove that the minimal polynomial of \(11^{1/3} + 2^{1/2}\) over \({\mathbf{Q}}\) has degree 6.
Spring 2013 #8 #algebra/qual/work
Let \(F\) be the field with 2 elements and \(K\) a splitting field of \(f(x) = x^6 + x^3 + 1\) over \(F\). You may assume that \(f\) is irreducible over \(F\).

Show that if \(r\) is a root of \(f\) in \(K\), then \(r^9 = 1\) but \(r^3\neq 1\).

Find \({ \operatorname{Gal}}(K/F)\) and express each intermediate field between \(F\) and \(K\) as \(F(\beta)\) for an appropriate \(\beta \in K\).
Galois Groups: Indirect Computations / Facts
Fall 2019 #7 #algebra/qual/completed
Let \(\zeta_n\) denote a primitive \(n\)th root of 1 \(\in {\mathbf{Q}}\). You may assume the roots of the minimal polynomial \(p_n(x)\) of \(\zeta_n\) are exactly the primitive \(n\)th roots of 1.
Show that the field extension \({\mathbf{Q}}(\zeta_n )\) over \({\mathbf{Q}}\) is Galois and prove its Galois group is \(({\mathbf{Z}}/n{\mathbf{Z}})^{\times}\).
How many subfields are there of \({\mathbf{Q}}(\zeta_{20} )\)?

Galois = normal + separable.

Separable: Minimal polynomial of every element has distinct roots.

Normal (if separable): Splitting field of an irreducible polynomial.

\(\zeta\) is a primitive root of unity \(\iff o(\zeta) = n\) in \({ \mathbf{F} }^{\times}\).

\(\phi(p^k) = p^{k1}(p1)\)

The lattice:
Let \(K = {\mathbf{Q}}(\zeta)\). Then \(K\) is the splitting field of \(f(x) = x^n  1\), which is irreducible over \({\mathbf{Q}}\), so \(K/{\mathbf{Q}}\) is normal. We also have \(f'(x) = nx^{n1}\) and \(\gcd(f, f') = 1\) since they can not share any roots.
Or equivalently, \(f\) splits into distinct linear factors \(f(x) = \prod_{k\leq n}(x\zeta^k)\).
Since it is a Galois extension, \({\left\lvert {{ \mathsf{Gal}}(K/{\mathbf{Q}})} \right\rvert} = [K: {\mathbf{Q}}] = \phi(n)\) for the totient function.
We can now define maps \begin{align*} \tau_j: K &\to K \\ \zeta &\mapsto \zeta^j \end{align*} and if we restrict to \(j\) such that \(\gcd(n, j) = 1\), this yields \(\phi(n)\) maps. Noting that if \(\zeta\) is a primitive root, then \((n, j) = 1\) implies that that \(\zeta^j\) is also a primitive root, and hence another root of \(\min(\zeta, {\mathbf{Q}})\), and so these are in fact automorphisms of \(K\) that fix \({\mathbf{Q}}\) and thus elements of \({ \mathsf{Gal}}(K/{\mathbf{Q}})\).
So define a map \begin{align*} \theta: {\mathbf{Z}}_n^{\times}&\to K \\ [j]_n &\mapsto \tau_j .\end{align*}
from the multiplicative group of units to the Galois group.
The claim is that this is a surjective homomorphism, and since both groups are the same size, an isomorphism.
Letting \(\sigma \in K\) be arbitrary, noting that \([K: {\mathbf{Q}}]\) has a basis \(\left\{{1, \zeta, \zeta^2, \cdots, \zeta^{n1}}\right\}\), it suffices to specify \(\sigma(\zeta)\) to fully determine the automorphism. (Since \(\sigma(\zeta^k) = \sigma(\zeta)^k\).)
In particular, \(\sigma(\zeta)\) satisfies the polynomial \(x^n  1\), since \(\sigma(\zeta)^n = \sigma(\zeta^n) = \sigma(1) = 1\), which means \(\sigma(\zeta)\) is another root of unity and \(\sigma(\zeta) = \zeta^k\) for some \(1\leq k \leq n\).
Moreover, since \(o(\zeta) = n \in K^{\times}\), we must have \(o(\zeta^k) = n \in K^{\times}\) as well. Noting that \(\left\{{\zeta^i}\right\}\) forms a cyclic subgroup \(H\leq K^{\times}\), then \(o(\zeta^k) = n \iff (n, k) = 1\) (by general theory of cyclic groups).
Thus \(\theta\) is surjective.
\begin{align*} \tau_j \circ \tau_k (\zeta) =\tau_j(\zeta^k) = \zeta^{jk} \implies \tau_{jk} = \theta(jk) = \tau_j \circ \tau_k .\end{align*}
We have \(K \cong {\mathbf{Z}}_{20}^{\times}\) and \(\phi(20) = 8\), so \(K \cong {\mathbf{Z}}_8\), so we have the following subgroups and corresponding intermediate fields:
 \(0 \sim {\mathbf{Q}}(\zeta_{20})\)
 \({\mathbf{Z}}_2 \sim {\mathbf{Q}}(\omega_1)\)
 \({\mathbf{Z}}_4 \sim {\mathbf{Q}}(\omega_2)\)
 \({\mathbf{Z}}_8 \sim {\mathbf{Q}}\)
For some elements \(\omega_i\) which exist by the primitive element theorem.
Fall 2018 #3 #algebra/qual/completed
Let \(F \subset K \subset L\) be finite degree field extensions. For each of the following assertions, give a proof or a counterexample.

If \(L/F\) is Galois, then so is \(K/F\).

If \(L/F\) is Galois, then so is \(L/K\).
 If \(K/F\) and \(L/K\) are both Galois, then so is \(L/F\).
 Every quadratic extension over \({\mathbf{Q}}\) is Galois.
Let \(L/K/F\).
False: Take \(L/K/F = {\mathbf{Q}}(\zeta_2, \sqrt[3] 2) \to {\mathbf{Q}}(\sqrt[3] 2) \to {\mathbf{Q}}\).
Then \(L/F\) is Galois, since it is the splitting field of \(x^3  2\) and \({\mathbf{Q}}\) has characteristic zero.
But \(K/F\) is not Galois, since it is not the splitting field of any irreducible polynomial.
True: If \(L/F\) is Galois, then \(L/K\) is normal and separable:

\(L/K\) is normal, since if \(\sigma: L \hookrightarrow\overline K\) lifts the identity on \(K\) and fixes \(L\), it also lifts the identity on \(F\) and fixes \(L\) (and \(\overline K = \overline F\)).

\(L/K\) is separable, since \(F[x] \subseteq K[x]\), and so if \(\alpha \in L\) where \(f(x) \coloneqq\min(\alpha, F)\) has no repeated factors, then \(f'(x) \coloneqq\min(\alpha, K)\) divides \(f\) and thus can not have repeated factors.
False: Use the fact that every quadratic extension is Galois, and take \(L/K/F = {\mathbf{Q}}(\sqrt[4] 2) \to {\mathbf{Q}}(\sqrt 2) \to {\mathbf{Q}}\).
Then each successive extension is quadratic (thus Galois) but \({\mathbf{Q}}(\sqrt[4] 2)\) is not the splitting field of any polynomial (noting that it does not split \(x^4  2\) completely.)
Spring 2018 #3 #algebra/qual/completed
Let \(K\) be a Galois extension of \({\mathbf{Q}}\) with Galois group \(G\), and let \(E_1 , E_2\) be intermediate fields of \(K\) which are the splitting fields of irreducible \(f_i (x) \in {\mathbf{Q}}[x]\).
Let \(E = E_1 E_2 \subset K\).
Let \(H_i = { \mathsf{Gal}}(K/E_i)\) and \(H = { \mathsf{Gal}}(K/E)\).

Show that \(H = H_1 \cap H_2\).

Show that \(H_1 H_2\) is a subgroup of \(G\).
 Show that \begin{align*} { \mathsf{Gal}}(K/(E_1 \cap E_2 )) = H_1 H_2 .\end{align*}

The Galois correspondence:
 \(H_1 \cap H_2 \rightleftharpoons E_1 E_2\),
 \(H_1 H_2 \rightleftharpoons E_1 \cap E_2\).
By the Galois correspondence, it suffices to show that the fixed field of \(H_1 \cap H_2\) is \(E_1 E_2\).
Let \(\sigma \in H_1 \cap H_2\); then \(\sigma \in \mathop{\mathrm{Aut}}(K)\) fixes both \(E_1\) and \(E_2\).
Not sure if this works – compositum is not literally product..?
Writing \(x \in E_1E_2\) as \(x=e_1 e_2\), we have \begin{align*} \sigma(x) = \sigma(e_1 e_2) = \sigma(e_1) \sigma(e_2) = e_1 e_2 =x, \end{align*}
so \(\sigma\) fixes \(E_1 E_2\).
That \(H_1 H_2 \subseteq G\) is clear, since if \(\sigma = \tau_1 \tau_2 \in H_1 H_2\), then each \(\tau_i\) is an automorphism of \(K\) that fixes \(E_i \supseteq {\mathbf{Q}}\), so each \(\tau_i\) fixes \({\mathbf{Q}}\) and thus \(\sigma\) fixes \({\mathbf{Q}}\).
All elements in this subset commute.

Let \(\sigma = \sigma_1 \sigma_2 \in H_1 H_2\).

Note that \(\sigma_1(e) = e\) for all \(e\in E_1\) by definition, since \(H_1\) fixes \(E_1\), and \(\sigma_2(e) \in E_1\) (?).

Then \begin{align*} \sigma_1(e) = e \quad \forall e \in E_1 \implies \sigma_1(\sigma_2(e)) = \sigma_2(e) \end{align*} and substituting \(e = \sigma_1(e)\) on the RHS yields \begin{align*} \sigma_1 \sigma_2(e) = \sigma_2 \sigma_1(e) ,\end{align*} where a similar proof holds for \(e\in E_2\) and thus for arbitrary \(x\in E_1 E_2\).
By the Galois correspondence, the subgroup \(H_1H_2 \leq G\) will correspond to an intermediate field \(E\) such that \(K/E/{\mathbf{Q}}\) and \(E\) is the fixed field of \(H_1 H_2\).
But if \(\sigma \in H_1 H_2\), then \(\sigma = \tau_1 \tau_2\) where \(\tau_i\) is an automorphism of \(K\) that fixes \(E_i\), and so \begin{align*} \sigma(x) = x \iff \tau_1\tau_2(x) = x &\iff \tau_2(x) = x \\ &~\&~ \\ \tau_1(x) = x &\iff x \in E_1 \cap E_2 .\end{align*} .
Fall 2017 #4 #algebra/qual/work

Let \(f (x)\) be an irreducible polynomial of degree 4 in \({\mathbf{Q}}[x]\) whose splitting field \(K\) over \({\mathbf{Q}}\) has Galois group \(G = S_4\).
Let \(\theta\) be a root of \(f(x)\). Prove that \({\mathbf{Q}}[\theta]\) is an extension of \({\mathbf{Q}}\) of degree 4 and that there are no intermediate fields between \({\mathbf{Q}}\) and \({\mathbf{Q}}[\theta]\).

Prove that if \(K\) is a Galois extension of \({\mathbf{Q}}\) of degree 4, then there is an intermediate subfield between \(K\) and \({\mathbf{Q}}\).
Spring 2017 #7 #algebra/qual/work
Let \(F\) be a field and let \(f(x) \in F[x]\).

Define what a splitting field of \(f(x)\) over \(F\) is.

Let \(F\) now be a finite field with \(q\) elements. Let \(E/F\) be a finite extension of degree \(n>0\). Exhibit an explicit polynomial \(g(x) \in F[x]\) such that \(E/F\) is a splitting field of \(g(x)\) over \(F\). Fully justify your answer.
 Show that the extension \(E/F\) in (b) is a Galois extension.
Spring 2016 #6 #algebra/qual/work
Let \(K\) be a Galois extension of a field \(F\) with \([K: F] = 2015\). Prove that \(K\) is an extension by radicals of the field \(F\).
 If \(N {~\trianglelefteq~}G\) is a normal subgroup and \(H\leq G\) is any subgroup containing \(N\), then \(N\) is normal in \(H\) since \(hNh^{1}\subseteq gNg^{1}= N\).
 In characteristic zero, a polynomial is solvable by radicals iff its Galois group is a solvable group.
Let \(G\coloneqq{ \mathsf{Gal}}(K/F)\), then it suffices to show that \(G\) is always a solvable group, i.e. any group of order \(n=2015\) is solvable. Factor \(2015 = 5\cdot 13\cdot 31\) – this is a \(pqr\) factorization, and in fact any group with exactly 3 prime factors (so \(n\) is squarefree in particular) will be solvable. Let \(p=5, q=13, r=31\) so that \(p<q<r\). We aim to construct a composition series whose successive quotients are simple groups. Applying Sylow 3 yields

\(n_p \divides qr, n_p \equiv 1 \operatorname{mod}p \implies n_5 \divides 13\cdot 31 = 403\) and \(n_5\equiv 1 \operatorname{mod}5\).
 So \(n_5 \in \left\{{1, 13, 31}\right\}\) by divisibility and imposing the congruence forces \(n_5\in \left\{{1, 31}\right\}\) since \(13\not\equiv 1 \operatorname{mod}5\).

\(n_q \divides pr, n_q \equiv 1 \operatorname{mod}q \implies n_{13} \divides 5\cdot 31 = 155\) and \(n_{13}\equiv 1 \operatorname{mod}13\).
 So \(n_{13} \in \left\{{1,5,31}\right\}\) by divisibility and the congruence imposes \(n_{13}\in \left\{{1}\right\}\) since \(5,31 \not\equiv 1\operatorname{mod}13\). In particular, there is one Sylow 13subgroup which is normal.

\(n_r \divides pq, n_r \equiv 1 \operatorname{mod}r \implies n_{31} \divides 5 \cdot 13 = 65\) and \(n_{31}\equiv 1 \operatorname{mod}31\).
 So \(n_{31}\in \left\{{1,5,13}\right\}\) by divisibility and the congruence imposes \(n_{31}\in \left\{{1}\right\}\) since \(5,13 \not\equiv 1\operatorname{mod}31\). In particular, the one Sylow 31subgroup is normal.
Let \(H_{13}, H_{31}\) be the two normal subgroups of \(G\). Taking the quotient \(\tilde G \coloneqq G/H_{31}\) yields a group of order \(5\cdot 13\), and a similar argument as above using the Sylow theorems shows that \(\tilde G\) has a normal subgroup of order 13. By the subgroup correspondence theorem, this yields a normal subgroup \(N_1{~\trianglelefteq~}G\) containing \(H_{31}\) which has order \(13\cdot 31\). So define \(N_2 \coloneqq H_{31}\) to obtain \begin{align*} G \trianglerighteq N_1 \trianglerighteq N_2 \coloneqq H_{31} \trianglerighteq 0 .\end{align*} Since the quotients \(N_i/N_{i+1}\) have prime power order, they are cyclic and thus simple. We know \(N_1\) is normal in \(G\) since it came from extending a normal group in a quotient, and we know \(N_2\) is normal in \(N_1\) since it was normal in all of \(G\). So \(G\) is solvable.
Fall 2015 #6 #algebra/qual/work

Let \(G\) be a finite group. Show that there exists a field extension \(K/F\) with \({ \operatorname{Gal}}(K/F) = G\).
You may assume that for any natural number \(n\) there is a field extension with Galois group \(S_n\).

Let \(K\) be a Galois extension of \(F\) with \({\left\lvert {{ \operatorname{Gal}}(K/F)} \right\rvert} = 12\). Prove that there exists an intermediate field \(E\) of \(K/F\) with \([E: F] = 3\).
 With \(K/F\) as in (b), does an intermediate field \(L\) necessarily exist satisfying \([L: F] = 2\)? Give a proof or counterexample.
Fall 2014 #1 #algebra/qual/work
Let \(f\in {\mathbf{Q}}[x]\) be an irreducible polynomial and \(L\) a finite Galois extension of \({\mathbf{Q}}\). Let \(f(x) = g_1(x)g_2(x)\cdots g_r(x)\) be a factorization of \(f\) into irreducibles in \(L[x]\).

Prove that each of the factors \(g_i(x)\) has the same degree.

Give an example showing that if \(L\) is not Galois over \({\mathbf{Q}}\), the conclusion of part (a) need not hold.
Spring 2013 #7 #algebra/qual/work
Let \(f(x) = g(x) h(x) \in {\mathbf{Q}}[x]\) and \(E,B,C/{\mathbf{Q}}\) be the splitting fields of \(f,g,h\) respectively.

Prove that \({ \operatorname{Gal}}(E/B)\) and \({ \operatorname{Gal}}(E/C)\) are normal subgroups of \({ \operatorname{Gal}}(E/{\mathbf{Q}})\).

Prove that \({ \operatorname{Gal}}(E/B) \cap{ \operatorname{Gal}}(E/C) = \left\{{1}\right\}\).

If \(B\cap C = {\mathbf{Q}}\), show that \({ \operatorname{Gal}}(E/B) { \operatorname{Gal}}(E/C) = { \operatorname{Gal}}(E/{\mathbf{Q}})\).

Under the hypothesis of (c), show that \({ \operatorname{Gal}}(E/{\mathbf{Q}}) \cong { \operatorname{Gal}}(E/B) \times { \operatorname{Gal}}(E/C)\).
 Use (d) to describe \({ \operatorname{Gal}}({\mathbf{Q}}[\alpha]/{\mathbf{Q}})\) where \(\alpha = \sqrt 2 + \sqrt 3\).
Fall 2012 #3 #algebra/qual/work
Let \(f(x) \in {\mathbf{Q}}[x]\) be an irreducible polynomial of degree 5. Assume that \(f\) has all but two roots in \({\mathbf{R}}\). Compute the Galois group of \(f(x)\) over \({\mathbf{Q}}\) and justify your answer.
\(p\)th Roots and \(x^{p^k}x\)
Spring 2021 #7 #algebra/qual/completed
Let \(p\) be a prime number and let \(F\) be a field of characteristic \(p\). Show that if \(a\in F\) is not a \(p\)th power in \(F\), then \(x^pa \in F[x]\) is irreducible.
 Contradiction: go to splitting field, apply Freshman’s dream.
 Use that this polynomial is ramified, and its only factors are \((xa)\).
 Suppose \(a\) is not a \(p\)th power in \(F\), then \(f(x) \coloneqq x^pa\) has no roots in \(F\).
 Toward a contradiction, suppose \(f\) is reducible in \(F[x]\).

In \(\operatorname{SF}(f)\), since \(\operatorname{ch}F = p\) we have \(f(x) = (x\zeta)^p\) for some \(\zeta = a^{1\over p}\).
 So if \(f\) is reducible in \(F[x]\), we have \(f(x) = p_1(x) p_2(x)\) where \(p(x) = (x\zeta)^q\in F[x]\) for some \(1\leq q < p\), since these are the only factors of \(f\).
 The claim is that \(\zeta\in F\) as well, which is a contradiction since \(\zeta\) is a \(p\)th root of \(a\).
 We have \(x^q\zeta^q \in F[x]\), so \(\zeta^q\in F\).

We know \(a = \zeta^p\in F\), and thus \(\zeta^{d} = \zeta\in F\) for \(d \coloneqq\gcd(p, n) = 1\). \(\contradiction\)
 Why this is true: write \(d = \gcd(p, n)\) in \({\mathbf{Z}}\) to obtain \(d = tp + sn\) for some \(t, s\).
 Then \(\zeta^d = \zeta^{tp+sn} = (\zeta^p)^t \cdot (\zeta^n)^s \in F\).
 By contrapositive, show that \(f(x) \coloneqq x^pa \in { \mathbf{F} }[x]\) reducible \(\implies a\) is a \(p\)th power in \({ \mathbf{F} }\).
 Eventually show \(a^\ell = b^p\) for some \(\ell\in {\mathbb{N}}\) and some \(b\in { \mathbf{F} }\), then \(\gcd(\ell, p) = 1\) forces \(b=a\) and \(\ell=p\).
 Use the fact that the constant term of any \(g\in { \mathbf{F} }[x]\) is actually in \({ \mathbf{F} }\).

Reducible: \(f\in { \mathbf{F} }[x]\) is reducible iff there exists \(g, h\in { \mathbf{F} }[x]\) nonconstant with \(f = g h\).
 Importantly, this factorization needs to happen in \({ \mathbf{F} }[x]\), since we can always find such factorizations in the splitting field \(\operatorname{SF}(f)[x]\).
 Bezout’s identity: \(\gcd(p, q) = d \implies\) there exist \(s,t\in {\mathbf{Z}}\) such that \begin{align*} sp + tq = d .\end{align*}

WTS: \(f(x) \coloneqq x^p  a\in { \mathbf{F} }[x]\) reducible \(\implies f\) has a root in the base field \({ \mathbf{F} }\).

Write \(f(x) = g(x) h(x)\) and factor \(f(x) = \prod_{i=1}^p (x r_i) \in \operatorname{SF}(f)[x]\) where the \(r_i\) are not necessarily distinct roots.

WLOG, \(g(x) = \prod_{i=1}^\ell (xr_i)\) for some \(1\leq \ell \leq p1\), i.e. rearrange the factors so that \(g\) is the first \(\ell\) of them.
 \(\ell \neq 1, p\) since \(f\) is reducible, making \(g, h\) nonconstant.

Set \(R_\ell \coloneqq\prod_{i=1}^\ell r_i\), which is the constant term in \(g\), so \(R_\ell \in { \mathbf{F} }\) since \(g\in { \mathbf{F} }[x]\).

Each \(r_i\) is a root of \(f\), so \(r_i^p  a = 0\) for all \(i\), so \(r_i^p = a\).

Trick: what is the \(p\)th power of \(R_\ell\)? \begin{align*} R_\ell^p &\coloneqq\qty{ \prod_{i=1}^\ell}^p \\ &= \prod_{i=1}^\ell r_i^p \\ &= \prod_{i=1}^\ell a \\ &= a^\ell ,\end{align*} so \(R_\ell^p = a^\ell\).

Use Bezout: \(\gcd(\ell, p) = 1\) since \(p\) is prime, so write \(tp + s\ell = 1\) for some \(t,s\in {\mathbf{Z}}\)

Use this to build a root of \(f\) that’s in \({ \mathbf{F} }\): write \begin{align*} a &= a^1\\ &= a^{tp + s\ell} \\ &= a^{tp} a^{s\ell} \\ &=a^{tp} (a^\ell)^s\\ &= a^{tp} (R_\ell^p)^s \\ &= (a^t R_\ell^s)^p \\ &\coloneqq\beta^p ,\end{align*} so \(a = \beta^p\).
 Check \(\beta\in { \mathbf{F} }\): use that \(R_\ell \in { \mathbf{F} }\) since it was a constant term of a polynomial in \({ \mathbf{F} }[x]\), \(a\in { \mathbf{F} }\) by assumption, and fields are closed under taking powers and products.
Fall 2019 #4 #algebra/qual/completed
Let \(F\) be a finite field with \(q\) elements. Let \(n\) be a positive integer relatively prime to \(q\) and let \(\omega\) be a primitive \(n\)th root of unity in an extension field of \(F\). Let \(E = F [\omega]\) and let \(k = [E : F]\).

Prove that \(n\) divides \(q^{k}1\).

Let \(m\) be the order of \(q\) in \({\mathbf{Z}}/n{\mathbf{Z}}^{\times}\). Prove that \(m\) divides \(k\).
 Prove that \(m = k\).
\todo[inline]{Revisit, tricky!}
 \({ \mathbf{F} }^{\times}\) is always cyclic for \({ \mathbf{F} }\) a field.
 Lagrange: \(H\leq G \implies {\sharp}H \divides {\sharp}G\).

Since \({\left\lvert {F} \right\rvert} = q\) and \([E:F] = k\), we have \({\left\lvert {E} \right\rvert} = q^k\) and \({\left\lvert {E^{\times}} \right\rvert} = q^k1\).

Noting that \(\zeta \in E^{\times}\) we must have \(n = o(\zeta) \divides {\left\lvert {E^{\times}} \right\rvert} = q^k1\) by Lagrange’s theorem.
 Rephrasing (a), we have \begin{align*} n \divides q^k1 &\iff q^k1 \cong 0 \operatorname{mod}n \\ &\iff q^k \cong 1 \operatorname{mod}n \\ &\iff m \coloneqq o(q) \divides k .\end{align*}

Since \(m\divides k \iff k = \ell m\), (claim) there is an intermediate subfield \(M\) such that \begin{align*} E \leq M \leq F \quad k = [F:E] = [F:M] [M:E] = \ell m ,\end{align*}
so \(M\) is a degree \(m\) extension of \(E\).

Now consider \(M^{\times}\).

By the argument in (a), \(n\) divides \(q^m  1 = {\left\lvert {M^{\times}} \right\rvert}\), and \(M^{\times}\) is cyclic, so it contains a cyclic subgroup \(H\) of order \(n\).

But then \(x\in H \implies p(x)\coloneqq x^n1 = 0\), and since \(p(x)\) has at most \(n\) roots in a field.

So \(H = \left\{{x \in M {~\mathrel{\Big\vert}~}x^n1 = 0}\right\}\), i.e. \(H\) contains all solutions to \(x^n1\) in \(E[x]\).

But \(\zeta\) is one such solution, so \(\zeta \in H \subset M^{\times}\subset M\).

Since \(F[\zeta]\) is the smallest field extension containing \(\zeta\), we must have \(F = M\), so \(\ell = 1\), and \(k = m\).
Spring 2019 #2 #algebra/qual/completed
Let \(F = { \mathbf{F} }_p\) , where \(p\) is a prime number.

Show that if \(\pi(x) \in F[x]\) is irreducible of degree \(d\), then \(\pi(x)\) divides \(x^{p^d}  x\).

Show that if \(\pi(x) \in F[x]\) is an irreducible polynomial that divides \(x^{p^n}  x\), then \(\deg \pi(x)\) divides \(n\).

Go to a field extension.
 Orders of multiplicative groups for finite fields are known.
 \({\mathbf{GF}}(p^n)\) is the splitting field of \(x^{p^n}  x \in { \mathbf{F} }_p[x]\).
 \(x^{p^d}  x \divides x^{p^n}  x \iff d \divides n\)
 \({\mathbf{GF}}(p^d) \leq {\mathbf{GF}}(p^n) \iff d\divides n\)
 \(x^{p^n}  x = \prod f_i(x)\) over all irreducible monic \(f_i\) of degree \(d\) dividing \(n\).
We can consider the quotient \(K = \displaystyle{\frac{{ \mathbf{F} }_p[x]}{\left\langle{\pi(x)}\right\rangle}}\), which since \(\pi(x)\) is irreducible is an extension of \({ \mathbf{F} }_p\) of degree \(d\) and thus a field of size \(p^d\) with a natural quotient map of rings \(\rho: { \mathbf{F} }_p[x] \to K\).
Since \(K^{\times}\) is a group of size \(p^d1\), we know that for any \(y \in K^{\times}\), we have by Lagrange’s theorem that the order of \(y\) divides \(p^d1\) and so \(y^{p^d} = y\).
So every element in \(K\) is a root of \(q(x) = x^{p^d}x\).
Since \(\rho\) is a ring morphism, we have
\begin{align*} \rho(q(x)) = \rho(x^{p^d}  x) &= \rho(x)^{p^d}  \rho(x) = 0 \in K \\ &\iff q(x) \in \ker \rho \\ &\iff q(x) \in \left\langle{\pi(x)}\right\rangle \\ &\iff \pi(x) \divides q(x) = x^{p^d}x ,\end{align*} where we’ve used that “to contain is to divide” in the last step.
\(\pi(x)\) divides \(x^{p^n}x \iff \deg \pi\) divides \(n\).
Let \(L \cong {\mathbf{GF}}(p^n)\) be the splitting field of \(\phi_n(x) \coloneqq x^{p^n}x\); then since \(\pi \divides \phi_n\) by assumption, \(\pi\) splits in \(L\). Let \(\alpha \in L\) be any root of \(\pi\); then there is a tower of extensions \({ \mathbf{F} }_p \leq { \mathbf{F} }_p(\alpha) \leq L\).
Then \({ \mathbf{F} }_p \leq { \mathbf{F} }_p(\alpha) \leq L\), and so \begin{align*} n &= [L: { \mathbf{F} }_p] \\ &= [L: { \mathbf{F} }_p(\alpha)]~[{ \mathbf{F} }_p(\alpha): { \mathbf{F} }_p] \\ &= \ell d ,\end{align*}
for some \(\ell \in {\mathbf{Z}}^{\geq 1}\), so \(d\) divides \(n\).
\(\impliedby\): If \(d\divides n\), use the fact (claim) that \(x^{p^n}  x = \prod f_i(x)\) over all irreducible monic \(f_i\) of degree \(d\) dividing \(n\). So \(f = f_i\) for some \(i\).
\(\star\) Fall 2016 #5 #algebra/qual/completed
How many monic irreducible polynomials over \({ \mathbf{F} }_p\) of prime degree \(\ell\) are there? Justify your answer.
Consider \(L \coloneqq{ \mathbf{F} }_p[x]/(x^{p^\ell}  x)\), this yields a field extension \(L/{ \mathbf{F} }_p\) with \([L: { \mathbf{F} }_p] = \ell\) and so \(L\cong { \mathbf{F} }_{p^\ell}\) is the splitting field of \(x^{p^\ell}  x\). Note that \(x^{p^\ell}x\) is the product of all monic irreducible polynomials in \({ \mathbf{F} }_p[x]\) of degree dividing \(\ell\), and since \(\ell\) is prime, the only such polynomials are of degrees \(1\) or \(\ell\) – this follows because any such polynomial would generate an intermediate extension \(L'\) with \(L/L'/{ \mathbf{F} }_p\), and multiplicativity in towers yields \begin{align*} [L:{ \mathbf{F} }_p] = [L: L'] \cdot [L' : { \mathbf{F} }_p] = \ell ,\end{align*} forcing either \([L: L'] = 1\) or \([L': { \mathbf{F} }_p] = 1\).
Let \(P\) be the desired number of monic irreducible degree \(\ell\) polynomials in \({ \mathbf{F} }_p[x]\). The idea is now to get a formula involving all monic irreducible (not necessarily degree \(\ell\)) polynomials and use it to solve for \(P\). By the above observation, we have a factorization \begin{align*} x^{p^\ell}  x = \prod_{i\in I} f_i(x) = \prod_{i\in I_1} f_i(x) \cdot \prod_{i\in I_2} g_i(x) ,\end{align*} where \(I\) is the set of all monic irreducible polynomials, and the above observation shows \(I = I_1{\textstyle\coprod}I_2\) where \(I_1\) are those of degree 1 and \(I_2\) are those of degree \(\ell\). Taking degrees of both sides yields \begin{align*} p^\ell = \sum_{i\in I_1} \deg f_i(x) + \sum_{i\in I_2} \deg g_i(x) = \sum_{i\in I_1} 1 + P\ell = {\sharp}I_1 + P\ell ,\end{align*} since each \(\deg f_i(x) = 1\) and there are \({\sharp}I_1\) many of them, and \(\deg g_i(x) = \ell\) and there are \(P\) of them. Rearranging yields \begin{align*} P\ell = p^\ell  {\sharp}I_1 \implies P = \ell^{1}\qty{p^\ell  {\sharp}I_1} ,\end{align*} and so it suffices to determine \({\sharp}I_1\), the number of monic linear irreducible polynomials in \({ \mathbf{F} }_p[x]\). These are all of the form \(x + a\) where \(a\in { \mathbf{F} }_p\), and there are \(p\) choices for \(a\), so the final count is \begin{align*} P = \ell^{1}\qty{p^\ell  p} .\end{align*}
\(\star\) Fall 2013 #7 #algebra/qual/work
Let \(F = { \mathbf{F} }_2\) and let \(\overline{F}\) denote its algebraic closure.

Show that \(\overline{F}\) is not a finite extension of \(F\).

Suppose that \(\alpha \in \overline{F}\) satisfies \(\alpha^{17} = 1\) and \(\alpha\neq 1\). Show that \(F(\alpha)/F\) has degree 8.
General Field Extensions
Spring 2020 #3 #algebra/qual/work
Let \(E\) be an extension field of \(F\) and \(\alpha\in E\) be algebraic of odd degree over \(F\).

Show that \(F(\alpha) = F(\alpha^2)\).

Prove that \(\alpha^{2020}\) is algebraic of odd degree over \(F\).
Spring 2012 #1 #algebra/qual/work
Suppose that \(F\subset E\) are fields such that \(E/F\) is Galois and \({\left\lvert {{ \operatorname{Gal}}(E/F)} \right\rvert} = 14\).

Show that there exists a unique intermediate field \(K\) with \(F\subset K \subset E\) such that \([K: F] = 2\).

Assume that there are at least two distinct intermediate subfields \(F \subset L_1, L_2 \subset E\) with \([L_i: F]= 7\). Prove that \({ \operatorname{Gal}}(E/F)\) is nonabelian.
Spring 2019 #8 #algebra/qual/completed
Let \(\zeta = e^{2\pi i/8}\).

What is the degree of \({\mathbf{Q}}(\zeta)/{\mathbf{Q}}\)?

How many quadratic subfields of \({\mathbf{Q}}(\zeta)\) are there?
 What is the degree of \({\mathbf{Q}}(\zeta, \sqrt[4] 2)\) over \({\mathbf{Q}}\)?

\(\zeta_n \coloneqq e^{2\pi i \over n}\), and \(\zeta_n^k\) is a primitive \(n\)th root of unity \(\iff \gcd(n, k) = 1\)
 In general, \(\zeta_n^k\) is a primitive \({n \over \gcd(n, k)}\)th root of unity.
 \(\deg \Phi_n(x) = \phi(n)\)

\(\phi(p^k) = p^k  p^{k1} = p^{k1}(p1)\)
 Proof: for a nontrivial gcd, the possibilities are \begin{align*} p, 2p, 3p, 4p, \cdots, p^{k2}p, p^{k1}p .\end{align*}
 \({ \mathsf{Gal}}({\mathbf{Q}}(\zeta)/{\mathbf{Q}}) \cong {\mathbf{Z}}/(n)^{\times}\)
Let \(K = {\mathbf{Q}}(\zeta)\).
 \(\zeta \coloneqq e^{2\pi i / 8}\) is a primitive \(8\)th root of unity
 The minimal polynomial of an \(n\)th root of unity is the \(n\)th cyclotomic polynomial \(\Phi_n\)
 The degree of the field extension is the degree of \(\Phi_8\), which is \begin{align*} \phi(8) = \phi(2^3) = 2^{31} \cdot (21) = 4 .\end{align*}
 So \([{\mathbf{Q}}(\zeta): {\mathbf{Q}}] = 4\).
 \({ \mathsf{Gal}}({\mathbf{Q}}(\zeta)/{\mathbf{Q}}) \cong {\mathbf{Z}}/(8)^{\times}\cong {\mathbf{Z}}/(4)\) by general theory
 \({\mathbf{Z}}/(4)\) has exactly one subgroup of index 2.
 Thus there is exactly one intermediate field of degree 2 (a quadratic extension).

Let \(L = {\mathbf{Q}}(\zeta, \sqrt[4] 2)\).

Note \({\mathbf{Q}}(\zeta) = {\mathbf{Q}}(i, \sqrt 2)\)

\({\mathbf{Q}}(i, \sqrt{2})\subseteq {\mathbf{Q}}(\zeta)\)
 \(\zeta_8^2 = i\), and \(\zeta_8 = \sqrt{2}^{1}+ i\sqrt{2}^{1}\) so \(\zeta_8 + \zeta_8 ^{1}= 2/\sqrt{2} = \sqrt{2}\).

\({\mathbf{Q}}(\zeta) \subseteq {\mathbf{Q}}(i, \sqrt{2})\):
 \(\zeta = e^{2\pi i / 8} = \sin(\pi/4) + i\cos(\pi/4) = {\sqrt 2 \over 2}\qty{1+i}\).

\({\mathbf{Q}}(i, \sqrt{2})\subseteq {\mathbf{Q}}(\zeta)\)

Thus \(L = {\mathbf{Q}}(i, \sqrt{2})(\sqrt[4]{2}) = {\mathbf{Q}}(i, \sqrt 2, \sqrt[4] 2) = {\mathbf{Q}}(i, \sqrt[4]{2})\).
 Uses the fact that \({\mathbf{Q}}(\sqrt 2) \subseteq {\mathbf{Q}}(\sqrt[4] 2)\) since \(\sqrt[4]{2}^2 = \sqrt{2}\)

Conclude \begin{align*} [L: {\mathbf{Q}}] = [L: {\mathbf{Q}}(\sqrt[4] 2)] ~[{\mathbf{Q}}(\sqrt[4] 2): {\mathbf{Q}}] = 2 \cdot 4 = 8 \end{align*} using the fact that the minimal polynomial of \(i\) over any subfield of \({\mathbf{R}}\) is always \(x^2 + 1\), so \(\min_{{\mathbf{Q}}(\sqrt[4] 2)}(i) = x^2 + 1\) which is degree 2.
Fall 2017 #3 #algebra/qual/work
Let \(F\) be a field. Let \(f(x)\) be an irreducible polynomial in \(F[x]\) of degree \(n\) and let \(g(x)\) be any polynomial in \(F[x]\). Let \(p(x)\) be an irreducible factor (of degree \(m\)) of the polynomial \(f(g(x))\).
Prove that \(n\) divides \(m\). Use this to prove that if \(r\) is an integer which is not a perfect square, and \(n\) is a positive integer then every irreducible factor of \(x^{2n}  r\) over \({\mathbf{Q}}[x]\) has even degree.
Spring 2015 #2 #algebra/qual/work
Let \({ \mathbf{F} }\) be a finite field.

Give (with proof) the decomposition of the additive group \(({ \mathbf{F} }, +)\) into a direct sum of cyclic groups.

The exponent of a finite group is the least common multiple of the orders of its elements. Prove that a finite abelian group has an element of order equal to its exponent.
 Prove that the multiplicative group \(({ \mathbf{F} }^{\times}, \cdot)\) is cyclic.
Spring 2014 #3 #algebra/qual/work
Let \(F\subset C\) be a field extension with \(C\) algebraically closed.

Prove that the intermediate field \(C_{\text{alg}} \subset C\) consisting of elements algebraic over \(F\) is algebraically closed.

Prove that if \(F\to E\) is an algebraic extension, there exists a homomorphism \(E\to C\) that is the identity on \(F\).