# Galois Theory

## 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_{n-1}(\zeta_n + \zeta_n^{-1}) &= \zeta_n^{n-1} + \zeta_n^{-(n-1)} = \zeta_n^{-1} + \zeta_n^{1} = \alpha_1 \\ \sigma_{n-2}(\zeta_n + \zeta_n^{-1}) &= \zeta_n^{n-2} + \zeta_n^{-(n-2)} = \zeta_n^{-2} + \zeta_n^{2} = \alpha_2 \\ \sigma_{n-3}(\zeta_n + \zeta_n^{-1}) &= \zeta_n^{n-3} + \zeta_n^{-(n-3)} = \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 $$\sigma_{k}(\alpha_1)=\sigma_{n-k}(\alpha_1)$$ and we pick up $$(n-1)/2$$ 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 non-perfect 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 $$f(x) = (x-t^{1\over p})^p$$ 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: $$G(x^2-2)$$

Compute the Galois group of $$x^2-2$$.

$${\mathbf{Z}}/2{\mathbf{Z}}$$?

### Exercise: $$G(x^p-2)$$

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 $$p(p-1)$$ 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 $$\min_{\zeta_3, {\mathbf{Q}}}(x) = (x^3-1)/(x-1) = x^2 + x + 1$$, 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 $$\sum_{k=0}^{n-1}\zeta_n^k = 0$$. 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 $$\min_{\sqrt 5, {\mathbf{Q}}}(x) = x^2-5$$ 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 $$f(x) = x^3-3x -3\in {\mathbf{Q}}[x]/{\mathbf{Q}}$$.

### 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 ad-hoc 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 $$f(x) = x^4-2 \in {\mathbf{Q}}[x]$$.

• 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 $$x^4-2$$ 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 $${ \mathbf{F} }^{\times}$$.

• $$\phi(p^k) = p^{k-1}(p-1)$$

• 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 $$f'(x) = nx^{n-1}$$ 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 $$\left\{{1, \zeta, \zeta^2, \cdots, \zeta^{n-1}}\right\}$$, 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 $$L$$, i-t 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 13-subgroup 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 31-subgroup 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 $$x^p-a \in F[x]$$ is irreducible.

• Contradiction: go to splitting field, apply Freshman’s dream.
• Use that this polynomial is ramified, and its only factors are $$(x-a)$$.

• Suppose $$a$$ is not a $$p$$th power in $$F$$, then $$f(x) \coloneqq x^p-a$$ 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 $$f(x) \coloneqq x^p-a \in { \mathbf{F} }[x]$$ 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, $$g(x) = \prod_{i=1}^\ell (x-r_i)$$ for some $$1\leq \ell \leq p-1$$, 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 $${\left\lvert {E^{\times}} \right\rvert} = q^k-1$$.

• Noting that $$\zeta \in E^{\times}$$ we must have $$n = o(\zeta) \divides {\left\lvert {E^{\times}} \right\rvert} = q^k-1$$ by Lagrange’s theorem.

• Rephrasing (a), we have \begin{align*} n \divides q^k-1 &\iff q^k-1 \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 $$x\in H \implies p(x)\coloneqq x^n-1 = 0$$, and since $$p(x)$$ has at most $$n$$ roots in a field.

• So $$H = \left\{{x \in M {~\mathrel{\Big\vert}~}x^n-1 = 0}\right\}$$, i.e. $$H$$ contains all solutions to $$x^n-1$$ 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 $$p^d-1$$, we know that for any $$y \in K^{\times}$$, we have by Lagrange’s theorem that the order of $$y$$ divides $$p^d-1$$ 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.
• $$\deg \Phi_n(x) = \phi(n)$$
• $$\phi(p^k) = p^k - p^{k-1} = p^{k-1}(p-1)$$
• Proof: for a nontrivial gcd, the possibilities are \begin{align*} p, 2p, 3p, 4p, \cdots, p^{k-2}p, p^{k-1}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^{3-1} \cdot (2-1) = 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 $$\zeta_8 + \zeta_8 ^{-1}= 2/\sqrt{2} = \sqrt{2}$$.
• $${\mathbf{Q}}(\zeta) \subseteq {\mathbf{Q}}(i, \sqrt{2})$$:
• $$\zeta = e^{2\pi i / 8} = \sin(\pi/4) + i\cos(\pi/4) = {\sqrt 2 \over 2}\qty{1+i}$$.
• 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 $$F$$.

#4 #algebra/qual/work #algebra/qual/completed #9 #10 #8 #7 #5 #3 #6 #2 #1