Problem 1

The splitting field of this polynomial is ${\mathbf{Q}}(\sqrt[3]2, \sqrt 3, \zeta_3)$ where $\zeta_3$ is a primitive third root of unity.

To get the degree of this extension, we extend fields in the indicated order. Since ${\mathbf{Q}}(\sqrt[3] 2, \sqrt 3)$ is totally real, the minimal polynomial of $\zeta$ over it still has degree $\phi(3) = 2$. A quick check also shows that $\sqrt 3$ is not contained in ${\mathbf{Q}}(\sqrt[3] 2)$, yielding another degree 2 extension, and finally a degree 3 extension.

Thus we have an extension of degree 12, and since we’ve constructed a Galois extension $L$ (a separable splitting field), if we define $G \coloneqq{ \mathsf{Gal}}({\mathbf{Q}}/L)$, we have ${\left\lvert {G} \right\rvert} = 12$. Since we know that the splitting field of ${\mathbf{Q}}(\sqrt[3] 2)/ {\mathbf{Q}}$ has Galois group $D_3$, we must have $D_3 \leq G$. This reduces the possibilities just $D_3 \times{\mathbf{Z}}_2 \cong D_6$.

We have the following subgroup diagram (Figure 1).

\

where we can simplify things by only considering conjugacy classes of subgroups, since these will correspond to conjugate field extensions (Figure 2).

\

We can explicitly identify the relevant automorphisms: $$\begin{align*} \begin{align*} \sigma: \sqrt[3] 2 \mapsto \zeta_3 \sqrt[3] 2 \\ \tau: \zeta_3 \mapsto \zeta_3^2 \\ \gamma: \sqrt 3 \mapsto -\sqrt 3 .\end{align*}$$ \end{align*} We can then present $G = \left\langle{\sigma, \gamma, \tau \mathrel{\Big|}\sigma^3 = \tau^2 = \gamma^2 = (\sigma\tau)^2 = [\sigma, \gamma] = [\tau, \gamma] = e}\right\rangle$, and obtain the following lattice:

\begin{align*} \begin{tikzcd} & & {<\sigma, \tau, \gamma>} & & & & \\ & & & & & & \\ <\tau> \times <\gamma> \arrow[rruu, dashed] & & {<\sigma, \tau>} \arrow[uu] & & {<\sigma, \tau\gamma>} \arrow[lluu] & & <\sigma> \times <\gamma> \arrow[lllluu] \\ & & & & & & \\ <\tau> \arrow[uu] \arrow[rruu, dashed] & & <\tau\gamma> \arrow[rruu, dashed] \arrow[lluu] & & <\gamma> \arrow[rruu] \arrow[lllluu] & & <\sigma> \arrow[uu] \arrow[lluu] \arrow[lllluu] \\ & & & & & & \\ & & & \arrow[llluu] \arrow[luu] \arrow[ruu] \arrow[rrruu] & & & \end{tikzcd} \end{align*}

which, up to conjugacy, fix the following intermediate field extensions (Figure 3).

\

$\hfill\blacksquare$

Problem 2

We can note that since $f$ has 4 roots, the Galois group $G$ of its splitting field will be a subgroup of $S_4$. Moreover, $G$ must be a transitive subgroup of $S_4$, i.e. the action of $G$ on the roots of $f$ should be transitive. This reduces the possibilities to $G \cong S^4, A^4, D^4, {\mathbf{Z}}_4, {\mathbf{Z}}_2^2$.

Since $f$ has exactly 2 real roots and thus a pair of roots that are complex conjugates, the automorphism given by complex conjugation is an element of $G$. But this corresponds to a 2-cycle $\tau = (ab)$, and we can then make the following conclusions:

• Not $A_4$: $A_4$ contains only even cycles, and $\tau$ is odd.
• Not $Z_4$: This subgroup is generated by a single 4-cycle $\sigma$, which up to conjugacy is $(1234)$, and $\sigma^n$ is not a 2-cycle for any $n$.
• Not ${\mathbf{Z}}_2^2$: In order to be transitive, this subgroup must be $\left\{{e, (12)(34), (13)(24), (14)(23)}\right\}$, which does not contain $\tau$.

The only remaining possibilities are $S^4$ and $D^4$. $\hfill\blacksquare$

Problem 3

Part 1

To see that $\phi(n)$ is even for all $n>2$, we can take a prime factorization of $n$ and write $$\begin{align*} \phi(n) = \phi\left( \prod_{i=1}^m p_i^{k_i}\right) = \prod_{i=1}^m \phi(p_i^{k_i}) = \prod_{i=1}^m p^{k_i - 1}(p - 1) = \prod_{i=1}^m p^{k_i - 1} \prod_{i=1}^m (p - 1) \end{align*}$$

where each $k_i \geq 1 \implies k_i-1 \geq 0$. But every prime power is odd, and a product of odd numbers is odd, so the first product is odd. It is also true that $p-1$ is even for every prime $p$, and the second term is a product of even terms and thus even. So $\phi(n)$ is the product of an even and an odd number, which is always even.

Part 2

Suppose $\phi(n) = 2$. Take a prime factorization of $n$, so we have $$\begin{align*} 2 = \phi(n) = \prod_{i=1}^m \phi(p_i^{k_i}) \end{align*}$$

Since the only factors of 2 are 1 and 2, we must have $\phi(p_i^{k_i}) = 2$ for exactly one $i$, and the rest must be equal to 1.

Consider the term that equals 2. We have $\phi(p_i^{k_i}) = p^{k_i - 1}(p-1) = 2$, so we must have either

• Case 1: $p-1 = 2$ and $p^{k_i - 1} = 1$, so $p=3$ and $k_i = 1$. So $3 \divides n$, but $3^\ell$ does not divide $n$ for any $\ell > 1$.
• Case 2: $p^{k_i - 1} = 2$ and $(p-1) = 1$, so $p=2$ and $k_i = 2$. Thus $2^2$ divides $n$ but $2^\ell$ does not for any $\ell > 2$.

In either case, it remains to check are whether the other factors where $\phi(p_j^{k_j}) = 1$ can contribute any other distinct divisors to $n$. We can note that $\phi(p_j^{k_j}) = 1$ iff $p^{k_j-1}(p-1) = 1$, so this forces $p=2$ and $k_j = 1$. So $n$ may or may not contain a single factor of 2, but by uniqueness of prime factorization, this can only happen in case 1. Note that this also forces $2\divides n$ but $2^2$ does not divide $n$.

In summary, we’ve found that $\phi(n) = 2$ implies that

• $3 \divides n$, 9 does not divide $n$, and
  • $2\divides n$, 4 does not divide $n$
  • $2$ does not divide $n$
• $2^2 \divides n, 2^3$ does not divide n.

This reduces the possibilities to the finite set $n \in \left\{{6,3,4}\right\}$, and $\phi(6) = \phi(3) = \phi(4) = 2$. $\hfill\blacksquare$

Problem 4

Note that since $\zeta(\zeta+\zeta^{-1}) = \zeta^2 + 1$, we have the relation $\zeta^2 - (\zeta+\zeta^{-1})\zeta + 1 = 0$. But then $$\begin{align*} f(x) = x^2 - (\zeta + \zeta^{-1})x + 1 \end{align*}$$

is a polynomial in ${\mathbf{Q}}(\zeta + \zeta^{-1})$ for which $f(\zeta) = 0$. Thus $g = \min(\zeta, {\mathbf{Q}}(\zeta + \zeta^{-1}))$ divides $f$, but since $\deg f = 2$ and ${\mathbf{Q}}(\zeta + \zeta^{-1})$ is totally real, $\zeta\not\in{\mathbf{Q}}(\zeta + \zeta^{-1})$. This means that $g$ can not be linear and must have degree at least 2, but the above argument shows that $g$ has degree at most 2, so it must be 2. Letting $m = [{\mathbf{Q}}(\zeta + \zeta^{-1}): {\mathbf{Q}}]$, we have $$\begin{align*} \begin{align*} [{\mathbf{Q}}(\zeta) : {\mathbf{Q}}] &= [{\mathbf{Q}}(\zeta): {\mathbf{Q}}(\zeta + \zeta^{-1})] [{\mathbf{Q}}(\zeta + \zeta^{-1}) : {\mathbf{Q}}] \\ \implies \phi(n) &= 2 m ,\end{align*}$$ \end{align*}

and so $m = \phi(n)/2$ as desired.

Problem 5

Suppose $F = K[\alpha_1, \cdots, \alpha_n]$ where $\alpha_1^{n_1} \in K$ for some $n_1$ and }or each $i$ we have $\alpha_i^{n_i} \in K[\alpha_1, \cdots, \alpha_{i-1}]$ for some powers $n_i$. We want to show that $F = E[\beta_1, \cdots \beta_m]$ where each $\beta_i$ satisfy a similar condition.

Let $A = \left\{{\alpha_i {~\mathrel{\Big\vert}~}\alpha_i \not\in E}\right\}$, then it is since $E \hookrightarrow F$, adjoining all elements of $A$ to $E$ will yield exactly $F$. Using the order of $\alpha_i$ given by the definition of $F$ as a radical extension, let $\beta_1$ be the $\alpha_i \in A$ with the smallest index $i$. Then by assumption, there is some $m_1$ such that $\beta^{m_1} \in K[\alpha_1, \cdots, \alpha_{i-1}] \subset F$, so we can construct $F_1 \coloneqq E[\beta_1]$ which will be a radical extension.

Inductively letting $A_2 = A \setminus\left\{{\beta_1}\right\}$ and repeating this process to construct $L_2$ will yield radical extensions at every step, and since $A$ is finite, there is some $n$ such that $L_n = L$. But then $L$ is a radical extension over $E$ as desired.

Problem 6

Part 1

Since $f$ is irreducible of degree $n$ and $u$ is a root of $f$, the minimal polynomial of $u$ over $K$ is in fact $f$, and thus the degree of the extension $K(u) / K$ is given by

$$\begin{align*} [K(u) : K] = \deg \min(u, K) = n. \end{align*}$$

To see that $K(u)$ is not Galois, we just note that since $f$ was irreducible, and $u$ was only one root of $f$, $K(u)$ is not the splitting field of $f$, and is thus not the splitting field of any other irreducible polynomial over $K$.

To see that $\mathrm{Aut}(K(u)/ K)$ is trivial, note that any $K{\hbox{-}}$automorphism of $K(u)$ can only send $u$ to one of its conjugates. But the only conjugate of $u$ in $K(u)$ is $u$ itself, so only the identity automorphism can occur.

Part 2

The normal closure $L$ of $K$ is defined as the smallest extension of $K$ such that if $\alpha$ is a root of any irreducible polynomial in $K[x]$ and $\alpha \in L$, then all of its conjugates are in $L$ as well. But this means any such polynomial splits in $L$. In particular, if $u\in L$, then $f$ splits in $L$, and so $L$ contains the splitting field $F$.

Part 3

By a theorem in class, this would force ${ \mathsf{Gal}}(E/K)$ to be solvable, which would imply that $S_n$ is solvable – but for $n\geq 5, S_n$ will not be solvable, a contradiction.

Qual Problems

Problem 1

Part 1

If $L/K$ is a finite field extension which is both separable and a splitting field of some polynomial in $K[x]$, then $[L: K] = {\left\lvert {{ \mathsf{Gal}}} \right\rvert}{L/ K}$.

Part 2

The extension ${\mathbf{Q}}(\zeta_{43})$ is the splitting field of the cyclotomic polynomial $\Phi_{43}(x) = \sum_{i=1}^42 x^i$, which is degree $\phi(43) = 42$ since $43$ is prime.

Moreover, the Galois group is isomorphic to ${\mathbf{Z}}_{43}^{\times}\cong {\mathbf{Z}}_{42}$.

Part 3

Since proper subfields will correspond to intermediate extensions which will correspond to subgroups of the Galois group, this problem is reduced to counting the number of distinct subgroups of ${\mathbf{Z}}_{42}$. This is a cyclic group, so there is exactly one subgroup of order $d$ for each $d$ dividing 42. Since $42 = 2*3*7$, we have

• A subgroup of order 2, corresponding to a field extension of degree 21,
• A subgroup of order 3, corresponding to a field extension of degree 14,
• A subgroup of order 6, corresponding to a field extension of degree 7,
• A subgroup of order 7, corresponding to a field extension of degree 6,
• A subgroup of order 14, corresponding to a field extension of degree 3,
• A subgroup of order 21, corresponding to a field extension of degree 2.

Problem 2

Part 1

A splitting field of $f$ over $F$ is an extension $L \geq F$ that contains every root of $f$, so that $f$ can be decomposed as a product of linear factors i.e. $f(x) = \prod_{i=1}^{\deg f} (x-\alpha_i)^{m_i}$ in $L[x]$.

Part 2

If $E \geq F$ is a finite extension, then it is algebraic and $E = F[\alpha_1, \cdots, \alpha_n]$. So we can let $g(x) = \prod_{i=1}^n (x-\alpha_i)$. By construction, each $\alpha_i$ is a root, and so $E$ is a splitting field for $g$.

Part 3

Since $E$ was shown to be a splitting field, it only remains to show that it is separable. But this follows from the fact that each $\alpha_i$ is a separable element, since their minimal polynomial over $F$ is $g$. So $E$ is a Galois extension.

Problem 3

Part 1

False: take $K \leq L \leq M$ as ${\mathbf{Q}}\leq {\mathbf{Q}}(\sqrt[3]2) \leq {\mathbf{Q}}(\sqrt[3]2, \zeta_3)$. Then $M$ is the splitting field of $x_3-2$, and in characteristic zero is thus Galois. But $L$ is not the splitting field of any irreducible polynomial in ${\mathbf{Q}}[x]$, so it is not Galois.

Part 2

This is true. By the Galois correspondence, it suffices to show that $H \coloneqq{ \mathsf{Gal}}(M/L)$ is a normal subgroup of $G \coloneqq{ \mathsf{Gal}}(M/ K)$. To that end, let $\phi \in G$, so $\phi: M \to M$ is a lift of $\operatorname{id}_K$. Then $H {~\trianglelefteq~}G$ iff $\phi H \phi^{-1}= H$. Letting $\sigma \in H$, we need to show that $$\begin{align*} (\phi^{-1}\circ \sigma \circ \phi)(L) = L, \end{align*}$$ i.e. that this composition is some automorphism of $M$ that fixes $L$.

Consider how this acts on elements of $L$. If $\ell \in L$, then $\ell = \sum k_i \ell_i$ since $L$ is a finite-degree extension, thus algebraic, thus spanned by some basis $\ell_i \in L$ as a vector space over $K$.

In particular, since $\phi$ is some $M{\hbox{-}}$automorphism, it restricts to an $L{\hbox{-}}$automorphism, which must send each $\ell_i$ to some conjugate $\ell_i'$. Similarly, $\phi^{-1}(\ell_i') = \ell_i$.

We thus have $$\begin{align*} \begin{align*} (\phi^{-1}\sigma \phi)(a) &= (\phi^{-1}\sigma \phi)(\sum k_i \ell_i) \\ &= (\phi^{-1}\sigma)(\sum k_i \phi(\ell_i)) \\ &= (\phi^{-1}\sigma)(\sum k_i \ell_i') \\ &= (\phi^{-1})(\sum k_i \sigma(\ell_i')) \\ &= (\phi^{-1})(\sum k_i \ell_i') \quad\text{since $\sigma$ fixes $L$}\\ &= \sum k_i \phi^{-1}(\ell_i') \\ &= \sum k_i \ell_i \\ ,\end{align*}$$ \end{align*}

and so this composite fixes $L$ as desired. This $H {~\trianglelefteq~}G$, which is what we wanted to show.