\chapter{Things Galois} %This chapter is mostly optional. %Read the first two sections and then decide %whether you want to read the rest of this chapter. \section{Motivation} \prototype{$\QQ(\sqrt2)$ and $\QQ(\cbrt{2})$.} The key idea in Galois theory is that of \emph{embeddings}, which give us another way to get at the idea of the ``conjugate'' we described earlier. Let $K$ be a number field. An \vocab{embedding} $\sigma \colon K \injto \CC$, is an \emph{injective field homomorphism}: it needs to preserve addition and multiplication, and in particular it should fix $1$. \begin{ques} Show that in this context, $\sigma(q) = q$ for any rational number $q$. \end{ques} \begin{example} [Examples of embeddings] \listhack \begin{enumerate}[(a)] \ii If $K = \QQ(i)$, the two embeddings of $K$ into $\CC$ are $z \mapsto z$ (the identity) and $z \mapsto \ol z$ (complex conjugation). \ii If $K = \QQ(\sqrt 2)$, the two embeddings of $K$ into $\CC$ are $a+b\sqrt 2 \mapsto a+b\sqrt 2$ (the identity) and $a+b\sqrt 2 \mapsto a-b\sqrt 2$ (conjugation). \ii If $K = \QQ(\cbrt 2)$, there are three embeddings: \begin{itemize} \ii The identity embedding, which sends $1 \mapsto 1$ and $\cbrt 2 \mapsto \cbrt 2$. \ii An embedding which sends $1 \mapsto 1$ and $\cbrt 2 \mapsto \omega \cbrt 2$, where $\omega$ is a cube root of unity. Note that this is enough to determine the rest of the embedding. \ii An embedding which sends $1 \mapsto 1$ and $\cbrt 2 \mapsto \omega^2 \cbrt 2$. \end{itemize} \end{enumerate} \end{example} I want to make several observations about these embeddings, which will form the core ideas of Galois theory. Pay attention here! \begin{itemize} \ii First, you'll notice some duality between roots: in the first example, $i$ gets sent to $\pm i$, $\sqrt 2$ gets sent to $\pm \sqrt 2$, and $\cbrt 2$ gets sent to the other roots of $x^3-2$. This is no coincidence, and one can show this occurs in general. Specifically, suppose $\alpha$ has minimal polynomial \[ 0 = c_n \alpha^n + c_{n-1} \alpha^{n-1} + \dots + c_1\alpha + c_0 \] where the $c_i$ are rational. Then applying any embedding $\sigma$ to both sides gives \begin{align*} 0 &= \sigma(c_n \alpha^n + c_{n-1} \alpha^{n-1} + \dots + c_1\alpha + c_0) \\ % &= \sigma(c_n \alpha^n) + \sigma(c_{n-1} \alpha^{n-1}) % + \dots + \sigma(c_1\alpha) + \sigma(c_0) \\ &= \sigma(c_n) \sigma(\alpha)^n + \sigma(c_{n-1}) \sigma(\alpha)^{n-1} + \dots + \sigma(c_1)\sigma(\alpha) + \sigma(c_0) \\ &= c_n \sigma(\alpha)^n + c_{n-1} \sigma(\alpha)^{n-1} + \dots + c_1\sigma(\alpha) + c_0 \end{align*} where in the last step we have used the fact that $c_i \in \QQ$, so they are fixed by $\sigma$. \emph{So, roots of minimal polynomials go to other roots of that polynomial.} \ii Next, I want to draw out a contrast between the second and third examples. Specifically, in example (b) where we consider embeddings $K = \QQ(\sqrt 2)$ to $\CC$. The image of these embeddings lands entirely in $K$: that is, we could just as well have looked at $K \to K$ rather than looking at $K \to \CC$. However, this is not true in (c): indeed $\QQ(\cbrt 2) \subset \RR$, but the non-identity embeddings have complex outputs! The key difference is to again think about conjugates. Key observation: \begin{moral} The field $K = \QQ(\cbrt 2)$ is ``deficient'' because the minimal polynomial $x^3-2$ has two other roots $\omega \cbrt{2}$ and $\omega^2 \cbrt{2}$ not contained in $K$. \end{moral} On the other hand $K = \QQ(\sqrt 2)$ is just fine because both roots of $x^2-2$ are contained inside $K$. Finally, one can actually fix the deficiency in $K = \QQ(\cbrt 2)$ by completing it to a field $\QQ(\cbrt 2, \omega)$. Fields like $\QQ(i)$ or $\QQ(\sqrt 2)$ which are ``self-contained'' are called \emph{Galois extensions}, as we'll explain shortly. \ii Finally, you'll notice that in the examples above, \emph{the number of embeddings from $K$ to $\CC$ happens to be the degree of $K$}. This is an important theorem, \Cref{thm:n_embeddings}. \end{itemize} In this chapter we'll develop these ideas in full generality, for any field other than $\QQ$. \section{Field extensions, algebraic closures, and splitting fields} \prototype{$\QQ(\cbrt 2)/\QQ$ is an extension, $\CC$ is an algebraic closure of any number field.} First, we define a notion of one field sitting inside another, in order to generalize the notion of a number field. \begin{definition} Let $K$ and $F$ be fields. If $F \subseteq K$, we write $K/F$ and say $K$ is a \vocab{field extension} of $F$. Thus $K$ is automatically an $F$-vector space (just like $\QQ(\sqrt 2)$ is automatically a $\QQ$-vector space). The \vocab{degree} is the dimension of this space, denoted $[K:F]$. If $[K:F]$ is finite, we say $K/F$ is a \vocab{finite (field) extension}. \end{definition} That's really all. There's nothing tricky at all. \begin{ques} What do you call a finite extension of $\QQ$? \end{ques} Degrees of finite extensions are multiplicative. \begin{theorem}[Field extensions have multiplicative degree] Let $F \subseteq K \subseteq L$ be fields with $L/K$, $K/F$ finite. Then \[ [L:K][K:F] = [L:F]. \] \end{theorem} \begin{proof} Basis bash: you can find a basis of $L$ over $K$, and then expand that into a basis $L$ over $F$. (Diligent readers can fill in details.) \end{proof} Next, given a field (like $\QQ(\cbrt2)$) we want something to embed it into (in our case $\CC$). So we just want a field that contains all the roots of all the polynomials: \begin{theorem}[Algebraic closures] Let $F$ be a field. Then there exists a field extension $\ol F$ containing $F$, called an \vocab{algebraic closure}, such that all polynomials in $\ol F[x]$ factor completely. \end{theorem} \begin{example} [$\CC$] $\CC$ is an algebraic closure of $\QQ$, $\RR$ and even itself. \end{example} \begin{abuse} Some authors also require the algebraic closure to be \emph{minimal by inclusion}: for example, given $\QQ$ they would want only $\ol\QQ$ (the algebraic numbers). It's a theorem that such a minimal algebraic closure is unique, and so these authors will refer to \emph{the} algebraic closure of $K$. I like $\CC$, so I'll use the looser definition. \end{abuse} \section{Embeddings into algebraic closures for number fields} Now that I've defined all these ingredients, I can prove: \begin{theorem}[The $n$ embeddings of a number field] \label{thm:n_embeddings} Let $K$ be a number field of degree $n$. Then there are exactly $n$ field homomorphisms $K \injto \CC$, say $\sigma_1, \dots, \sigma_n$ which fix $\QQ$. \end{theorem} \begin{remark} Note that a nontrivial homomorphism of fields is necessarily injective (the kernel is an ideal). This justifies the use of ``$\injto$'', and we call each $\sigma_i$ an \vocab{embedding} of $K$ into $\CC$. \end{remark} \begin{proof} This is actually kind of fun! Recall that any irreducible polynomial over $\QQ$ has distinct roots (\Cref{lem:irred_complex}). We'll adjoin elements $\alpha_1, \alpha_2, \dots, \alpha_m$ one at a time to $\QQ$, until we eventually get all of $K$, that is, \[ K = \QQ(\alpha_1, \dots, \alpha_n). \] Diagrammatically, this is \begin{center} \begin{tikzcd} \QQ \ar[r, hook] \ar[d, hook, "\id"'] & \QQ(\alpha_1) \ar[r, hook] \ar[d, hook, "\tau_1"'] & \QQ(\alpha_1, \alpha_2) \ar[r, hook] \ar[d, hook, "\tau_2"'] & \dots \ar[r, hook] & K \ar[d, hook, "\tau_m=\sigma"] \\ \CC \ar[r] & \CC \ar[r] & \CC \ar[r] & \dots \ar[r] & \CC \end{tikzcd} \end{center} First, we claim there are exactly \[ [\QQ(\alpha_1) : \QQ] \] ways to pick $\tau_1$. Observe that $\tau_1$ is determined by where it sends $\alpha_1$ (since it has to fix $\QQ$). Letting $p_1$ be the minimal polynomial of $\alpha_1$, we see that there are $\deg p_1$ choices for $\tau_1$, one for each (distinct) root of $p_1$. That proves the claim. Similarly, given a choice of $\tau_1$, there are \[ [\QQ(\alpha_1, \alpha_2) : \QQ(\alpha_1)] \] ways to pick $\tau_2$. (It's a little different: $\tau_1$ need not be the identity. But it's still true that $\tau_2$ is determined by where it sends $\alpha_2$, and as before there are $[\QQ(\alpha_1, \alpha_2) : \QQ(\alpha_1)]$ possible ways.) Multiplying these all together gives the desired $[K:\QQ]$. \end{proof} \begin{remark} The primitive element theorem actually implies that $m = 1$ is sufficient; we don't need to build a whole tower. This simplifies the proof somewhat. \end{remark} It's common to see expressions like ``let $K$ be a number field of degree $n$, and $\sigma_1, \dots, \sigma_n$ its $n$ embeddings'' without further explanation. The relation between these embeddings and the Galois conjugates is given as follows. \begin{theorem}[Embeddings are evenly distributed over conjugates] \label{thm:conj_distrb} Let $K$ be a number field of degree $n$ with $n$ embeddings $\sigma_1$, \dots, $\sigma_n$, and let $\alpha \in K$ have $m$ Galois conjugates over $\QQ$. Then $\sigma_j(\alpha)$ is ``evenly distributed'' over each of these $m$ conjugates: for any Galois conjugate $\beta$, exactly $\frac nm$ of the embeddings send $\alpha$ to $\beta$. \end{theorem} \begin{proof} In the previous proof, adjoin $\alpha_1 = \alpha$ first. \end{proof} So, now we can define the trace and norm over $\QQ$ in a nice way: given a number field $K$, we set \[ \Tr_{K/\QQ}(\alpha) = \sum_{i=1}^n \sigma_i(\alpha) \quad\text{and}\quad \Norm_{K/\QQ}(\alpha) = \prod_{i=1}^n \sigma_i(\alpha) \] where $\sigma_i$ are the $n$ embeddings of $K$ into $\CC$. \section{Everyone hates characteristic 2: separable vs irreducible} \prototype{$\QQ$ has characteristic zero, hence irreducible polynomials are separable.} Now, we want a version of the above theorem for any field $F$. If you read the proof, you'll see that the only thing that ever uses anything about the field $\QQ$ is \Cref{lem:irred_complex}, where we use the fact that \begin{quote} \itshape Irreducible polynomials over $F$ have no double roots. \end{quote} Let's call a polynomial with no double roots \vocab{separable}; thus we want irreducible polynomials to be separable. We did this for $\QQ$ in the last chapter by taking derivatives. Should work for any field, right? Nope. Suppose we took the derivative of some polynomial like $2x^3 + 24x + 9$, namely $6x^2 + 24$. In $\CC$ it's obvious that the derivative of a nonconstant polynomial $f'$ isn't zero. But suppose we considered the above as a polynomial in $\FF_3$, i.e.\ modulo $3$. Then the derivative is zero. Oh, no! We have to impose a condition that prevents something like this from happening. \begin{definition} For a field $F$, the \vocab{characteristic} of $F$ is the smallest positive integer $p$ such that, \[ \underbrace{1_F + \dots + 1_F}_{\text{$p$ times}} = 0 \] or zero if no such integer $p$ exists. \end{definition} \begin{example}[Field characteristics] Old friends $\RR$, $\QQ$, $\CC$ all have characteristic zero. But $\FF_p$, the integers modulo $p$, is a field of characteristic $p$. \end{example} \begin{exercise} Let $F$ be a field of characteristic $p$. Show that if $p > 0$ then $p$ is a prime number. (A proof is given next chapter.) \end{exercise} With the assumption of characteristic zero, our earlier proof works. \begin{lemma}[Separability in characteristic zero] Any irreducible polynomial in a characteristic zero field is separable. \end{lemma} Unfortunately, this lemma is false if the ``characteristic zero'' condition is dropped. \begin{remark} The reason it's called \emph{separable} is (I think) this picture: I have a polynomial and I want to break it into irreducible parts. Normally, if I have a double root in a polynomial, that means it's not irreducible. But in characteristic $p > 0$ this fails. So inseparable polynomials are strange when you think about them: somehow you have double roots that can't be separated from each other. \end{remark} We can get this to work for any field extension in which separability is not an issue. \begin{definition} A \vocab{separable extension} $K/F$ is one in which every irreducible polynomial in $F$ is separable (for example, if $F$ has characteristic zero). A field $F$ is \vocab{perfect} if any finite field extension $K/F$ is separable. \end{definition} In fact, as we see in the next chapter: \begin{theorem} [Finite fields are perfect] Suppose $F$ is a field with finitely many elements. Then it is perfect. \end{theorem} Thus, we will almost never have to worry about separability since every field we see in the Napkin is either finite or characteristic $0$. So the inclusion of the word ``separable'' is mostly a formality. Proceeding onwards, we obtain \begin{theorem}[The $n$ embeddings of any separable extension] Let $K/F$ be a separable extension of degree $n$ and let $\ol F$ be an algebraic closure of $F$. Then there are exactly $n$ field homomorphisms $K \injto \ol F$, say $\sigma_1, \dots, \sigma_n$, which fix $F$. \end{theorem} In any case, this lets us define the trace for \emph{any} separable normal extension. \begin{definition} Let $K/F$ be a separable extension of degree $n$, and let $\sigma_1$, \dots, $\sigma_n$ be the $n$ embeddings into an algebraic closure of $F$. Then we define \[ \Tr_{K/F}(\alpha) = \sum_{i=1}^n \sigma_i(\alpha) \quad\text{and}\quad \Norm_{K/F}(\alpha) = \prod_{i=1}^n \sigma_i(\alpha). \] When $F = \QQ$ and the algebraic closure is $\CC$, this coincides with our earlier definition! \end{definition} \section{Automorphism groups and Galois extensions} \prototype{$\QQ(\sqrt 2)$ is Galois but $\QQ(\cbrt 2)$ is not.} We now want to get back at the idea we stated at the beginning of this section that $\QQ(\cbrt 2)$ is deficient in a way that $\QQ(\sqrt 2)$ is not. First, we define the ``internal'' automorphisms. \begin{definition} Suppose $K/F$ is a finite extension. Then $\Aut(K/F)$ is the set of \emph{field isomorphisms} $\sigma : K \to K$ which fix $F$. In symbols \[ \Aut(K/F) = \left\{ \sigma : K \to K \mid \text{$\sigma$ is identity on $F$} \right\}. \] This is a group under function composition! \end{definition} Note that this time, we have a condition that $F$ is fixed by $\sigma$. (This was not there before when we considered $F = \QQ$, because we got it for free.) \begin{example}[Old examples of automorphism groups] Reprising the example at the beginning of the chapter in the new notation, we have: \begin{enumerate}[(a)] \ii $\Aut(\QQ(i) / \QQ) \cong \Zc 2$, with elements $z \mapsto z$ and $z \mapsto \ol z$. \ii $\Aut(\QQ(\sqrt 2) / \QQ) \cong \Zc 2$ in the same way. \ii $\Aut(\QQ(\cbrt 2) / \QQ)$ is the trivial group, with only the identity embedding! \end{enumerate} \end{example} \begin{example}[Automorphism group of $\QQ(\sqrt2,\sqrt3)$] Here's a new example: let $K = \QQ(\sqrt2, \sqrt3)$. It turns out that $\Aut(K/\QQ) = \{1, \sigma, \tau, \sigma\tau\}$, where \[ \sigma : \begin{cases} \sqrt2 &\mapsto -\sqrt2 \\ \sqrt3 &\mapsto \sqrt3 \end{cases} \quad\text{and}\quad \tau : \begin{cases} \sqrt2 &\mapsto \sqrt2 \\ \sqrt3 &\mapsto -\sqrt3. \end{cases} \] In other words, $\Aut(K/\QQ)$ is the Klein Four Group. \end{example} First, let's repeat the proof of the observation that these embeddings shuffle around roots (akin to the first observation in the introduction): \begin{lemma} [Root shuffling in $\Aut(K/F)$] Let $f \in F[x]$, suppose $K/F$ is a finite extension, and assume $\alpha \in K$ is a root of $f$. Then for any $\sigma \in \Aut(K/F)$, $\sigma(\alpha)$ is also a root of $f$. \label{lem:root_shuffle} \end{lemma} \begin{proof} Let $f(x) = c_n x^n + c_{n-1}x^{n-1} + \dots + c_0$, where $c_i \in F$. Thus, \[ 0 = \sigma(f(\alpha)) = \sigma\left( c_n\alpha^n + \dots + c_0 \right) = c_n\sigma(\alpha)^n + \dots + c_0 = f(\sigma(\alpha)). \qedhere \] \end{proof} In particular, taking $f$ to be the minimal polynomial of $\alpha$ we deduce \begin{moral} An embedding $\sigma \in \Aut(K/F)$ sends an $\alpha \in K$ to one of its various Galois conjugates (over $F$). \end{moral} Next, let's look again at the ``deficiency'' of certain fields. Look at $K = \QQ(\cbrt 2)$. So, again $K / \QQ$ is deficient for two reasons. First, while there are three maps $\QQ(\cbrt 2) \injto \CC$, only one of them lives in $\Aut(K/\QQ)$, namely the identity. In other words, $\left\lvert \Aut(K/\QQ) \right\rvert$ is \emph{too small}. Secondly, $K$ is missing some Galois conjugates ($\omega \cbrt 2$ and $\omega^2 \cbrt 2$). The way to capture the fact that there are missing Galois conjugates is the notion of a splitting field. \begin{definition} Let $F$ be a field and $p(x) \in F[x]$ a polynomial of degree $n$. Then $p(x)$ has roots $\alpha_1, \dots, \alpha_n$ in an algebraic closure of $F$. The \vocab{splitting field} of $F$ is defined as $F(\alpha_1, \dots, \alpha_n)$. \end{definition} In other words, the splitting field is the smallest field in which $p(x)$ splits. \begin{example}[Examples of splitting fields] \listhack \begin{enumerate}[(a)] \ii The splitting field of $x^2 - 5$ over $\QQ$ is $\QQ(\sqrt 5)$. This is a degree $2$ extension. \ii The splitting field of $x^2+x+1$ over $\QQ$ is $\QQ(\omega)$, where $\omega$ is a cube root of unity. This is a degree $3$ extension. % In particular, the splitting field of $x^3-2$ is a degree \emph{six} extension. \ii The splitting field of $x^2+3x+2 = (x+1)(x+2)$ is just $\QQ$! There's nothing to do. \end{enumerate} \end{example} \begin{example} [Splitting fields: a cautionary tale] The splitting field of $x^3 - 2$ over $\QQ$ is in fact \[ \QQ( \cbrt 2, \omega ) \] and not just $\QQ(\cbrt 2)$! One must really adjoin \emph{all} the roots, and it's not necessarily the case that these roots will generate each other. To be clear: \begin{itemize} \ii For $x^2-5$, we adjoin $\sqrt 5$ and this will automatically include $-\sqrt 5$. \ii For $x^2+x+1$, we adjoin $\omega$ and get the other root $\omega^2$ for free. \ii But for $x^3-2$, if we adjoin $\cbrt 2$, we do NOT get $\omega\cbrt2$ and $\omega^2\cbrt2$ for free. Indeed, $\QQ(\cbrt 2) \subset \RR$! \end{itemize} Note that in particular, the splitting field of $x^3-2$ over $\QQ$ is \emph{degree six}, not just degree three. \end{example} In general, \textbf{the splitting field of a polynomial can be an extension of degree up to $n!$}. The reason is that if $p(x)$ has $n$ roots and none of them are ``related'' to each other, then any permutation of the roots will work. Now, we obtain: \begin{theorem}[Galois extensions are splitting] For finite extensions $K/F$, $\left\lvert \Aut(K/F) \right\rvert$ divides $[K:F]$, with equality if and only if $K$ is the \emph{splitting field} of some separable polynomial with coefficients in $F$. \label{thm:Galois_splitting} \end{theorem} The proof of this is deferred to an optional section at the end of the chapter. If $K/F$ is a finite extension and $\left\lvert \Aut(K/F) \right\rvert = [K:F]$, we say the extension $K/F$ is \vocab{Galois}. In that case, we denote $\Aut(K/F)$ by $\Gal(K/F)$ instead and call this the \vocab{Galois group} of $K/F$. \begin{example} [Examples and non-examples of Galois extensions] \listhack \begin{enumerate}[(a)] \ii The extension $\QQ(\sqrt2) / \QQ$ is Galois, since it's the splitting field of $x^2-2$ over $\QQ$. The Galois group has order two, $\sqrt 2 \mapsto \pm \sqrt 2$. \ii The extension $\QQ(\sqrt2, \sqrt 3) / \QQ$ is Galois, since it's the splitting field of $(x^2-5)^2-6$ over $\QQ$. As discussed before, the Galois group is $\Zc 2 \times \Zc 2$. \ii The extension $\QQ(\cbrt{2}) / \QQ$ is \emph{not} Galois. \end{enumerate} \end{example} %Here is some more intuition on what $[K:F]$ actually measures: suppose $K$ is a splitting field %of some $(x-\alpha_1) \dots (x-\alpha_n)$, meaning $K = F(\alpha_1, \dots, \alpha_n)$. %Then a permutation $\sigma \in \Aut(K/F)$ is determined by where it sends each $\alpha_i$. %The dimension of $[K:F]$ measures how much ``redundancy'' there is among the $\alpha_i$. %For example, in the case of \[ (x-\sqrt5)(x+\sqrt5) = x^2-5 \]the $\sqrt 5$ and $-\sqrt 5$ were redundant, %in the sense that knowing $\sigma(\sqrt 5)$ tells you $\sigma(-\sqrt 5) = -\sigma(\sqrt 5)$. %But in the $x^3-2$ case, knowing $\sigma(\cbrt{2})$ does \emph{not} tell you where %$\omega\cbrt{2}$ should go; this is reflected in the fact that $[K:F]$ and $\Aut(K/F)$ %are both six rather than three. To explore $\QQ(\cbrt 2)$ one last time: \begin{example} [Galois closures, and the automorphism group of $\QQ(\cbrt2, \omega)$] Let's return to the field $K = \QQ(\cbrt 2, \omega)$, which is a field with $[K:\QQ] = 6$. Consider the two automorphisms: \[ \sigma: \begin{cases} \cbrt 2 &\mapsto \omega \cbrt 2 \\ \omega &\mapsto \omega \end{cases} \quad\text{and}\quad \tau: \begin{cases} \cbrt 2 &\mapsto \cbrt 2 \\ \omega &\mapsto \omega^2. \end{cases} \] Notice that $\sigma^3 = \tau^2 = \id$. From this one can see that the automorphism group of $K$ must have order $6$ (it certainly has order $\le 6$; now use Lagrange's theorem). So, $K/\QQ$ is Galois! Actually one can check explicitly that \[ \Gal(K/\QQ) \cong S_3 \] is the symmetric group on $3$ elements, with order $3! = 6$. \end{example} This example illustrates the fact that given a non-Galois field extension, one can ``add in'' missing conjugates to make it Galois. This is called taking a \vocab{Galois closure}. \section{Fundamental theorem of Galois theory} After all this stuff about Galois Theory, I might as well tell you the fundamental theorem, though I won't prove it. Basically, it says that if $K/F$ is Galois with Galois group $G$, then: \begin{moral} Subgroups of $G$ correspond exactly to fields $E$ with $F \subseteq E \subseteq K$. \end{moral} To tell you how the bijection goes, I have to define a fixed field. \begin{definition} Let $K$ be a field and $H$ a subgroup of $\Aut(K/F)$. We define the \vocab{fixed field} of $H$, denoted $K^H$, as \[ K^H \defeq \left\{ x \in K : \sigma(x)=x \; \forall \sigma \in H \right\}. \] \end{definition} \begin{ques} Verify quickly that $K^H$ is actually a field. \end{ques} Now let's look at examples again. Consider $K = \QQ(\sqrt2, \sqrt3)$, where \[ G = \Gal(K/\QQ) = \{\id, \sigma, \tau, \sigma\tau\} \] is the Klein four group (where $\sigma(\sqrt2) = -\sqrt 2$ but $\sigma(\sqrt 3) = \sqrt 3$; $\tau$ goes the other way). \begin{ques} Let $H = \{\id, \sigma\}$. What is $K^H$? \end{ques} In that case, the diagram of fields between $\QQ$ and $K$ matches exactly with the subgroups of $G$, as follows: \begin{center} \begin{tikzcd} & \QQ(\sqrt2, \sqrt3) \ar[ld, dash] \ar[d, dash] \ar[rd, dash] & \\ \QQ(\sqrt2) \ar[rd, dash] & \QQ(\sqrt6) \ar[d, dash] & \QQ(\sqrt3) \ar[ld, dash] \\ & \QQ & \end{tikzcd} \begin{tikzcd} & \{\id\} \ar[ld, dash] \ar[d, dash] \ar[rd, dash] & \\ \{\id,\tau\} \ar[rd, dash] & \{\id,\sigma\tau\} \ar[d, dash] & \{\id,\sigma\} \ar[ld, dash] \\ & G & \end{tikzcd} \end{center} We see that subgroups correspond to fixed fields. That, and much more, holds in general. \begin{theorem}[Fundamental theorem of Galois theory] Let $K/F$ be a Galois extension with Galois group $G = \Gal(K/F)$. \begin{enumerate}[(a)] \ii There is a bijection between field towers $F \subseteq E \subseteq K$ and subgroups $H \subseteq G$: \[ \left\{ \begin{array}{c} K \\ \mid \\ E \\ \mid \\ F \end{array} \right\} \iff \left\{ \begin{array}{c} 1 \\ \mid \\ H \\ \mid \\ G \end{array} \right\} \] The bijection sends $H$ to its fixed field $K^H$, and hence is inclusion reversing. \ii Under this bijection, we have $[K:E] = \left\lvert H \right\rvert$ and $[E:F] = [G:H]$. \ii $K/E$ is always Galois, and its Galois group is $\Gal(K/E) = H$. \ii $E/F$ is Galois if and only if $H$ is normal in $G$. If so, $\Gal(E/F) = G/H$. \end{enumerate} \end{theorem} \begin{exercise} Suppose we apply this theorem for \[ K = \QQ(\cbrt2, \omega). \] Verify that the fact $E = \QQ(\cbrt 2)$ is not Galois corresponds to the fact that $S_3$ does not have normal subgroups of order $2$. \end{exercise} \section\problemhead \begin{sproblem}[Galois group of the cyclotomic field] Let $p$ be an odd rational prime and $\zeta_p$ a primitive $p$th root of unity. Let $K = \QQ(\zeta_p)$. Show that \[ \Gal(K/\QQ) \cong (\ZZ/p\ZZ)^\ast. \] \begin{hint} Look at the image of $\zeta_p$. \end{hint} \begin{sol} It's just $\Zc{p-1}$, since $\zeta_p$ needs to get sent to one (any) of the $p-1$ primitive roots of unity. \end{sol} \end{sproblem} \begin{problem}[Greek constructions] Prove that the three Greek constructions \begin{enumerate}[(a)] \ii doubling the cube, \ii squaring the circle, and \ii trisecting an angle \end{enumerate} are all impossible. (Assume $\pi$ is transcendental.) \begin{hint} Repeated quadratic extensions have degree $2$, so one can only get powers of two. \end{hint} \end{problem} \begin{problem} [China Hong Kong Math Olympiad] \yod Prove that there are no rational numbers $p$, $q$, $r$ satisfying \[ \cos\left( \frac{2\pi}{7} \right) = p + \sqrt{q} + \sqrt[3]{r}. \] \begin{sol} A similar (but not identical) problem is solved here: \url{https://aops.com/community/c6h149153p842956}. \end{sol} \end{problem} \begin{problem} Show that the only automorphism of $\RR$ is the identity. Hence $\Aut(\RR)$ is the trivial group. \begin{hint} Hint: $\sigma(x^2) = \sigma(x)^2 \ge 0$ plus Cauchy's Functional Equation. \end{hint} \end{problem} \begin{problem}[Artin's primitive element theorem] \yod Let $K$ be a number field. Show that $K \cong \QQ(\gamma)$ for some $\gamma$. \label{prob:artin_primitive_elm} \begin{hint} By induction, suffices to show $\QQ(\alpha, \beta) = \QQ(\gamma)$ for some $\gamma$ in terms of $\alpha$ and $\beta$. For all but finitely many rational $\lambda$, the choice $\gamma = \alpha + \lambda \beta$ will work. \end{hint} \begin{sol} \url{http://www.math.cornell.edu/~kbrown/6310/primitive.pdf} \end{sol} \end{problem} \section{(Optional) Proof that Galois extensions are splitting} We prove \Cref{thm:Galois_splitting}. First, we extract a useful fragment from the fundamental theorem. \begin{theorem}[Fixed field theorem] \label{thm:fixed_field_theorem} Let $K$ be a field and $G$ a subgroup of $\Aut(K)$. Then $[K:K^G] = \left\lvert G \right\rvert$. \end{theorem} The inequality itself is not difficult: \begin{exercise} Show that $[K:F] \ge |\Aut(K/F)|$, and that equality holds if and only if the set of elements fixed by all $\sigma \in \Aut(K/F)$ is exactly $F$. (Use \Cref{thm:fixed_field_theorem}.) \end{exercise} The equality case is trickier. The easier direction is when $K$ is a splitting field. Assume $K = F(\alpha_1, \dots, \alpha_n)$ is the splitting field of some separable polynomial $p \in F[x]$ with $n$ distinct roots $\alpha_1, \dots, \alpha_n$. Adjoin them one by one: \begin{center} \begin{tikzcd} F \ar[r, hook] \ar[d, "\id"'] & F(\alpha_1) \ar[r, hook] \ar[d, "\tau_1"'] & F(\alpha_1, \alpha_2) \ar[r, hook] \ar[d, "\tau_2"'] & \dots \ar[r, hook] & K \ar[d, "\tau_n=\sigma"] \\ F \ar[r, hook] & F(\alpha_1) \ar[r, hook] & F(\alpha_1, \alpha_2) \ar[r, hook] & \dots \ar[r, hook] & K \end{tikzcd} \end{center} (Does this diagram look familiar?) Every map $K \to K$ which fixes $F$ corresponds to an above commutative diagram. As before, there are exactly $[F(\alpha_1) : F]$ ways to pick $\tau_1$. (You need the fact that the minimal polynomial $p_1$ of $\alpha_1$ is separable for this: there need to be exactly $\deg p_1 = [F(\alpha_1) : F]$ distinct roots to nail $p_1$ into.) Similarly, given a choice of $\tau_1$, there are $[F(\alpha_1, \alpha_2) : F(\alpha_1)]$ ways to pick $\tau_2$. Multiplying these all together gives the desired $[K:F]$. \bigskip Now assume $K/F$ is Galois. First, we state: \begin{lemma} Let $K/F$ be Galois, and $p \in F[x]$ irreducible. If any root of $p$ (in $\ol F$) lies in $K$, then all of them do, and in fact $p$ is separable. \end{lemma} \begin{proof} Let $\alpha \in K$ be the prescribed root. Consider the set \[ S = \left\{ \sigma(\alpha) \mid \sigma \in \Gal(K/F) \right\}. \] (Note that $\alpha \in S$ since $\Gal(K/F) \ni \id$.) By construction, any $\tau \in \Gal(K/F)$ fixes $S$. So if we construct \[ \tilde p(x) = \prod_{\beta \in S} (x - \beta), \] then by Vieta's Formulas, we find that all the coefficients of $\tilde p$ are fixed by elements of $\sigma$. By the \emph{equality case} we specified in the exercise, it follows that $\tilde p$ has coefficients in $F$! (This is where we use the condition.) Also, by \Cref{lem:root_shuffle}, $\tilde p$ divides $p$. Yet $p$ was irreducible, so it is the minimal polynomial of $\alpha$ in $F[x]$, and therefore we must have that $p$ divides $\tilde p$. Hence $p = \tilde p$. Since $\tilde p$ was built to be separable, so is $p$. \end{proof} Now we're basically done -- pick a basis $\omega_1$, \dots, $\omega_n$ of $K/F$, and let $p_i$ be their minimal polynomials; by the above, we don't get any roots outside $K$. Consider $P = p_1 \dots p_n$, removing any repeated factors. The roots of $P$ are $\omega_1$, \dots, $\omega_n$ and some other guys in $K$. So $K$ is the splitting field of $P$.