$\require{physics}\newcommand{\cbrt}[1]{\sqrt[3]{#1}}\newcommand{\sgn}{\text{sgn}}\newcommand{\ii}[1]{\textit{#1}}\newcommand{\eps}{\varepsilon}\newcommand{\EE}{\mathbb E}\newcommand{\PP}{\mathbb P}\newcommand{\Var}{\mathrm{Var}}\newcommand{\Cov}{\mathrm{Cov}}\newcommand{\pperp}{\perp\kern-6pt\perp}\newcommand{\LL}{\mathcal{L}}\newcommand{\pa}{\partial}\newcommand{\AAA}{\mathscr{A}}\newcommand{\BBB}{\mathscr{B}}\newcommand{\CCC}{\mathscr{C}}\newcommand{\DDD}{\mathscr{D}}\newcommand{\EEE}{\mathscr{E}}\newcommand{\FFF}{\mathscr{F}}\newcommand{\WFF}{\widetilde{\FFF}}\newcommand{\GGG}{\mathscr{G}}\newcommand{\HHH}{\mathscr{H}}\newcommand{\PPP}{\mathscr{P}}\newcommand{\Ff}{\mathcal{F}}\newcommand{\Gg}{\mathcal{G}}\newcommand{\Hh}{\mathbb{H}}\DeclareMathOperator{\ess}{ess}\newcommand{\CC}{\mathbb C}\newcommand{\FF}{\mathbb F}\newcommand{\NN}{\mathbb N}\newcommand{\QQ}{\mathbb Q}\newcommand{\RR}{\mathbb R}\newcommand{\ZZ}{\mathbb Z}\newcommand{\KK}{\mathbb K}\newcommand{\SSS}{\mathbb S}\newcommand{\II}{\mathbb I}\newcommand{\conj}[1]{\overline{#1}}\DeclareMathOperator{\cis}{cis}\newcommand{\abs}[1]{\left\lvert #1 \right\rvert}\newcommand{\norm}[1]{\left\lVert #1 \right\rVert}\newcommand{\floor}[1]{\left\lfloor #1 \right\rfloor}\newcommand{\ceil}[1]{\left\lceil #1 \right\rceil}\DeclareMathOperator*{\range}{range}\DeclareMathOperator*{\nul}{null}\DeclareMathOperator*{\Tr}{Tr}\DeclareMathOperator*{\tr}{Tr}\newcommand{\id}{1\!\!1}\newcommand{\Id}{1\!\!1}\newcommand{\der}{\ \mathrm {d}}\newcommand{\Zc}[1]{\ZZ / #1 \ZZ}\newcommand{\Zm}[1]{\left(\ZZ / #1 \ZZ\right)^\times}\DeclareMathOperator{\Hom}{Hom}\DeclareMathOperator{\End}{End}\newcommand{\GL}{\mathbb{GL}}\newcommand{\SL}{\mathbb{SL}}\newcommand{\SO}{\mathbb{SO}}\newcommand{\OO}{\mathbb{O}}\newcommand{\SU}{\mathbb{SU}}\newcommand{\U}{\mathbb{U}}\newcommand{\Spin}{\mathrm{Spin}}\newcommand{\Cl}{\mathrm{Cl}}\newcommand{\gr}{\mathrm{gr}}\newcommand{\gl}{\mathfrak{gl}}\newcommand{\sl}{\mathfrak{sl}}\newcommand{\so}{\mathfrak{so}}\newcommand{\su}{\mathfrak{su}}\newcommand{\sp}{\mathfrak{sp}}\newcommand{\uu}{\mathfrak{u}}\newcommand{\fg}{\mathfrak{g}}\newcommand{\hh}{\mathfrak{h}}\DeclareMathOperator{\Ad}{Ad}\DeclareMathOperator{\ad}{ad}\DeclareMathOperator{\Rad}{Rad}\DeclareMathOperator{\im}{im}\renewcommand{\BB}{\mathcal{B}}\newcommand{\HH}{\mathcal{H}}\DeclareMathOperator{\Lie}{Lie}\DeclareMathOperator{\Mat}{Mat}\DeclareMathOperator{\span}{span}\DeclareMathOperator{\proj}{proj}$ Suppose you know a lot about commutative algebra and linear algebra but know absolutely nothing about the theory of fields. What's the fastest way to get up to speed? Well, I'm in this situation. Cook! Basically, you care a lot about field extensions and polynomials. If $K / F$, then $K$ is an $F$-vector space with dimension called the ==**degree**== $[K:F]$. We usually care about the case where this degree is finite. In this case, $F\subset K\subset L$ implies $[L: F] = [L:K][K:F]$ trivially. $\alpha\in K$ is ==**algebraic over $F$**== if the substitution ring homomorphism $\varphi: F[x]\to K$ is not injective; else it is ==**transcendental**==. A lot of the time if we don't care about $K$, we will just say that $\alpha$ is ==**algebraic over $F$**== and leave $K$ implicit -- $\alpha$ is just some operator over a $F$-vector space which also has field structure. Lol! If you're transcendental, then you create the field of rational functions $F(x)$ so there isn't more to say. If you're algebraic, then since $F[x]$ is a PID (by Euclidean algorithm) we obtain that the kernel of $\varphi$ is $\braket{f}$ for some monic $f$ called the ==**irreducible polynomial for $\alpha$ over $F$**==. $\alpha$ acts on $F[\alpha]$ as a $F$-vector space, and this $f$ is also the minimal polynomial. A basis is $1, \alpha, \dots, \alpha^{n-1}$. $F(\alpha) = F[\alpha]$, and $[F(\alpha) : F] = n$. Hence, finite field extensions are generated by finitely many algebraic elements, and the set of algebraic elements over $F\subset K$ forms a subfield. >[!idea] Deep idea >Here's a statement: **All of the roots of an irreducible polynomial over $F$ which splits in $K$ are symmetric**. This is not quite true. However, it is true pointwise; the symmetry is *transitive*; it may just not be all of $S_n$. Given algebraic $\alpha,\beta$, $F(\alpha)$ and $F(\beta)$ are isomorphic field extensions iff $\alpha$ and $\beta$ have the same irreducible polynomials. We want to re-iterate the PID property of $F[x]$, which will be used endlessly to prove basic theorems; if $(x - \alpha) \mid g\in F[x]$, then $f\mid g$. Every field has a characteristic $p$ which is either $0$ or a prime number; there are no homomorphisms between these fields, hence the category of fields fractures into separate universes. There is a unique minimal field (initial element) for each $p$, called the prime subfields: $\QQ$ and $\FF_p$. Thus the "absolute theory of fields" (idk what this means, maybe just fields by themselves without context) is reduce to studying field **extensions**. Given $f\in F[x]$, we can adjoin a root via $K = F[x] / (f)$. This guy is a field iff $f$ is irreducible, so let's make sure that's the case. Then the image of $x$ is a factor of $f$. One can repeat until $f$ splits over $K$. >[!idea] Small pitfall. >Consider like $x^3 - 2$; you can have $\QQ[\sqrt[3]{2}]$, thus having just a single element doesn't mean the whole thing splits. Finally, there's a silly trick: given $f\in F[x]$, the derivative $f'$ makes sense formally, and can recognize multiple roots. Notably, if $f$ is **irreducible**, then $f$ has multiple roots in $K/F$ iff $f'$ is the zero polynomial. ### Extended Example: Finite Fields These are kind of cool. Our prime subfield is $F = \FF_p$, and $K$ is some $r$-dimensional vector space. Hence the order of the field is $q = p^r$. Lagrange says the elements of $K$ are the roots of $x^q - x$ (they comprise $q$ distinct roots!) The structure theorem for abelian groups says that $K^\times$ has to be cyclic, else everything is a root of some $x^\ell - x$ for $\ell < q$ which kills you. We now turn to the question of existence, by letting $L$ be minimal such that $x^q - x$ splits over $L$. The derivative is $-1$, hence there are no multiple roots anywhere, and whenever working mod $p$ we have $(x+y)^q = x^q + y^q$. Thus we're done. All such fields are isomorphic; pick any generator $\alpha$ of $K^\times$, then $K = F(\alpha)$ with some minimal polynomial $f$ of degree $r$. Of course, $f | x^q -x$, hence any order-$q$ finite field $K'$ also has some $\alpha'\in K'$ satisfying $f(\alpha') = 0$. By dimension-counting $F[\alpha'] = K'$, thus the map $K\to K'$ via $\alpha\to \alpha'$ is an isomorphism (through $F[x] / (f)$). By dimensionality, if $\FF_q \subset\FF_{q'}$ then $r\mid r'$. Conversely, if $r' = rs$, then $x^q - 1 \mid x^{q'} - 1$ thus $\FF_q\subset \FF_{q'}$ as desired. This enumerates all possible pairs of finite extensions of finite fields. Notably, $K/F$ admits a primitive element; any $\FF_{q'}$ is of the form $\FF_q(\alpha)$ where $\alpha$ is simply a multiplicative generator of $\FF_{q'}^\times$. Notably, the irreducible polynomials in $F[x]$ with degree dividing $r$ are in bijection with the irreducible factors of $x^q - x$ over $\FF_p$, that is, the irreducible polynomials over $\FF_p$ which split completely over $\FF_q$. To summarize: >[!theorem] Classification of Finite Fields >All finite fields look like >$ >\prod_{y\in \FF_q} (x - y) = x^q - x,\qquad q = p^r >$ >up to isomorphism. ### Extended Example: Characteristic Zero **Primitive Elements of Finite Extensions:** Let $F$ have char $0$. Then any extension $K$ has a primitive element. The proof reduces to the case where $K = F(\alpha,\beta)$ is generated by two guys. Your intuition here should be that for $c\in F$, $\alpha + c\beta$ is a primitive element unless $c$ is fine-tuned; e.g. in $\QQ(\sqrt{2} + \sqrt{3})$ I should be able to take basically any $\sqrt{2} + c\sqrt{3}$ and generate everything. All I have to do is create $\beta$ inside $F(\alpha + c\beta)$. It's really hard to do this "by hand;" instead we use number theory. This is some of the ingenuity they were talking about I guess. Start with the minimal polynomials $f(x) = (x-\alpha)p(x)$ and $g(x) = (x-\beta)q(x)$ for $\alpha,\beta$; these guys have coefficients in $F$. Then, consider the polynomial $h(x) = f((\alpha + c\beta) - cx)$. Crucially: 1. This is another polynomial with root $\beta$. 2. This polynomial lives in $F(\alpha + c\beta)[x]$. 3. Almost surely, $(g, h) = (x - \beta)$ exactly! This is easiest to show by working in some massive field extension $L$ where $f,g$ split completely into $\prod(x - \alpha_i)$ and $\prod (x - \beta_i)$; then one simply observes that $\frac{\alpha - \alpha_i}{c} + \beta$ and $\beta_i$ collide for exactly one value of $c$. What does this mean? Well, for cofinitely many $c$, I can simply write down $h(x)\in F(\alpha + c\beta)[x]$, $g(x)\in F(\alpha + c\beta)[x]$, and perform arithmetic all inside $F(\alpha + c\beta)$ to reduce the expression to $(x - \alpha)$. This is exactly the desired. # Galois Theory Crash-course ### What is a Galois Extension? Henceforth, $\char F = 0$. An isomorphism/automorphism of finite field extensions restricts to the identity on the base field; the Galois group is the set of automorphisms of $K/F$. A ==**Galois extension**== is one such that the degree of the extension equals the order of the group $G(K/F)$. This is the simplest characterization of a Galois extension as a "maximally symmetric" field extension; we will exhibit two other natural equivalent notions. >[!idea] Take the minimal polynomial $f$ of a primitive element $\gamma$. $f$ has roots $\gamma_i\in K$ (it won't split completely unless $K$ is Galois). $f$ is preserved by any automorphism hence the set $\{\gamma_i\}$ is preserved under $G(K/L)$. To each $\gamma_i$, there is a unique automorphism sending $\gamma\to \gamma_i$. Hence, you can just "follow the roots of $f$." >[!Claim] Any two splitting fields of $f$ over $F$ are isomorphic. Indeed, consider $K_1,K_2$. Then $K_1/F$ has primitive element $\gamma$ with minimal polynomial $g\in F[x]$. One then considers a splitting field for $g$ over $K_2$, which we call $L$. Then, pick any root $\gamma'\in L$ of $g$, and $K = F(\gamma)$ will be isomorphic to $F(\gamma')$ via $\gamma\to \gamma'$. There can only be one splitting field for $g$ over $F$ inside $L$, hence $K_2 = F(\gamma')$. ### Splitting Fields A **splitting field** is a field extension like $\QQ[\cbrt{2}, \cbrt{2}\omega, \cbrt{2}\omega^2]$, as opposed to a generic extension like $\QQ[\cbrt{2}]$. This introduces some Galois-theory flavors, but actually everything here works even if $F, K$ are not characteristic $0$ (I never use the existence of a primitive element). >[!definition] Splitting Field >$K/F$ is a ==**splitting field**== if, for all irreducible $g\in F[x]$ which admit a root $\beta\in K$, $g$ completely factorizes ("splits") over $K$. > [!idea]- Pedantic, an obvious trick. > Note, by the way, that this construction can be reversed; it is equivalent to the following. $K/F$ is a splitting field iff for all $\beta \in K$, the minimal polynomial of $\beta$ splits over $K$. It was written in the above (cumbersome) manner because this is how we like to think of splitting fields; given a polynomial, if it touches $K$ then it splits apart. > > After all, for any irreducible $g\in F[x]$ such that $g(\beta) = 0$, $g$ is by definition a minimal polynomial for $\beta$. Conversely, any minimal polynomial for $\beta$ is irreducible in $F[x]$. >[!theorem] Splitting Field Characterization >$K/F$ is a splitting field iff there exists a $f\in F[x]$ which splits over $K$ into $\prod (x - \alpha_i)$ such that $F[\alpha_1,\dots, \alpha_n] = K$. $\implies$ is clear: one starts by picking any $\alpha\in K\setminus F$, writing down its minimal polynomial $f_1$ over $F$, and then splitting it, yielding some larger $L = F[\alpha, \dots]$. Then, pick some $\beta\in K\setminus L$, write down its minimal polynomial $f_2$ over $F$ (again!), and split it, etc. For degree reasons, this process terminates. $f_1f_2\dots\in F[x]$ splits over $K$ as desired, and $\alpha,\beta\dots$ generate $K$. $\impliedby$ is a fun exercise. Write $\beta = p(\alpha_1,\dots, \alpha_n)$ where $p\in F[x_1,\dots]$. Then let $h = \prod_\sigma \left(x - p(\alpha_{\sigma(i)\dots})\right)$. By symmetry, all coefficients of $h$ are symmetric polynomials in $\alpha$, hence $h\in F[x]$. Then $\gcd(g,h)$ is nonzero thus $g\mid h$ and we conclude. > [!idea] Galois Extension $\iff$ Splitting Field > Indeed, if you're a splitting field, then characteristic of primitive has degree $[K:F]$, all of whose roots exist, and we also know it's $G(K/F)$. > Conversely, if you're Galois, then the characteristic of the primitive has exactly $[K:F] = G(K/F)$ roots hence the primitive polynomial splits. But then you're exactly the splitting field for $f$. ### Fixed Fields Let $K$ be a field, let $H$ be a finite subgroup of the field automorphisms of $K$. $F = K^H$ is the stabilizer or ==**fixed field**== of $H$. Clearly $H\subset G(K/F)$. Given any $\beta \in K$, we obtain a whole orbit $\{\sigma(\beta)\}$ of symmetric terms; then clearly $\prod (x - \sigma(\beta))$ lies in $F[x]$ by symmetry and is irreducible; hence $K$ is algebraic. All elements of $K$ have order dividing $\abs{H}$ hence $[K:F]$ is finite, but that means I can find a primitive element thus $[K:F] = \abs{H}$. >[!proof]- Huh >Well look, it's symmetric. Then, if you claim to have a factor $\prod_{\sigma \in \tilde{H}} (x - \sigma(\beta))$, note that by definition of orbit I can find a $h\in H$ which changes $\sigma$, but this is supposed to preserve the polynomial so you die. Using this, let's look at an arbitrary fresh finite extension $K/F$. $F\subset K^{G(K/F)}$ but this might be strict, hence we obtain that $G(K/F) \mid [K:F]$. Turning right around, note that $H\subset G(K/F)$ hence $H = G(K/F)$ and $K/F$ is Galois in the fixed-field case. Hence: > [!idea] Fixed fields $\iff$ Galois Extensions > $F = K^{G(K/F)}$ iff $\abs{G(K/F)} = [K : F]$. ### Intermission: Reminders on the progress. So, again: Galois Extensions have the following overpowered properties: - $\iff$ $K$ is the splitting field of any $f$ over $F$. (This is the most useful way to show something is a splitting field!) - $\iff$ *All* irreducible $f\in F[x]$ which touch $K$ split over $K$. - $\iff$ $[K:F] = \abs{G(K / F)}$ - $\iff$ $F = K^{G(K/F)}$. (This is the most useful property of a splitting field!) - $\implies$ The characteristic polynomial of any $\beta\in K$ is exactly $\prod_{\sigma\in G(K/F)} (x - \sigma(\beta))$. - Use the trick: take a primitive $\gamma_1$ and track its roots $\{\gamma_i\}$. $G(K/F)$ is completely characterized by where you send $\gamma_1$. Now, suppose I'm looking at some bum-ahh field $\QQ[\cbrt{2}]$ and wishing I was a Galois extension. Have no fear! Any finite extension lies in a splitting field, because you can just take the minimal polynomial your primitive and split it. Similarly, suppose I'm looking at some $F\subset L\subset K$, and I know that $K/F$ is Galois. Then $K/L$ is also Galois, because $K$ is the splitting field for $L$ over the exact same polynomial for $F$, $f\subset F[x]\subset L[x]$. Final preparation: suppose I've got arbitrary finite extensions $F\subset L\subset K$ and pick primitive $e$ for $L$ over $F$ plus primitive $a$ for $K$ over $L$. If reflection $e_i\in L$, there's a unique automorphism $\psi$ for $L/F$ sending $e\to e_i$. The question is, can this guy extend to an automorphism for $K/F$? The answer is yes, and I call it polynomial bootstrapping; any guy in $K$ can be written as a polynomial $L[a]$, and all I'm going to do is move all the coefficients on $L$. Food for thought (this will be clarified in the Galois theorem!!): this extension is un-natural, because our choice of $a$ mattered, and of course there are other automorphisms which simply don't look like a polynomial automorphism (not all elements of $K/F$ preserve $L$ as a subspace!) ### Main Galois Theorem We're ready for the finale (of this crash-course). Fix some Galois extension $K/F$, and I will completely characterize all intermediate fields $F\subset L\subset K$. >[!theorem] >The intermediate fields $L$ are in bijection with subgroups of $G(K/F)$ via $L\leftrightarrow G(K/L)$. Indeed, any $L$ maps to such a subgroup, and any subgroup maps to such an $L$ (via the fixed field). The fixed field of $G(K/L)$ is exactly $L$, and the Galois group of $K^H$ is exactly $H$. Notable corollary: there are finitely many intermediate fields! But here's the really cool part. You'll notice that in the main theorem, we didn't really use $F$ at all; it just sort of exists to keep everything finite and grounded. >[!theorem] >$L/F$ is also Galois iff $H\trianglelefteq G$, and if so, $G(L/F) \cong G(K/F) / G(K/L)$! This one is *really* cool, but also reminiscent of [[Regular Covering Spaces]]. There should be an algebraic geometric reason why these are the same, and I hope to learn it by the end of this summer! (Narrator: did this mf finish? time will tell 🤡). Let's deploy some mf trackers: $e$ primitive on $L/F$ such that $e_1,\dots,e_n\in K$ with irreducible polynomial $f$. For any $\psi\in G$, the stabilizer of $\psi(e)$ is exactly $\psi H \psi^{-1}$. Thus, $\psi(e) \in L = K^H$ iff $\psi H \psi^{-1} = H$. $G$ is transitive because $e\to e_i$ works in $L$, and we can polystrap to $K$, thus $L/F$ is Galois iff $\{e_1,\dots, e_n\}\in L$ iff $\psi H\psi^{-1} = H$ for all $\psi\in G$ iff $H\trianglelefteq G$. By polystrap, $G(L/F)G(K/L)\subset G(K/F)$. By size reasons which apply iff $L/F$ Galois, we have equality, and by staring at the polystrap, this is semi-direct. Hence we conclude. Fun places to go after this: - [[Kronecker-Weber Theorem]] - [[Kummer Extensions]] # Function Fields - Intro Algebraic Geometry ### Plane Algebraic Curves A ==**plane algebraic curve**== or ==**riemann surface**== is the the locus $X$ of zeros in $\CC^2$ (more correctly, $P^2$ where $P$ is the projective complex plane) of some polynomial $f(t,x)$. This is a natural object of study even outside algebraic geometry, and you already have immense intuition about it; whenever you discuss a branch cut of $\sqrt{x}$ or $\log(x)$, you're visualizing the entire manifold $t^2 - x = 0$ and $e^t - x = 0$. It is a plane, because as a topological space it has dimension $4 - 2 = 2$. Obviously, the projection $\pi:X\to T$ should induce an $n$-sheeted covering space, modulo a finite number of numerical singularities; this is called a ==**branched covering**==, and the minimal set of points you must omit are called ==**branch points**==. From now on, this notion of cofinite equivalence classes is implicit. >[!theorem] >Let $f(t,x)$ be irreducible in $\CC[t,x]$, with degree $n > 0$ in $x$. The Riemann surface of $f$ is an $n$-sheeted branched covering of $P$ via projection onto $t$. The topological part follows "obviously from smoothness" assuming the finiteness of the pre-image is obtained. Fix some $t_0$; we want to show $f(t_0, x)$ probably has $n$ roots. This guy is $a_n(t)x^n+\dots +a_0(t)$ and $a_n(t_0)x^n+\dots+a_0(t_0)$ has at most $n$ roots. - If the degree of $f(t_0, x)$ is less than $n$, then $a_n(t_0)$ vanished so this occurs almost nowhere. - If $f(t_0, x)$ has a multiple root, we can start waffling slightly. This occurs in $(t,x)$ space at the common zeros of $f$ and $\frac{\pa f}{\pa x}$. $f$ is assumed irreducible, so this should follow from a weak form of ==**Bezout's theorem**==, proved here: >[!claim] Baby Bezout >Two nonzero polynomials $f,g$, relatively prime in $\CC[t,x]$, have only finitely many common roots in $P^2$. >[!idea]- >The main idea of this argument is that $\CC[t,x]$ is good because things are literal multivariable polynomials, but a bit sad because its not a PID, so being relatively prime does not give you an algebraic equality. So you drop to the Euclidean $\CC(t)[x]$, clear denominators, and cite Gauss' lemma. >[!proof]- Gauss lift $\CC(t)[x]$ > Let $A = \CC[t]$ for clarity. Pick out primitives $f\to af'$, $g\to bg'$ where $(a,b) = 1$. Then, over $F[x]$ which is Euclidean, $f',g'$ have a (wlog primitive) GCD which we'll call $h$; this $h$ must be $1$. Hence, $1 = rf' + sg'$ for some $r,s\in F$. Hence $ab = rbf + sgh$, and we can clear all the denominators to get$u(t) = r_1(t,x)f(t,x) + s_1(t,x)g(t,x).$Hence, if $(t_0, x_0)$ is a root, then $t_0$ is a root of $u$. This occurs for finitely many $t_0$. The entire argument is symmetric in $t,x$, hence we conclude. [[Basic Number Theory Blitz]] > [!idea]- Finishing the proof: "obviously from smoothness" > Well, the derivative is zero, implicit function theorem it.