# General Field Theory

The most useful tricks of the trade:

• $${\sharp}{\mathbf{G}}_m({\mathbf{GF}}(p^k)) = p^k-1$$, since every element is invertible except 0. You can use this to cook up strong numerical constraints on orders of elements. E.g. if $$a^{17}=1$$ in some finite field of size $$p^k$$, $$o(a)$$ divides 17 and $$o(a)$$ divides $$p^{k}-1$$, so $$o(a)$$ divides $$\gcd(17, p^{k}-1)$$.
• Multiplicativity in towers can force numerical divisibility constraints. E.g. if $$\alpha$$ is a root of any irreducible $$f$$, take the tower $$\operatorname{SF}(\alpha, k)/k(\alpha)/k$$: then the degree of $$\min_{\alpha, k}(x)\in k[x]$$ divides the degree of the extension $$[\operatorname{SF}(\alpha, k) : k]$$.

## Basics: Polynomials

For $${ \mathbf{F} }$$ a field, a polynomial $$f\in { \mathbf{F} }[x]$$ is reducible if and only if $$f$$ can be factored as $$f(x) = g(x) h(x)$$ for some $$g, h\in { \mathbf{F} }[x]$$ with $$\deg g, \deg h \geq 1$$ (so $$g, h$$ are nonconstant). $$f$$ is irreducible if $$f$$ is not reducible.

For $$R$$ a UFD, a polynomial $$p\in R[x]$$ is primitive iff the greatest common divisors of its coefficients is a unit.

Let $$R$$ be a UFD and $$F$$ its field of fractions. Then a primitive $$p\in R[x]$$ (so e.g. $$p$$ monic) is irreducible in $$R[x] \iff p$$ is irreducible in $$F[x]$$.

More precisely, if $$p = AB$$ is reducible in $$F[x]$$, then there exist $$r,s\in F$$ such that $$rA, sB\in R[x]$$ and $$p = (rA)(sB)$$ is a factorization in $$R[x]$$.

A primitive polynomial $$p\in {\mathbf{Q}}[x]$$ is irreducible $$\iff p$$ is irreducible in $${\mathbf{Z}}[x]$$.

## Definitions

The characteristic of a ring $$R$$ is the smallest integer $$p$$ such that $$\sum_{k=1}^p 1 = 0$$.

If $$\operatorname{ch}k = p$$ then $$(a+b)^p = a^p + b^p$$ and $$(ab)^p = a^p b^p$$.

For $$H \leq \mathop{\mathrm{Aut}}_{{ \mathsf{Fields}_{k}}}(L)$$, \begin{align*} L^H \coloneqq\left\{{ \ell \in L {~\mathrel{\Big\vert}~}\sigma(l) = \ell }\right\} .\end{align*}

The prime subfield of a field $$F$$ is the subfield generated by $$1$$.

The prime subfield of any field is isomorphic to either $${\mathbf{Q}}$$ or $${ \mathbf{F} }_p$$ for some $$p$$.

\begin{align*} \mathop{\mathrm{Aut}}(L/k) = \left\{{\sigma: L \to L {~\mathrel{\Big\vert}~} {\left.{{\sigma}} \right|_{{k}} } = \operatorname{id}_k}\right\} .\end{align*}

Let $$k$$ denote a field, and $$L/k$$ extension. Every field morphism is an embedding (injection). An embedding of $$k{\hbox{-}}$$algebras $$L\hookrightarrow L'$$ will refer to any ring morphism over $$k$$, i.e. a field morphism that restricts to the identity on $$k$$: More generally, we can ask for lifts of any map $$\sigma: k\to k'$$: Most often, we’ll take $$\sigma: k\to k$$ to be the identity.

The following are equivalent:

• $$k$$ is a perfect field.

• If $$\operatorname{ch}k > 0$$, the Frobenius is an automorphism of $$k$$, so $$k^p = k$$.

• Every finite extension $$F/k$$ is separable.

• Every irreducible polynomial $$p\in k[x]$$ is separable.

Example of a non-perfect field: $${ \mathbf{F} }_p(t)$$. Use that $$f(x) \coloneqq x^p - t$$ is irreducible in $${ \mathbf{F} }_p(t)[x]$$ but not separable.

$$k$$ is perfect (using the irreducible implies separable condition) if either

• $$\operatorname{ch}k = 0$$ or
• $$\operatorname{ch}k = p > 0$$ and $$k^p = k$$.

For $$\operatorname{ch}k = 0$$, use that irreducible implies separable.

For $$\operatorname{ch}k = p$$, show that $$k^p\neq k \iff$$ irreducible does not imply separable, so there exists an inseparable irreducible.

• Supposing $$k^p\neq k$$, choose $$a\in k$$ not a $$p$$th power.

• Note that $$f(x) \coloneqq x^p-a$$ has only one root in $$\overline{k}$$: in a splitting field, any root $$r$$ satisfies $$r^p=a$$, so \begin{align*} x^p - a = x^p - r^p = (x-r)^p .\end{align*}

• Note $$f$$ is irreducible: its only possible divisors are $$(x-r)^m$$ for $$m \leq p$$. Expanding yields \begin{align*} (x-r)^m = \sum_{k=0}^m {m\choose k} x^{m-k} (-r)^{k} = x^m + {m\choose 1} x^{m-1} (-r)^m + \cdots ,\end{align*} so the coefficient of $$x^{m-1}$$ is $$-mr \in k$$.

• Thus if $$(x-r)^m$$ has a nontrivial divisor in $$k[x]$$ then $$m$$ must be in $$k^{\times}$$, forcing $$r\in k$$. But then $$r^p = a\in k$$, $$\contradiction$$.

Let $$K/k$$ be an extension.

\begin{align*} [K: k] = \dim_{{ \mathsf{Vect}}_k} K \end{align*} is the dimension of $$K$$ as a $$k{\hbox{-}}$$vector space. Automorphisms of fields over $$K$$ are defined as

\begin{align*} \mathop{\mathrm{Aut}}_{{ \mathsf{Fields}_{k}}}(K) \coloneqq\mathop{\mathrm{Aut}}(K/k) \coloneqq\left\{{ \sigma: K \to K' {~\mathrel{\Big\vert}~}{ \left.{{\sigma}} \right|_{{k}} } = \operatorname{id}_k }\right\} , \end{align*} so lifts of the identity on $$k$$, and \begin{align*} \left\{{K:k}\right\} \coloneqq{\sharp}\mathop{\mathrm{Aut}}(K/k) .\end{align*}

If $$K/k$$ is finite, normal, and separable, \begin{align*} { \mathsf{Gal}}(K/k) \coloneqq\mathop{\mathrm{Aut}}(K/k) ,\end{align*} where in general \begin{align*} \left\{{K: k}\right\} \leq [K: k] \end{align*} with equality when $$L/k$$ is Galois.

• All fields are simple rings (no proper nontrivial ideals).
• Thus every field morphism is either zero or injective.
• The characteristic of any field $$k$$ is either 0 or $$p$$ a prime.
• If $$L/k$$ is algebraic, then $$\min(\alpha, L)$$ divides $$\min(\alpha, k)$$.

Let $$L/F/k$$ be a finite tower of field extensions. \begin{align*} [L : k] = [L: F][F: k] .\end{align*}

## Finite Fields

$${\mathbf{GF}}(p^n)\cong \frac{{ \mathbf{F} }_p}{(f)}$$ where $$f \in { \mathbf{F} }_p[x]$$ is any irreducible of degree $$n$$, and $${\mathbf{GF}}(p^n) \cong { \mathbf{F} }[\alpha] \cong \mathop{\mathrm{span}}_{ \mathbf{F} }\left\{{1, \alpha, \cdots, \alpha^{n-1}}\right\}$$ for any root $$\alpha$$ of $$f$$.

Every finite field $$F$$ is isomorphic to a unique field of the form $${\mathbf{GF}}(p^n)$$ and if $$\operatorname{ch}F = p$$, it has prime subfield $${ \mathbf{F} }_p$$.

$${\mathbf{GF}}(p^\ell) \leq {\mathbf{GF}}(p^k) \iff \ell$$ divides $$k$$.

$${\mathbf{GF}}(p^n)$$ is the splitting field of $$\rho(x) = x^{p^n} - x$$, and the elements are exactly the roots of $$\rho$$.

Every element is a root by Cauchy’s theorem, and the $$p^n$$ roots are distinct since its derivative is identically $$-1$$.

\todo[inline]{todo}


Let $$\rho_n \coloneqq x^{p^n} - x$$. Then $$f(x) \divides \rho_n(x) \iff \deg f \divides n$$ and $$f$$ is irreducible.

$$x^{p^n} - x = \prod f_i(x)$$ over all irreducible monic $$f_i \in { \mathbf{F} }_p[x]$$ of degree $$d$$ dividing $$n$$.

$$\impliedby$$:

• Suppose $$f$$ is irreducible of degree $$d$$.
• Then $$f \divides x^{p^d} - x$$, by considering $$F[x]/\left\langle{f}\right\rangle$$.
• Thus $$x^{p^d} - x \divides x^{p^n} - x \iff d\divides n$$.

$$\implies$$:

• $$\alpha \in {\mathbf{GF}}(p^n) \iff \alpha^{p^n} - \alpha = 0$$, so every element is a root of $$\phi_n$$ and $$\deg \min(\alpha, { \mathbf{F} }_p) \divides n$$ since $${ \mathbf{F} }_p(\alpha)$$ is an intermediate extension.

• So if $$f$$ is an irreducible factor of $$\phi_n$$, $$f$$ is the minimal polynomial of some root $$\alpha$$ of $$\phi_n$$, so $$\deg f \divides n$$.

• $$\phi_n'(x) = p^nx^{p^{n-1}} \neq 0$$, so $$\phi_n$$ is squarefree and thus has no repeated factors. So $$\phi_n$$ is the product of all such irreducible $$f$$.

If $${ \mathbf{F} }$$ is a finite field then $$F\neq \overline{F}$$.

If $$k = \left\{{ a_1, a_2, \cdots a_n }\right\}$$ then define the polynomial \begin{align*} f(x) \coloneqq 1 +\prod_{j=1}^n (x-a_j) \in k[x] .\end{align*} This has no roots in $$k$$.

## Cyclotomic Polynomials

\begin{align*} \phi(n) \coloneqq{\sharp}\left\{{ k\leq n {~\mathrel{\Big\vert}~}\gcd(k ,n ) = 1}\right\} .\end{align*}

• $$\phi(p) = p-1$$, because every number $$k\leq p-1$$ is coprime to $$p$$.
• $$\phi(p^k) = p^{k} - p^{k-1}$$, since there are $$p^k$$ total numbers less than $$p^k$$, most of which are coprime to $$p$$. The ones to remove are those dividing $$p^k$$: the only divisors of $$p^k$$ are $$p^\ell$$ for $$0\leq \ell \leq k$$, and $$\gcd(p^k, m) = p^\ell$$ whenever $$m=tp$$ for $$t = 1,2,3,\cdots,p^{k-1}$$ (i.e. $$m$$ is divisible by some power of $$p$$, so the $$p^{k-1}$$ multiples of $$p$$ are possible).
• $$\phi$$ is multiplicative (arithmetically, so only on prime powers!)

\begin{align*} \phi(1) &= 1 \\ \phi(2) &= 1 \\ \phi(3) &= 2 \\ \phi(4) &= 2 \\ \phi(6) &= 2 \\ \phi(8) &= 4 \\ .\end{align*}

Let $$\zeta_n = e^{2\pi i/n}$$, then the $$n$$th cyclotomic polynomial is given by \begin{align*} \Phi_{n}(x)=\prod_{k=1 \atop (j, n)=1}^{n}\left(x- \zeta_n^k\right) \in {\mathbf{Z}}[x] ,\end{align*}

which is a product over primitive roots of unity. It is the unique irreducible polynomial which is a divisor of $$x^n - 1$$ but not a divisor of $$x^k-1$$ for any $$k<n$$.

Note that $$\deg \Phi_n(x) = \phi(n)$$ for $$\phi$$ the totient function.

Any subfield of $$\operatorname{SF}(x^n-1)$$ is a cyclotomic field.

Computing $$\Phi_n$$:

• \begin{align*} \Phi_{n}(z)=\prod_{\substack{ d \divides n \\ d > 0} }\left(z^{d}-1\right)^{\mu\left(\frac{n}{d}\right)} \end{align*} where \begin{align*} \mu(n) \equiv\left\{ \begin{array}{ll}{0} & {\text { if } n \text { has one or more repeated prime factors }} \\ {1} & {\text { if } n=1} \\ {(-1)^{k}} & {\text { if } n \text { is a product of } k \text { distinct primes, }}\end{array}\right. \end{align*}

• \begin{align*} x^{n}-1=\prod_{d | n} \Phi_{d}(x) \implies \Phi_n(x) = \qty{x^n-1} \qty{\prod_{d | n \atop d < n} \Phi_{d}(x)}^{-1}, \end{align*} so just use polynomial long division.

\begin{align*} \Phi_{p}(x) &= x^{p-1}+x^{p-2}+\cdots+x+1 \\ \Phi_{2 p}(x) &= x^{p-1}-x^{p-2}+\cdots-x+1 \\ \\ k\divides n \implies \Phi_{n}(x) &= \Phi_{n\over k}\left(x^{k}\right) \\ \\ \Phi_1(z) &= z-1 \\ \Phi_2(z) &= z+1 \\ \Phi_4(z) &= z^2+1 \\ \Phi_6(z) &= z^2 -z + 1 \\ \Phi_8(z) &= z^4+1 .\end{align*}

The splitting field of $$x^m-1$$ is $${\mathbf{Q}}(\zeta_m)$$ for $$\zeta_m$$ any primitive root of unity, and \begin{align*} { \mathsf{Gal}}({\mathbf{Q}}(\zeta_m)_{/{\mathbf{Q}}}) \cong ({\mathbf{Z}}/m{\mathbf{Z}})^{\times} .\end{align*}

If $$K_{/{\mathbf{Q}}}$$ is an abelian extension, then $$K \subseteq {\mathbf{Q}}(\zeta_m)$$ for some $$m$$.

## Misc

\todo[inline]{todo}


## Exercises

If $$f\in k[x]^{{\mathrm{irr}}}$$ with $$\operatorname{ch}k = p$$, then there is a unique separable $$g\in k[x]^{{\mathrm{irr}}}$$ such that $$f(x) = g(x^{p^k})$$ for some unique $$k$$.

Show \begin{align*} x^\ell - 1 \divides x^m-1 \iff \ell\divides m .\end{align*}

$$\implies$$

• Write $$m = \ell q + r$$ with $$0\leq r < \ell$$.
• Write \begin{align*} p(x) = {x^m-1 \over x^\ell - 1} = {x^{lq+r} -1 \over x^\ell - 1} = x^r{x^{lq} - 1 \over x^\ell - 1} + {x^r - 1 \over x^\ell - 1} = q(x) + {x^r-1 \over x^\ell - 1} ,\end{align*} where $$p,q$$ are polynomial by divisibility.
• So the remaining ratio must be polynomial, but since $$r<\ell$$ is strict this forces $$r=0$$. Thus $$\ell \divides m$$.
\todo[inline]{I don't like this proof!}


$$\impliedby$$:

• Write $$m = \ell q + r$$, then $$r=0$$ by divisibility.
• Then $$x^m-1 = x^{\ell q} - 1 \coloneqq z^q-1$$ where $$z\coloneqq x^\ell$$.
• Use that $$z-1 \divides z^q - 1$$, so $$x^{\ell}-1 \divides x{\ell q} -1 = x^m-1$$.

Show that if $$f \in { \mathbf{F} }_p[x]^{{\mathrm{irr}}}$$ is degree $$d$$, \begin{align*} f \divides x^{p^n}-x \iff d\divides n .\end{align*}

$$\impliedby$$:

• If $$d\divides n$$, $$x^d-1 \divides x^n-1$$ by a previous exercise, and so $$p^d-1 \divides p^n-1$$.
• So $$x^{p^d-1} \divides x^{p^n-1}$$, now multiply through by $$x$$.
• Claim: $$f \divides x^{p^d-1}$$, from which the result immediately follows.
• For $$\alpha$$ any root of $$f$$, $${ \mathbf{F} }_p(\alpha)$$ is a finite field of size $$p^d$$ since $$[{ \mathbf{F} }_p(\alpha):{ \mathbf{F} }_p] = d$$.
• So $${ \mathbf{F} }_p(\alpha)\cong {\mathbf{GF}}(p^d)$$, which is the splitting field of $$x^{p^d}-x$$.
• Thus $$\alpha$$ is a root of $$x^{p^d}-x$$. Iterating over all roots yields the divisibility statement.

$$\implies$$:

• If $$f\divides g_n(x) \coloneqq x^{p^n}-x$$, then every root $$\alpha$$ of $$f$$ is a root of $$g_n$$.
• So $${ \mathbf{F} }_p(\alpha) \subseteq {\mathbf{GF}}(p^n)$$.
• The result follows from the computation \begin{align*} n &= [{\mathbf{GF}}(p^n) : { \mathbf{F} }_p] \\ &= [{\mathbf{GF}}(p^n) : { \mathbf{F} }_p(\alpha)] \cdot [{ \mathbf{F} }_p(\alpha) : { \mathbf{F} }_p] \\ &= kd .\end{align*}