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:
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\(k\) is a perfect field.
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If \(\operatorname{ch}k > 0\), the Frobenius is an automorphism of \(k\), so \(k^p = k\).
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Every finite extension \(F/k\) is separable.
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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.
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Supposing \(k^p\neq k\), choose \(a\in k\) not a \(p\)th power.
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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*}
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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 \(-mr \in k\).
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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.
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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 \(x^{p^d} - x \divides x^{p^n} - x \iff d\divides n\).
\(\implies\):
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\(\alpha \in {\mathbf{GF}}(p^n) \iff \alpha^{p^n} - \alpha = 0\), 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.
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So if \(f\) is an irreducible factor of \(\phi_n\), \(f\) is the minimal polynomial of some root \(\alpha\) of \(\phi_n\), so \(\deg f \divides n\).
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\(\phi_n'(x) = p^nx^{p^{n-1}} \neq 0\), 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 \(\Phi_n\):
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\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*}
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\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*}