Polynomial rings and Gauss’s lemma
For a field \(k\), \(k[x]\) is a Euclidean domain with respect to degree, hence a PID and a UFD, and its prime elements are the irreducible polynomials. A polynomial in \(\ZZ[x]\) is primitive if the greatest common divisor of its coefficients is \(1\), and by Gauss’s lemma a product of primitive polynomials is primitive.
Polynomial ring $K[x]$ (Tag FD-CI4NB)
Let \(K\) be a commutative ring. The polynomial ring in the variable \(x\) over \(K\) is \[ K[x] \coloneqq \theset{\sum_{i=0}^n a_ix^i \st n\in \ZZ^{\geq 0},\ a_0,\ldots,a_n\in K}, \] where two polynomials are equal when their coefficients of \(x^i\) agree for every \(i\geq0\), with missing coefficients taken to be \(0\). Addition is coefficientwise, and multiplication is \(\bigl(\sum_i a_ix^i\bigr)\bigl(\sum_j b_jx^j\bigr)=\sum_k\bigl(\sum_{i+j=k}a_ib_j\bigr)x^k\).
Field of rational functions $K(x)$ (Tag FD-24RNF)
Let \(K\) be a field. The field of rational functions in the variable \(x\) over \(K\) is the field of fractions of the polynomial ring \(K[x]\): \[ K(x)\coloneqq\theset{\frac{P(x)}{Q(x)} \st P,Q \in K[x],\ Q\neq 0}. \]
\(\CC(x)\) is the field of meromorphic functions on the Riemann sphere \(\hat\CC=\CC\cup\theset{\infty}\). Every rational function is meromorphic on \(\hat\CC\). Conversely, a meromorphic function \(h\) on the compact surface \(\hat\CC\) has finitely many poles; subtracting from \(h\) its principal parts at the poles in \(\CC\), which are rational functions, and the polynomial principal part at \(\infty\) leaves a holomorphic function on \(\hat\CC\), which is constant by Liouville’s theorem, so \(h\in\CC(x)\).
Reducible and irreducible polynomials (Tag D-BVMTZ)
Let \(k\) be a field and let \(f\in k[x]\) be nonconstant. The polynomial \(f\) is reducible if there exist \(g, h\in k[x]\) with \(\deg g, \deg h \geq 1\) and \(f = gh\). The polynomial \(f\) is irreducible if it is not reducible.
Irreducible polynomial in terms of factor degrees (Tag FD-KPW3H)
Let \(K\) be a field. A polynomial \(p\in K[x]\) is irreducible if \(p\) is nonconstant and there are no \(q,r\in K[x]\) with \(p=qr\), \(\deg q<\deg p\), and \(\deg r<\deg p\).
Irreducible polynomial over a field (Tag FD-GHY34)
Let \(K\) be a field. A polynomial \(p\in K[x]\) is irreducible if \(\deg p \geq 1\) and whenever \(p= qr\) with \(q,r\in K[x]\), \(q\) or \(r\) is constant.
Every polynomial of degree \(1\) is irreducible: if \(p=qr\), then \(\deg q+\deg r=1\), so one of \(q,r\) has degree \(0\).
Primitive polynomials (Tag D-4VC6X)
Let \(R\) be a unique factorization domain. A polynomial \(p\in R[x]\) is primitive if a greatest common divisor of its coefficients is a unit.
Gauss's lemma (Tag T-JEZZY)
Let \(R\) be a unique factorization domain with field of fractions \(F\), and let \(p\in R[x]\) be primitive. Then \(p\) is irreducible in \(R[x]\) if and only if \(p\) is irreducible in \(F[x]\).
More precisely, if \(p=AB\) with \(A,B\in F[x]\) nonconstant, then there exist \(r,s\in F^\times\) with \(rs=1\) such that \(rA,sB\in R[x]\), so \(p=(rA)(sB)\) is a factorization of \(p\) in \(R[x]\) into nonconstant polynomials.
Every monic polynomial in \(R[x]\) is primitive, since its leading coefficient \(1\) is a unit; so a monic \(p\in\ZZ[x]\) is irreducible in \(\ZZ[x]\) if and only if it is irreducible in \(\QQ[x]\).
Gauss' lemma (Tag FT-OXN3Y)
Let \(R\) be a unique factorization domain with field of fractions \(F\), and let \(p\in R[x]\). If \(p=gh\) with \(g,h\in F[x]\) and \(\deg g,\deg h\ge1\), then there are \(g',h'\in R[x]\) with \(\deg g'=\deg g\), \(\deg h'=\deg h\), and \(p=g'h'\). In particular, if \(p\) is reducible in \(F[x]\), then \(p\) is reducible in \(R[x]\).
If \(R\) is a unique factorization domain, then so is \(R[x]\); the proof uses Gauss’ lemma to reduce factorization in \(R[x]\) to factorization in \(R\) and in the principal ideal domain \(F[x]\). By induction, \(R[x_1,\ldots,x_n]\) is a unique factorization domain, and so is the polynomial ring over \(R\) in any set of variables, since each polynomial involves finitely many variables.
A primitive polynomial \(p\in \ZZ[x]\) of positive degree is irreducible in \(\ZZ[x]\) if and only if it is irreducible in \(\QQ[x]\).
Standard factorizations and root counts
Factorization of $x^n - y^n$ (Tag FF-UC7SQ)
For \(n\ge1\), in \(\ZZ[x,y]\), \[ x^n-y^n=(x-y)\qty{x^{n-1}+x^{n-2}y+\cdots+xy^{n-2}+y^{n-1}}=(x-y)\sum_{k=0}^{n-1}x^{n-1-k}y^k. \]
This is FF-ED3CD over the commutative ring \(R=\ZZ[y]\) with \(a=y\).
Factorization of $x^n - a^n$ (Tag FF-ED3CD)
Let \(R\) be a commutative ring, let \(a\in R\), and let \(n\ge1\). In \(R[x]\), \[ x^n-a^n=(x-a)\sum_{k=0}^{n-1}a^kx^{n-1-k}. \]
Expanding, \[ (x-a)\sum_{k=0}^{n-1}a^kx^{n-1-k}=\sum_{k=0}^{n-1}a^kx^{n-k}-\sum_{k=1}^{n}a^kx^{n-k}, \] and the two sums cancel except for the terms \(x^n\) and \(-a^n\).
Factorization of $x^n + y^n$ for odd $n$ (Tag FF-2AKVH)
Let \(n\ge1\). If \(n\) is odd, then in \(\ZZ[x,y]\) \[ x^n+y^n=(x+y)\sum_{k=0}^{n-1}(-1)^kx^{n-1-k}y^k . \] If \(n\) is even, then \(x+y\) does not divide \(x^n+y^n\) in \(\ZZ[x,y]\).
For odd \(n\), \((-y)^n=-y^n\), so substituting \(a=-y\) in \(x^n-a^n=(x-a)\sum_{k=0}^{n-1}a^kx^{n-1-k}\) (FF-ED3CD) gives the factorization.
Since \(x+y\) is monic in \(x\) over \(\ZZ[y]\), division by \(x+y\) in \((\ZZ[y])[x]\) shows that \(x+y\) divides \(f\in\ZZ[x,y]\) if and only if \(f(-y,y)=0\). For even \(n\), \((-y)^n+y^n=2y^n\ne0\).
Factorization of $x^n + a^n$ for odd $n$ (Tag FF-HAMDC)
Let \(R\) be a commutative ring, let \(a\in R\), and let \(n\ge1\). If \(n\) is odd, then in \(R[x]\) \[ x^n+a^n=(x+a)\sum_{k=0}^{n-1}(-a)^kx^{n-1-k}. \] If \(n\) is even and \(2a^n\ne0\) in \(R\), for example if \(R\) is a field of characteristic other than \(2\) and \(a\ne0\), then \(x+a\) does not divide \(x^n+a^n\) in \(R[x]\).
For odd \(n\), \((-a)^n=-a^n\), so the factorization is FF-ED3CD applied to \(-a\). Since \(x+a\) is monic, division by \(x+a\) in \(R[x]\) leaves the constant remainder \(f(-a)\) for every \(f\in R[x]\). For even \(n\) the remainder of \(x^n+a^n\) is \((-a)^n+a^n=2a^n\ne0\).
Descartes' rule of signs (Tag FF-HK72Z)
Let \(p\in\RR[x]\) be a nonzero polynomial, and let \(V(p)\) be the number of sign changes in the sequence of its nonzero coefficients, listed in order of degree. The number \(r^+\) of positive real roots of \(p\), counted with multiplicity, satisfies \[ r^+\le V(p)\qquad\text{and}\qquad V(p)-r^+\in2\ZZ. \] The negative real roots of \(p\) are the negatives of the positive real roots of \(p(-x)\), so their number \(r^-\), counted with multiplicity, satisfies \(r^-\le V(p(-x))\) and \(V(p(-x))-r^-\in2\ZZ\).
Consequently, if \(p(0)\ne0\) and \(\deg p=d\), then \(p\) has at least \(d-V(p)-V(p(-x))\) nonreal complex roots, counted with multiplicity.
Let \(f(x)=x^3+x^2-x-1\). Its coefficient signs are \(+,+,-,-\), so \(V(f)=1\) and \(f\) has exactly one positive root. The polynomial \(f(-x)=-x^3+x^2+x-1\) has signs \(-,+,+,-\), so \(V(f(-x))=2\) and \(f\) has \(2\) or \(0\) negative roots; in fact \(f=(x-1)(x+1)^2\) has the negative root \(-1\) with multiplicity \(2\).
Let \(g(x)=x^3-1\). Then \(V(g)=1\), so \(g\) has exactly one positive root, and \(g(-x)=-x^3-1\) has \(V(g(-x))=0\), so \(g\) has no negative root. Hence \(g\) has at least \(3-1-0=2\) nonreal roots, so exactly \(2\).
Characteristic, automorphisms, and perfect fields
The characteristic of a field is \(0\) or a prime \(p\), and its prime subfield is \(\QQ\) or \(\FF_p\) respectively. In characteristic \(p\), \((a+b)^p = a^p+b^p\), so \(a\mapsto a^p\) is a field homomorphism. For a group \(G\) of automorphisms of a field \(L\), the fixed field \(L^G\) is the subfield of elements fixed by every element of \(G\). A field \(k\) is perfect if every irreducible polynomial over \(k\) is separable; equivalently, \(k\) has characteristic \(0\), or characteristic \(p\) and \(a\mapsto a^p\) is surjective on \(k\).
Characteristic (Tag D-JNCUB)
Let \(R\) be a ring. The characteristic of \(R\) is the smallest integer \(n\geq 1\) such that \(\sum_{k=1}^n 1_R = 0\) if such an integer exists, and \(0\) otherwise.
Freshman's dream (Tag PR-3X3TO)
Let \(k\) be a field of characteristic \(p > 0\). For all \(a, b \in k\), \[ (a+b)^p = a^p + b^p \quad\text{and}\quad (ab)^p = a^p b^p. \]
Fixed field (Tag D-5FG7E)
Let \(L/k\) be a field extension and let \(H \leq \Aut_{\Fieldsover{k}}(L)\) be a subgroup of the automorphisms of \(L\) fixing \(k\). The fixed field of \(H\) is \[ L^H \coloneqq \theset{ \ell \in L \st \sigma(\ell) = \ell \text{ for all } \sigma\in H }. \]
Prime subfield (Tag D-EOCCU)
Let \(F\) be a field. The prime subfield of \(F\) is the intersection of all subfields of \(F\), that is, the smallest subfield of \(F\) containing \(1\).
Prime subfields are $\QQ$ or $\FF_p$ (Tag T-U3EZL)
Let \(F\) be a field. Its prime subfield is isomorphic to \(\QQ\) if \(\ch F=0\), and to \(\FF_p\) if \(\ch F=p>0\), in which case \(p\) is prime.
Automorphism group of a field extension (Tag D-MN47W)
Let \(L/k\) be a field extension. The automorphism group of \(L/k\) is the group, under composition, \[ \Aut(L/k) \coloneqq \theset{\sigma\colon L \to L \text{ a field automorphism} \suchthat \restrictionof{\sigma}{k} = \id_k}. \]
Embeddings and lifts of field homomorphisms (Tag D-KGF4K)
Let \(k\) be a field and let \(L/k\) and \(L'/k\) be field extensions. A \(k\)-embedding \(L\embeds L'\) is a ring homomorphism \(\tau\colon L\to L'\) that restricts to the identity on \(k\):
Here \(\iota\) and \(\iota'\) are the inclusions, and the triangle commutes: \(\tau\circ\iota=\iota'\).
More generally, let \(\sigma\colon k\to k'\) be a homomorphism of fields and let \(L/k\) and \(L'/k'\) be field extensions with inclusions \(\iota\colon k\to L\) and \(\iota'\colon k'\to L'\). A lift of \(\sigma\) to \(L\to L'\) is a ring homomorphism \(\tau\colon L\to L'\) such that \(\tau\circ\iota=\iota'\circ\sigma\):
A lift of \(\sigma=\id_k\) is a \(k\)-embedding.
Every ring homomorphism \(\tau\colon L\to L'\) between fields is injective: \(\ker\tau\) is an ideal of \(L\) not containing \(1\), and the only such ideal of a field is \(0\).
Perfect fields (Tag D-KQFIV)
A field \(k\) is perfect if either \(k\) has characteristic \(0\), or \(k\) has characteristic \(p>0\) and the Frobenius map \(k\to k\), \(a\mapsto a^p\), is surjective, that is, \(k^p = k\).
For a field \(k\), the following are equivalent:
An irreducible \(f\in k[x]\) has a repeated root if and only if \(\gcd(f,f')\neq 1\), which for irreducible \(f\) happens if and only if \(f'=0\). In characteristic \(0\) this never happens for nonconstant \(f\), so (1), (2), and (3) all hold.
Let \(\characteristic k=p>0\). If \(k^p=k\) and \(f'=0\), then \(f(x)=\sum_i a_ix^{pi}\) with \(a_i=b_i^p\) for some \(b_i\in k\), so \(f=\bigl(\sum_i b_ix^i\bigr)^p\) is not irreducible; hence (1) implies (2). If \(a\in k\setminus k^p\), then \(x^p-a\) is irreducible over \(k\) and equals \((x-\alpha)^p\) over a splitting field, where \(\alpha^p=a\), so it is not separable; hence (2) implies (1). Finally, (2) implies (3) because the minimal polynomial of each element of \(F\) is irreducible over \(k\), and (3) implies (2) because a root \(\alpha\) of an irreducible \(f\) generates the finite extension \(k(\alpha)/k\), in which \(\alpha\) is separable exactly when \(f\) is separable.
Perfect field (Tag FD-YYLYR)
A field \(k\) is perfect if every irreducible polynomial in \(k[x]\) is separable.
A field \(k\) is perfect if and only if either \(\ch k=0\), or \(\ch k=p>0\) and the Frobenius map \(F\colon k\to k\), \(F(a)=a^p\), is surjective, hence an automorphism. In particular, every field of characteristic zero and every finite field is perfect.
An irreducible \(f\in k[x]\) has a repeated root if and only if \(\gcd(f,f')\ne1\), and since \(f\) is irreducible and \(\deg f'<\deg f\), this happens if and only if \(f'=0\).
If \(\ch k=0\), then \(f'\ne0\) for every nonconstant \(f\), so \(k\) is perfect.
Let \(\ch k=p>0\). Then \(f'=0\) if and only if \(f(x)=\sum_i a_ix^{pi}\) for some \(a_i\in k\). If \(F\) is surjective, choose \(b_i\in k\) with \(b_i^p=a_i\); then \(f=\qty{\sum_ib_ix^i}^p\) is reducible, so no irreducible polynomial has \(f'=0\), and \(k\) is perfect. If \(F\) is not surjective, choose \(a\in k\) that is not a \(p\)th power, and let \(\beta\) be a root of \(x^p-a\) in a splitting field, so \(x^p-a=(x-\beta)^p\). A monic factor of \(x^p-a\) in \(k[x]\) of degree \(j\) with \(0<j<p\) is \((x-\beta)^j\), whose coefficient of \(x^{j-1}\) is \(-j\beta\); since \(j\) is a unit in \(k\), this would give \(\beta\in k\) and \(a=\beta^p\). So \(x^p-a\) is irreducible and not separable, and \(k\) is not perfect.
The Frobenius map is an injective ring homomorphism, so on a finite field it is surjective.
Algebraically closed field (Tag FD-3GZPO)
A field \(k\) is algebraically closed if every nonconstant polynomial \(f\in k[x]\) has a root in \(k\).
Characterization of perfect fields (Tag PR-IK6AM)
A field \(k\) is perfect, that is, every irreducible polynomial in \(k[x]\) is separable, if and only if either
\(\ch k = 0\), or
\(\ch k = p > 0\) and \(k^p = k\).
Cyclotomic polynomials
The \(n\)th cyclotomic polynomial \(\Phi_n(x)=\prod(x-\zeta)\), the product over the primitive \(n\)th roots of unity \(\zeta\in\CC\), has integer coefficients and satisfies \[ x^n-1=\prod_{d\mid n}\Phi_d(x). \] Over \(\QQ\), \(\Phi_n\) is irreducible of degree \(\phi(n)\), where \(\phi\) is Euler's totient function, so \([\QQ(\zeta_n):\QQ]=\phi(n)\) for a primitive \(n\)th root of unity \(\zeta_n\). By the Kronecker–Weber theorem, every finite abelian extension of \(\QQ\) is contained in a cyclotomic field.
Cyclotomic polynomials (Tag D-BLV6F)
Let \(n\geq 1\) and \(\zeta_n \coloneqq e^{2\pi i/n}\in\CC\). The \(n\)th cyclotomic polynomial is the product over the primitive \(n\)th roots of unity \[ \Phi_{n}(x) \coloneqq \prod_{\substack{1\leq k\leq n \\ \gcd(k, n)=1}}\left(x- \zeta_n^k\right) \in \CC[x]. \]
Let \(n\geq 1\). Then \(\Phi_n(x)\) lies in \(\ZZ[x]\), is monic of degree \(\phi(n)\), where \(\phi\) is Euler's totient function, and is irreducible in \(\QQ[x]\). Moreover, \(\Phi_n\) is the unique monic irreducible polynomial in \(\QQ[x]\) that divides \(x^n - 1\) and does not divide \(x^k-1\) for any \(1\leq k<n\).
Cyclotomic field (Tag D-IPR4B)
Let \(n\geq 1\) and let \(\zeta_n\in\CC\) be a primitive \(n\)th root of unity. The cyclotomic field of \(n\)th roots of unity is \(\QQ(\zeta_n)\), the splitting field of \(x^n-1\) over \(\QQ\) [@DF04].
More generally, for a field \(k\) whose characteristic does not divide \(n\), the splitting field of \(x^n-1\) over \(k\) is \(k(\zeta)\) for any primitive \(n\)th root of unity \(\zeta\) in it.
Splitting field of $x^m-1$ over $\QQ$ and its Galois group (Tag PR-DCK6S)
Let \(m \geq 1\) and let \(\zeta_m\) be a primitive $m$th root of unity in \(\CC\). The splitting field of \(x^m - 1\) over \(\QQ\) is the cyclotomic field \(\QQ(\zeta_m)\), and \[ \Gal(\QQ(\zeta_m)/\QQ) \cong \unitsof{(\ZZ/m\ZZ)}. \]
Kronecker--Weber theorem (Tag T-QX5QU)
Let \(K\subseteq\CC\) be a finite Galois extension of \(\QQ\) with \(\Gal(K/\QQ)\) abelian. Then there is an integer \(m\geq 1\) such that \(K \subseteq \QQ(\zeta_m)\), where \(\zeta_m\coloneqq e^{2\pi i/m}\).
Elementary symmetric functions (Tag D-FK47C)
Let \(R\) be a commutative ring and let \(x_1, \ldots, x_n\) be indeterminates over \(R\). The elementary symmetric functions in \(x_1, \ldots, x_n\) are \(e_0 \coloneqq 1\) and \[ e_k \coloneqq \sum_{1\leq i_1 < i_2 < \cdots < i_k\leq n} x_{i_1} x_{i_2}\cdots x_{i_k} \in R[x_1, \ldots, x_n], \qquad 1 \leq k \leq n. \]
In \(R[x_1,\ldots,x_n][t]\), \[ \prod_{i=1}^n (t - x_i) = \sum_{k=0}^n (-1)^k e_k t^{n-k}. \] Consequently, if a monic polynomial of degree \(n\) over a commutative ring factors as \(\prod_{i=1}^n(t-r_i)\), its coefficient of \(t^{n-k}\) is \((-1)^k e_k(r_1,\ldots,r_n)\).
Let \(R\) be a commutative ring. Every symmetric polynomial in \(R[x_1, \ldots, x_n]\) is equal to \(g(e_1, \ldots, e_n)\) for a unique polynomial \(g\in R[y_1,\ldots,y_n]\).
See [@DF04].
Exercises
$x^\ell-1$ divides $x^m-1$ iff $\ell$ divides $m$ (Tag E-LUR7G)
Show \[ x^\ell - 1 \divides x^m-1 \iff \ell\divides m .\]
Solution
\(\implies\)
- Write \(m = \ell q + r\) with \(0\leq r < \ell\).
- Write \[ p(x) = {x^m-1 \over x^\ell - 1} = {x^{lq+r} -1 \over x^\ell - 1} = x^r{x^{lq} - 1 \over x^\ell - 1} + {x^r - 1 \over x^\ell - 1} = q(x) + {x^r-1 \over x^\ell - 1} ,\] 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 \(\ell \divides m\).
\(\impliedby\):
- Write \(m = \ell q + r\), then \(r=0\) by divisibility.
- Then \(x^m-1 = x^{\ell q} - 1 \da z^q-1\) where \(z\da x^\ell\).
- Use that \(z-1 \divides z^q - 1\), so \(x^{\ell}-1 \divides x^{\ell q} -1 = x^m-1\).
Irreducible polynomials in characteristic $p$ are $g(x^{p^k})$ for unique separable $g$ (Tag E-6OUJV)
If \(f\in k[x]^{\irr}\) with \(\ch k = p\), then there is a unique separable \(g\in k[x]^{\irr}\) such that \(f(x) = g(x^{p^e})\) for some unique \(e \ge 0\).
Solution
Let \(f\in k[x]\) be irreducible and \(\operatorname{char}k=p>0\).
For any polynomial \(h(x)=\sum a_i x^i\), \[ h'(x)=0 \quad\Longleftrightarrow\quad a_i=0\text{ whenever }p\nmid i \quad\Longleftrightarrow\quad h(x)=h_1(x^p) \] for some \(h_1\in k[x]\). If \(h\) is irreducible and \(h=h_1(x^p)\), then \(h_1\) is irreducible, since any factorization of \(h_1\) would induce one of \(h\).
Choose \(e\ge0\) maximal such that \[ f(x)=g(x^{p^e}) \] for some \(g\in k[x]\). Such an \(e\) exists because \(e=0\) works and \(p^e\mid\deg f\), so \(e\) is bounded. The polynomial \(g\) is irreducible. By maximality, \(g'(x)\ne0\); since \(g\) is irreducible, this is equivalent to \(g\) being separable. Thus the required representation exists.
For uniqueness, suppose \[ f(x)=g_1(x^{p^{e_1}})=g_2(x^{p^{e_2}}) \] with \(g_1,g_2\) separable. Assume \(e_1<e_2\). Comparing coefficients shows that every exponent occurring in \(g_1\) is divisible by \(p^{e_2-e_1}\), hence in particular by \(p\). Thus \(g_1'(x)=0\), contradicting separability. Therefore \(e_1=e_2\), and then equality of the two substituted polynomials forces \(g_1=g_2\).
Hence both \(e\) and the separable irreducible polynomial \(g\) are unique.
$x^{\ell}-1$ divides $x^{m}-1$ if and only if $\ell$ divides $m$ (Tag E-OB3LO)
Show \[ x^\ell - 1 \divides x^m-1 \iff \ell\divides m .\]
Solution
\(\implies\)
- Write \(m = \ell q + r\) with \(0\leq r < \ell\).
- Write \[ p(x) = {x^m-1 \over x^\ell - 1} = {x^{lq+r} -1 \over x^\ell - 1} = x^r{x^{lq} - 1 \over x^\ell - 1} + {x^r - 1 \over x^\ell - 1} = q(x) + {x^r-1 \over x^\ell - 1} ,\] 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 \(\ell \divides m\).
\(\impliedby\):
- Write \(m = \ell q + r\), then \(r=0\) by divisibility.
- Then \(x^m-1 = x^{\ell q} - 1 \da z^q-1\) where \(z\da x^\ell\).
- Use that \(z-1 \divides z^q - 1\), so \(x^{\ell}-1 \divides x^{\ell q} -1 = x^m-1\).
An irreducible $f\in\mathbb{F}_p[x]$ of degree $d$ divides $x^{p^n}-x$ if and only if $d$ divides $n$ (Tag E-PHSV5)
Show that if \(f \in \FF_p[x]^{\irr}\) is degree \(d\), \[ f \divides x^{p^n}-x \iff d\divides n .\]
Solution
Let \(\alpha\) be a root of \(f\) in an algebraic closure \(\overline{\FF}_p\); then \(f\) is (up to a unit) the minimal polynomial of \(\alpha\) and \(\FF_p(\alpha)\cong\GF(p^d)\).
\(\impliedby\):
- If \(d\divides n\), then \(x^d-1 \divides x^n-1\) (E-LUR7G), and evaluating at \(x=p\) gives \(p^d-1 \divides p^n-1\).
- Applying E-LUR7G again, \(x^{p^d-1}-1 \divides x^{p^n-1}-1\); multiplying by \(x\) gives \(x^{p^d}-x\divides x^{p^n}-x\).
- Every element \(a\) of the field \(\FF_p(\alpha)\) of order \(p^d\) satisfies \(a^{p^d}=a\), so \(\alpha\) is a root of \(x^{p^d}-x\) and \(f\divides x^{p^d}-x\).
- Hence \(f \divides x^{p^d}-x \divides x^{p^n}-x\).
\(\implies\):
- If \(f\divides x^{p^n}-x\), then \(\alpha^{p^n}=\alpha\), so \(\alpha\) lies in the subfield \(\GF(p^n)=\{a\in\overline{\FF}_p:a^{p^n}=a\}\), and \(\FF_p(\alpha) \subseteq \GF(p^n)\).
- By the tower law, \[n = [\GF(p^n) : \FF_p] = [\GF(p^n) : \FF_p(\alpha)] \cdot [\FF_p(\alpha) : \FF_p] = [\GF(p^n) : \FF_p(\alpha)]\cdot d,\] so \(d\divides n\).