For fields \(K\subseteq L\subseteq M\), \([M:K]=[M:L][L:K]\). An element \(\alpha\in L\) is algebraic over \(K\) if and only if \([K(\alpha):K]\) is finite, and then \([K(\alpha):K]\) is the degree of the minimal polynomial of \(\alpha\). An algebraic extension \(L/K\) is normal if every irreducible polynomial in \(K[x]\) with a root in \(L\) splits into linear factors in \(L[x]\); a finite extension is normal if and only if it is the splitting field of a polynomial in \(K[x]\). It is separable if the minimal polynomial of every element of \(L\) has distinct roots. A polynomial \(f\) has a repeated root in its splitting field if and only if \(\gcd(f,f')\ne1\), and every algebraic extension of a perfect field, such as a field of characteristic \(0\) or a finite field, is separable. For each prime power \(q\) there is a field \(\FF_q\) with \(q\) elements, unique up to isomorphism, and every finite extension of a finite field is normal and separable.
Subfield $K(\alpha)$ generated by an element of an extension field (Tag FD-J3HIA)
Let \(F\) be a field, \(K\subseteq F\) a subfield, and \(\alpha\in F\). The subfield generated by \(K\) and \(\alpha\), written \(K(\alpha)\), is the smallest subfield of \(F\) containing \(K\) and \(\alpha\), that is, the intersection of all subfields of \(F\) that contain \(K\cup\theset{\alpha}\).
Subring $K[\alpha]$ generated by an element of an extension field (Tag FD-3V3ZW)
Let \(F\) be a field, \(K\subseteq F\) a subfield, and \(\alpha\in F\). The subring generated by \(K\) and \(\alpha\), written \(K[\alpha]\), is the smallest subring of \(F\) containing \(K\) and \(\alpha\), that is, the intersection of all subrings of \(F\) that contain \(K\cup\theset{\alpha}\).
\(K[\alpha]=\theset{g(\alpha) \st g\in K[x]}\), the image of the evaluation homomorphism \(K[x]\to F\), \(g\mapsto g(\alpha)\): this image is a subring containing \(K\) and \(\alpha\), and every such subring contains each \(g(\alpha)\).
Subfield $K(\alpha)$ generated by an element of an extension field (Tag FD-QPNDA)
Let \(F\) be a field, \(K\subseteq F\) a subfield, and \(\alpha\in F\). The subfield generated by \(K\) and \(\alpha\), written \(K(\alpha)\), is the smallest subfield of \(F\) containing \(K\) and \(\alpha\), that is, the intersection of all subfields of \(F\) that contain \(K\cup\theset{\alpha}\).
Algebraic field extension (Tag FD-NS5RF)
A field extension \(L/k\) is algebraic if every \(\alpha \in L\) is algebraic over \(k\), that is, a root of some nonzero polynomial \(f\in k[x]\).
Algebraic element of a field extension (Tag FD-2EVYB)
Let \(L/K\) be a field extension. An element \(\alpha \in L\) is algebraic over \(K\) if there exists a nonzero polynomial \(g\in K[x]\) with \(g(\alpha) = 0\).
Finite extensions are algebraic (Tag T-NGBVC)
Let \(L/K\) be a field extension with \([L:K]<\infty\). Then every \(\alpha\in L\) is algebraic over \(K\).
Let \(n=[L:K]\). The \(n+1\) elements \(1,\alpha,\ldots,\alpha^n\) of the \(n\)-dimensional \(K\)-vector space \(L\) are linearly dependent, so there are \(c_0,\ldots,c_n\in K\), not all zero, with \(\sum_{i=0}^n c_i\alpha^i=0\). Thus \(\alpha\) is a root of the nonzero polynomial \(\sum_{i=0}^n c_ix^i\in K[x]\).
An algebraic extension of an algebraic extension is algebraic (Tag PR-ABSJX)
Let \(k \subseteq K \subseteq L\) be fields, with the extensions not necessarily finite. If \(L/K\) and \(K/k\) are algebraic, then \(L/k\) is algebraic.
Tower law for field degrees (Tag PR-ZO73V)
Let \(k\subseteq F\subseteq L\) be field extensions with \([L:F]\) and \([F:k]\) finite. Then \([L:k]\) is finite and \[ [L : k] = [L: F]\,[F: k]. \]
Let \(\alpha_1,\ldots,\alpha_m\) be a basis of \(F\) over \(k\) and \(\beta_1,\ldots,\beta_n\) a basis of \(L\) over \(F\). Every \(x\in L\) is \(\sum_j c_j\beta_j\) with \(c_j\in F\), and each \(c_j=\sum_i a_{ij}\alpha_i\) with \(a_{ij}\in k\), so the \(\alpha_i\beta_j\) span \(L\) over \(k\). If \(\sum_{i,j}a_{ij}\alpha_i\beta_j=0\) with \(a_{ij}\in k\), then linear independence of the \(\beta_j\) over \(F\) gives \(\sum_ia_{ij}\alpha_i=0\) for each \(j\), and linear independence of the \(\alpha_i\) over \(k\) gives \(a_{ij}=0\). So the \(mn\) products \(\alpha_i\beta_j\) form a basis of \(L\) over \(k\).
Simple extensions (Tag D-PUOGJ)
A field extension \(L/k\) is simple if there exists \(\alpha\in L\) such that \(L = k(\alpha)\).
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 \(0\).
Distinguished classes of field extensions (Tag D-JMATC)
A class \(\mathcal{S}\) of field extensions is distinguished if it satisfies the following two conditions.
- For every tower of fields \(k\subseteq K\subseteq L\), \(L/k \in \mathcal{S}\) if and only if \(L/K \in \mathcal{S}\) and \(K/k\in \mathcal{S}\):
- If \(K/k\in \mathcal{S}\) and \(L/k\) is an extension such that \(K\) and \(L\) are subfields of a common field, then \(LK/L \in \mathcal{S}\), where \(LK\) is the compositum. The extension \(LK/L\) is the lift of \(K/k\) to \(L\):
Let \(\mathcal{S}\) be a distinguished class of field extensions. If \(L/k\in\mathcal{S}\) and \(K/k\in \mathcal{S}\), with \(K\) and \(L\) subfields of a common field, then \(LK/k \in \mathcal{S}\):
By the lifting condition applied to \(K/k\in\mathcal{S}\) and \(L/k\), the extension \(LK/L\) lies in \(\mathcal{S}\). Applying the tower condition to \(k\subseteq L\subseteq LK\), with \(LK/L\in\mathcal{S}\) and \(L/k\in\mathcal{S}\), gives \(LK/k\in\mathcal{S}\).
Splitting field of a polynomial (Tag FD-LHTRR)
Let \(k\) be a field and \(f\in k[x]\) a nonzero polynomial. A splitting field of \(f\) over \(k\) is a field extension \(F/k\) such that \(f\) splits into linear factors in \(F[x]\) and \(f\) does not split into linear factors in \(E[x]\) for any proper subfield \(E\subsetneq F\) containing \(k\).
Equivalently, \(f\) splits in \(F[x]\) as \(f=c\prod_{i=1}^n(x-\alpha_i)\) and \(F=k(\alpha_1,\ldots,\alpha_n)\): a subfield of \(F\) containing \(k\) over which \(f\) splits contains every root \(\alpha_i\), and \(k(\alpha_1,\ldots,\alpha_n)\) is such a subfield.
Normal field extension (Tag D-LZTAK)
Let \(L/k\) be an algebraic field extension. The extension \(L/k\) is normal if every irreducible polynomial \(f\in k[x]\) that has a root in \(L\) splits into linear factors in \(L[x]\).
Equivalently, for every \(\alpha\in L\), all roots of the minimal polynomial of \(\alpha\) over \(k\) in an algebraic closure of \(L\), the conjugates of \(\alpha\) over \(k\), lie in \(L\). Thus an irreducible \(f\in k[x]\) either splits in \(L[x]\) or has no root in \(L\).
Normal field extension (Tag FD-JJFZ3)
An algebraic field extension \(L/K\) is normal if every irreducible polynomial \(p \in K[x]\) either has no roots in \(L\) or splits into linear factors in \(L[x]\).
Characterization of finite normal extensions as splitting fields (Tag PR-TZN4M)
Let \(k\) be a field and \(L/k\) a field extension. Then \(L/k\) is finite and normal if and only if \(L\) is the splitting field over \(k\) of some nonzero polynomial \(f\in k[x]\).
Fix an algebraic closure \(\bar k\) containing \(L\).
Suppose \(L/k\) is finite and normal, and write \(L=k(\beta_1,\ldots,\beta_r)\). Let \(f\) be the product of the minimal polynomials of the \(\beta_i\) over \(k\). By normality each factor splits in \(L\), and \(L\) is generated over \(k\) by roots of \(f\), so \(L\) is the splitting field of \(f\).
Conversely, let \(L=k(\alpha_1,\ldots,\alpha_n)\), where \(\alpha_1,\ldots,\alpha_n\) are the roots of \(f\) in \(L\) and \(f\) splits in \(L\). Then \(L/k\) is finite. A \(k\)-embedding \(\sigma\colon L\to\bar k\) sends each \(\alpha_i\) to a root of \(f\), and \(f\) splits in \(L\), so \(\sigma\) permutes \(\theset{\alpha_1,\ldots,\alpha_n}\) and \(\sigma(L)=L\). By PR-OZYUC, \(L/k\) is normal.
Normal closure of a field extension (Tag D-XD5NG)
Let \(K/k\) be an algebraic field extension. A normal closure of \(K/k\) is a field extension \(N/K\) such that \(N/k\) is normal and no proper subfield of \(N\) containing \(K\) is normal over \(k\).
A normal closure exists: inside an algebraic closure \(\bar K\) of \(K\), take the subfield generated over \(k\) by all roots in \(\bar K\) of the minimal polynomials over \(k\) of the elements of \(K\). It is unique up to an isomorphism that restricts to the identity on \(K\).
Compositum and intersection of normal extensions (Tag C-4GQK3)
Let \(k\) be a field, let \(\Omega\) be an algebraic closure of \(k\), and let \(E_1, E_2 \subseteq \Omega\) be extensions of \(k\). If \(E_1/k\) and \(E_2/k\) are normal, then the compositum \(E_1E_2/k\) and the intersection \((E_1 \cap E_2)/k\) are normal.
If $L/k$ is normal, then $L/K$ is normal for every intermediate field $K$ (Tag PR-MHOVR)
Let \(L/k\) be a finite normal extension and let \(K\) be a field with \(k \subseteq K \subseteq L\). Then \(L/K\) is normal.
Characterization of normal algebraic extensions (Tag PR-OZYUC)
Let \(k\) be a field, let \(L/k\) be an algebraic extension, and let \(\bar{k}\) be an algebraic closure of \(k\) containing \(L\). Then \(L/k\) is normal if and only if every \(k\)-embedding \(\sigma\colon L\to \bar{k}\) satisfies \(\sigma(L) = L\). In that case every \(k\)-embedding \(\sigma\colon L\to\bar k\) restricts to a \(k\)-automorphism of \(L\):
Suppose \(L/k\) is normal and let \(\sigma\colon L\to\bar k\) be a \(k\)-embedding. For \(\alpha\in L\) with minimal polynomial \(m_\alpha\in k[x]\), \(\sigma(\alpha)\) is a root of \(m_\alpha\), and all roots of \(m_\alpha\) lie in \(L\), so \(\sigma(L)\subseteq L\). Moreover \(\sigma\) injects the finite set of roots of \(m_\alpha\) in \(L\) into itself, hence permutes it, so \(\alpha\) lies in \(\sigma(L)\); thus \(\sigma(L)=L\).
Conversely, suppose every \(k\)-embedding \(L\to\bar k\) has image \(L\). Let \(f\in k[x]\) be irreducible with a root \(\alpha\in L\), and let \(\beta\in\bar k\) be any root of \(f\). There is a \(k\)-embedding \(k(\alpha)\to\bar k\) with \(\alpha\mapsto\beta\), and since \(L/k(\alpha)\) is algebraic and \(\bar k\) is algebraically closed, it extends to a \(k\)-embedding \(\sigma\colon L\to\bar k\). Then \(\beta=\sigma(\alpha)\in\sigma(L)=L\), so \(f\) splits in \(L\) and \(L/k\) is normal.
Separable polynomial (Tag D-ZT46D)
Let \(k\) be a field with algebraic closure \(\bar{k}\). A nonzero polynomial \(f \in k[x]\) is separable if \(f\) has no repeated roots in \(\bar{k}\).
Separable field extension (Tag D-JGYLA)
Let \(L/k\) be a field extension. An element \(\alpha\in L\) that is algebraic over \(k\) is separable over \(k\) if its minimal polynomial over \(k\) is separable. An algebraic extension \(L/k\) is separable if every \(\alpha\in L\) is separable over \(k\).
An algebraic extension \(L/k\) is separable if and only if every finite subextension \(k\subseteq L'\subseteq L\) is separable over \(k\).
If \(L/k\) is separable, every element of a subextension \(L'\) is an element of \(L\), hence separable over \(k\). Conversely, each \(\alpha\in L\) lies in the finite subextension \(k(\alpha)\), which is separable by hypothesis, so \(\alpha\) is separable over \(k\).
Separable field extension (Tag FD-6WSIA)
Let \(L/K\) be an algebraic field extension. \(L/K\) is separable if for every \(\alpha \in L\) the minimal polynomial \(m_\alpha\in K[x]\) of \(\alpha\) over \(K\) is separable.
Let \(K\) be a field and \(f\in K[x]\) irreducible. Then \(f\) is separable if and only if its formal derivative \(f'\) is nonzero.
Let \(\bar K\) be an algebraic closure of \(K\). A root \(\beta\in\bar K\) of \(f\) is repeated if and only if \(f'(\beta)=0\), since writing \(f=(x-\beta)g\) gives \(f'(\beta)=g(\beta)\). If \(f'=0\), every root of \(f\) is repeated, so \(f\) is not separable. If \(f'\neq0\) and \(\beta\) is a repeated root, then \(f\), being irreducible, is a scalar multiple of the minimal polynomial of \(\beta\) over \(K\), so \(f\) divides \(f'\); this is impossible because \(\deg f'<\deg f\) and \(f'\neq0\).
Separable degree (Tag D-WB4M5)
Let \(L/k\) be an algebraic field extension, let \(\bar{k}\) be an algebraic closure of \(k\), and fix an embedding \(\sigma\colon k\embeds \bar{k}\). The separable degree \([L:k]_s\) is the cardinality of the set of field embeddings \(\sigma'\colon L\to \bar{k}\) that extend \(\sigma\), that is, with \(\sigma'|_k=\sigma\):
Here \(\iota\colon k\embeds L\) is the inclusion.
Characterizations of separability for finite extensions (Tag PR-3VQBI)
Let \(L/k\) be a finite field extension, and let \([L:k]_s\) be its separable degree, the number of \(k\)-embeddings \(L \to \bar{k}\) into an algebraic closure \(\bar{k}\) of \(k\). The following are equivalent:
\(L/k\) is separable.
\(L = k(\alpha)\) for some \(\alpha \in L\) separable over \(k\).
\(L = k(S)\) for some set \(S \subseteq L\) of elements separable over \(k\).
\([L:k]_s = [L:k]\).
A polynomial is separable if and only if $\gcd(f, f') = 1$ (Tag PR-ENHVC)
Let \(k\) be a field and \(f \in k[x]\) a nonzero polynomial. Then \(f\) is separable if and only if \(\gcd(f, f') = 1\) in \(k[x]\), that is, if and only if \(f\) and \(f'\) have no common root in an algebraic closure \(\bar{k}\). Moreover, the repeated roots of \(f\) in \(\bar{k}\) are exactly the roots of \(\gcd(f, f')\).
The derivative detects separability for irreducible polynomials (Tag PR-TLBPS)
Let \(k\) be a field and let \(f\in k[x]\) be nonzero.
\(f\) is separable if and only if \(\gcd(f, f') = 1\).
If \(f\) is irreducible, then \(f\) is separable if and only if \(f'\neq 0\).
If \(\ch k = 0\), every irreducible \(f\) is separable.
If \(\ch k = p>0\), an irreducible \(f\) is inseparable if and only if \(f(x) = g(x^p)\) for some \(g\in k[x]\).
A root \(\alpha\) of \(f\) in a splitting field is repeated if and only if \(f'(\alpha)=0\), so \(f\) has a repeated root if and only if \(f\) and \(f'\) have a common root, if and only if \(\gcd(f,f')\neq1\); the gcd does not change under field extension.
If \(f\) is irreducible, \(\gcd(f,f')\) is \(1\) or an associate of \(f\), and it is an associate of \(f\) if and only if \(f\divides f'\), which, since \(\deg f'<\deg f\), holds if and only if \(f'=0\).
If \(\ch k=0\) and \(\deg f\geq1\), then \(f'\) has degree \(\deg f-1\), so \(f'\neq0\).
Write \(f=\sum_i a_ix^i\). Then \(f'=\sum_i ia_ix^{i-1}=0\) if and only if \(p\divides i\) whenever \(a_i\neq0\), that is, if and only if \(f(x)=g(x^p)\) for some \(g\in k[x]\).
A reducible polynomial with nonzero derivative can be inseparable: \(f(x) = x^2(x-1)\in\QQ[x]\) has \(f'=3x^2-2x\neq 0\) and the repeated root \(0\).
Inseparable irreducible polynomials are polynomials in $x^p$ (Tag C-C2GYX)
Let \(k\) be a field of characteristic \(p > 0\), and let \(f\in k[x]\) be irreducible. Then \(f\) is inseparable if and only if \(f(x) = g(x^p)\) for some \(g\in k[x]\). Moreover, there are a unique integer \(n \geq 0\) and an irreducible separable polynomial \(g\in k[x]\) with \(f(x) = g(x^{p^n})\), and \(f\) is inseparable if and only if \(n \geq 1\).
A polynomial has a repeated root in a splitting field if and only if it shares a root with its derivative. Since \(f\) is irreducible, \(f\) shares a root with \(f'\) if and only if \(f \divides f'\), and since \(\deg f' < \deg f\) this holds if and only if \(f' = 0\). Writing \(f = \sum_i a_i x^i\), we have \(f' = \sum_i i a_i x^{i-1}\), which vanishes if and only if \(a_i = 0\) whenever \(p \nmid i\), that is, if and only if \(f(x) = g(x^p)\) for some \(g\in k[x]\).
For the second statement, let \(n\) be the largest integer with \(f\in k[x^{p^n}]\); it exists because \(\deg f \geq 1\). Write \(f(x) = g(x^{p^n})\). A factorization of \(g\) into nonconstant polynomials would give one of \(f\), so \(g\) is irreducible, and \(g\notin k[x^p]\) by maximality of \(n\), so \(g\) is separable by the first statement. Conversely, if \(f(x) = g(x^{p^m})\) with \(g\) separable and irreducible, then \(g\notin k[x^p]\) by the first statement, so \(f\notin k[x^{p^{m+1}}]\) and \(m = n\). Finally, \(n\geq 1\) if and only if \(f\in k[x^p]\), which by the first statement holds if and only if \(f\) is inseparable.
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.
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 \(k^p = k\).
Algebraic extensions of perfect fields are separable (Tag PR-25FLW)
Let \(k\) be a field with \(\ch k=0\) or \(k\) finite. Then every algebraic extension \(L/k\) is separable.
Such a field is perfect. For \(\alpha\in L\), the minimal polynomial of \(\alpha\) over \(k\) is irreducible in \(k[x]\), hence separable, so \(\alpha\) is separable over \(k\).
Primitive element theorem (Tag T-3HPWT)
Let \(L/k\) be a finite separable field extension. Then \(L/k\) is simple: there exists \(\alpha\in L\) with \(L=k(\alpha)\).
Separability is transitive in towers (Tag PR-YCTNC)
Let \(k\subseteq K\subseteq L\) be algebraic field extensions. Then \(L/k\) is separable if and only if \(L/K\) and \(K/k\) are both separable:
Separability of a compositum (Tag PR-KFQJG)
Let \(k\) be a field and \(E, F\) algebraic extensions of \(k\) contained in a common field, with compositum \(EF\). Then \(E/k\) and \(F/k\) are separable if and only if \(EF/k\) is separable.
A field with $p^n$ elements is a splitting field of $x^{p^n} - x$ (Tag PR-32R3E)
Let \(p\) be a prime, let \(n \geq 1\), and let \(F\) be a field with \(p^n\) elements, whose prime subfield is \(\FF_p\). Then \(F\) is a splitting field of \(\rho(x) = x^{p^n} - x\) over \(\FF_p\), and the elements of \(F\) are exactly the \(p^n\) roots of \(\rho\) in \(F\).
Construction of $\GF(p^n)$ as $\FF_p[x]/(f)$ (Tag T-QBQLM)
Let \(p\) be a prime, \(n\geq 1\), and \(f \in \FF_p[x]\) an irreducible polynomial of degree \(n\), and let \(\alpha\) be a root of \(f\) in an extension field of \(\FF_p\). Then \[ \GF(p^n)\cong \frac{\FF_p[x]}{(f)} \cong \FF_p[\alpha] = \spanof_{\FF_p}\theset{1, \alpha, \ldots, \alpha^{n-1}}. \]
Subfields of the finite field $\FF_{p^k}$ (Tag PR-Q2IFX)
Let \(p\) be a prime and \(k,\ell\geq 1\). The field \(\FF_{p^k}\) contains a subfield with \(p^\ell\) elements if and only if \(\ell\mid k\). In that case the subfield is unique, equal to \(\theset{x\in\FF_{p^k}\suchthat x^{p^\ell}=x}\).
If \(F\subseteq\FF_{p^k}\) is a subfield with \(p^\ell\) elements, then \(\FF_{p^k}\) is an \(F\)-vector space of some dimension \(d\), so \(p^k=(p^\ell)^d\) and \(\ell\mid k\).
Conversely, suppose \(\ell\mid k\). Every element of \(\FF_{p^k}\) is a root of \(x^{p^k}-x\), which has degree \(p^k\), so \(x^{p^k}-x=\prod_{a\in\FF_{p^k}}(x-a)\). Since \(\ell\mid k\), \(p^\ell-1\) divides \(p^k-1\), so \(x^{p^\ell-1}-1\) divides \(x^{p^k-1}-1\) and \(x^{p^\ell}-x\) divides \(x^{p^k}-x\). Hence \(x^{p^\ell}-x\) has \(p^\ell\) distinct roots in \(\FF_{p^k}\). They form the fixed set of the field automorphism \(x\mapsto x^{p^\ell}\), which is a subfield with \(p^\ell\) elements. Every element of a subfield with \(p^\ell\) elements satisfies \(x^{p^\ell}=x\), so such a subfield is contained in, and hence equal to, this set.
Prime subfield and isomorphism type of a finite field (Tag PR-ADLQN)
Let \(F\) be a finite field. Then \(\ch F = p\) for a prime \(p\), the prime subfield of \(F\) is isomorphic to \(\FF_p\), and \(F \cong \FF_{p^n}\) for the unique integer \(n = [F:\FF_p] \geq 1\), so that \(\abs{F} = p^n\).
Finite fields are not algebraically closed (Tag PR-DEG36)
A finite field \(F\) is not algebraically closed.
Write \(F = \theset{a_1, \ldots, a_q}\) and let \(f(x) \coloneqq 1 + \prod_{i=1}^q (x - a_i) \in F[x]\). Then \(f\) is nonconstant, and \(f(a_j) = 1 \neq 0\) for every \(j\), so \(f\) has no root in \(F\).
Freshman's dream (Tag PR-3X3TO)
Let \(k\) be a field of characteristic \(p > 0\). For all \(a, b \in k\), \[ (a+b)^p = a^p + b^p \quad\text{and}\quad (ab)^p = a^p b^p. \]
Quadratic extensions of $\FF_p$ are unique up to isomorphism (Tag PR-F2N4L)
Let \(p\) be a prime. Any two field extensions \(E/\FF_p\) and \(E'/\FF_p\) of degree \(2\) are isomorphic.
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 \(1\leq k<n\).
Primitive root of unity (Tag FD-HM4P2)
Let \(F\) be a field and \(n\geq 1\) an integer. An element \(\zeta\in F\) is a primitive \(n\)th root of unity if \(\zeta^n = 1\) and \(\zeta^m\neq 1\) for \(1\leq m<n\), that is, \(\zeta\) has order \(n\) in \(F^\times\).
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 \[ \Gal(\QQ(\zeta_m)/\QQ) \cong \unitsof{(\ZZ/m\ZZ)}. \]
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.
Computing cyclotomic polynomials $\Phi_n$ (Tag PR-JLDJ6)
Let \(n \geq 1\) and let \(\Phi_n\) be the \(n\)th cyclotomic polynomial. Then \[ x^{n}-1=\prod_{\substack{d \divides n \\ d > 0}} \Phi_{d}(x), \qquad\text{so}\qquad \Phi_n(x) = \qty{x^n-1} \inverseof{\qty{\prod_{\substack{d \divides n \\ 0 < d < n}} \Phi_{d}(x)}}, \] and \[ \Phi_{n}(x)=\prod_{\substack{ d \divides n \\ d > 0} }\left(x^{d}-1\right)^{\mu\left(n/d\right)}, \] where \(\mu\) is the Möbius function: \[ \mu(m) = \begin{cases} 1 & \text{if } m = 1, \\ (-1)^{k} & \text{if } m \text{ is a product of } k \text{ distinct primes}, \\ 0 & \text{if } p^2 \divides m \text{ for some prime } p. \end{cases} \]
The first formula computes \(\Phi_n\) recursively: divide \(x^n - 1\) by the product of the \(\Phi_d\) with \(d \divides n\), \(d < n\), using polynomial long division.
Cyclotomic polynomial $\Phi_p$ for a prime $p$ (Tag FF-MUJDE)
For a prime \(p\), the cyclotomic polynomial \(\Phi_p\) is \[ \Phi_p(x)=\frac{x^p-1}{x-1}=1+x+x^2+\cdots+x^{p-1}. \]
The roots of \(x^p-1\) are the \(p\)th roots of unity, and every one of them other than \(1\) is primitive because \(p\) is prime. So \(x^p-1=(x-1)\Phi_p(x)\), and dividing by \(x-1\) with FF-ED3CD at \(a=1\) gives the sum.
Cyclotomic polynomial $\Phi_{2p}$ for an odd prime $p$ (Tag FF-UACKH)
For an odd prime \(p\), the cyclotomic polynomial \(\Phi_{2p}\) is \[ \Phi_{2p}(x)=\Phi_p(-x)=1-x+x^2-\cdots+x^{p-1}. \]
Since \(p\) is odd, \(-1\) has order \(2\) and \(\zeta\mapsto-\zeta\) is a bijection from the primitive \(p\)th roots of unity to the primitive \(2p\)th roots of unity. Hence \[ \Phi_{2p}(x)=\prod_\zeta(x+\zeta)=(-1)^{p-1}\prod_\zeta(-x-\zeta)=\Phi_p(-x), \] the products running over the primitive \(p\)th roots of unity \(\zeta\), because \(p-1\) is even. Substituting \(-x\) in \(\Phi_p(x)=1+x+\cdots+x^{p-1}\) (FF-MUJDE) gives the sum.
For \(p=2\) the formula fails: \(\Phi_4(x)=x^2+1\), whereas \(\Phi_2(-x)=1-x\).