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    Separability

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    Every algebraic extension of a field of characteristic \(0\) or of a finite field is separable; inseparable extensions occur only over imperfect fields of characteristic \(p\).

    Separable polynomial

    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

    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

    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 \(f'\neq0\).

    What does it mean for a field extension $L/K$ to be separable?
    What does it mean for a field extension $L/K$ to be separable, in terms of minimal polynomials?

    A polynomial is separable if and only if $\gcd(f, f') = 1$

    A polynomial is separable exactly when it is coprime to its derivative; their gcd records repeated roots.Suggest a slogan

    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

    Separability is equivalent to coprimality with the derivative; for irreducibles, to a nonzero derivative.Suggest a slogan

    Let \(k\) be a field and let \(f\in k[x]\) be nonzero.

    1. \(f\) is separable if and only if \(\gcd(f, f') = 1\).

    2. If \(f\) is irreducible, then \(f\) is separable if and only if \(f'\neq 0\).

    3. If \(\ch k = 0\), every irreducible \(f\) is separable.

    4. If \(\ch k = p>0\), an irreducible \(f\) is inseparable if and only if \(f(x) = g(x^p)\) for some \(g\in k[x]\).

    1. 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.

    2. 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 \(f'=0\).

    3. If \(\ch k=0\) and \(\deg f\geq1\), then \(f'\) has degree \(\deg f-1\), so \(f'\neq0\).

    4. 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 $x^p$

    In characteristic $p$, an irreducible polynomial is $g(x^{p^n})$ for a unique $n\geq 0$ and a separable irreducible $g$.Suggest a slogan

    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.

    (The derivative test)

    A polynomial \(f\) over a field is separable if and only if \(\gcd(f, f') = 1\). An irreducible \(f\) is inseparable if and only if \(f' = 0\), which in characteristic \(p\) happens if and only if \(f(x) = g(x^p)\) for a polynomial \(g\). For example, over \(\FF_p(t)\) the polynomial \(x^p - t\) is irreducible and \(x^p - t = (x - t^{1/p})^p\) over \(\FF_p(t^{1/p})\), so \(\FF_p(t^{1/p})/\FF_p(t)\) is inseparable.

    Permanence and the Galois condition

    A finite extension \(L/K\) is separable if and only if the number of \(K\)-embeddings of \(L\) into an algebraic closure of \(K\) equals \([L:K]\). Every algebraic extension of a perfect field is separable. If \(K\subseteq L\subseteq M\), then \(M/K\) is separable if and only if \(M/L\) and \(L/K\) are separable, and a compositum of separable extensions is separable.

    Characterizations of separability for finite extensions

    A finite extension is separable exactly when separable generators exist and the separable degree equals the degree.Suggest a slogan

    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 \(k\).

    • \([L:k]_s = [L:k]\).

    Separable splitting fields are Galois

    A finite separable splitting field is Galois, with degree equal to the number of base-field automorphisms.Suggest a slogan

    Let \(L/k\) be a finite field extension, with separable degree \([L:k]_s\).

    1. If \(L/k\) is separable, then \([L: k] = [L:k]_s\).

    2. If \(L/k\) is separable and \(L\) is the splitting field of a polynomial in \(k[x]\), then \[ [L:k] = \abs{\Aut_{\Fieldsover{k}}(L)}, \] so \(L/k\) is Galois with \(\Gal(L/k)=\Aut_{\Fieldsover{k}}(L)\).

    1. Fix an algebraic closure \(\bar k\) containing \(L\). Since \(L/k\) is a finite splitting field, it is normal (PR-TZN4M), so every \(k\)-embedding \(L\to\bar k\) has image \(L\) (PR-OZYUC) and is a \(k\)-automorphism of \(L\). Hence \([L:k]_s=\abs{\Aut_{\Fieldsover{k}}(L)}\), and (a) gives \([L:k]=\abs{\Aut_{\Fieldsover{k}}(L)}\).

    Irreducible polynomials over a perfect field are separable

    Over a perfect field, irreducible polynomials have no repeated roots.Suggest a slogan

    Let \(k\) be a perfect field, for instance a field of characteristic \(0\) or a finite field. Every irreducible polynomial \(f \in k[x]\) has distinct roots in a splitting field of \(f\) over \(k\).

    Over an imperfect field, an irreducible polynomial can have a repeated root: for \(k = \FF_p(t)\), the polynomial \(x^p - t\) is irreducible, and in a splitting field it factors as \((x - t^{1/p})^p\).

    Algebraic extensions of perfect fields are separable

    In characteristic zero or over a finite field, every algebraic extension is separable.Suggest a slogan

    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 \(k\).

    Separable degree

    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.

    Separability is transitive in towers

    An algebraic tower is separable from bottom to top exactly when both stages are separable.Suggest a slogan

    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

    A compositum is separable exactly when both constituent algebraic extensions are separable.Suggest a slogan

    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 finite extension is Galois if and only if it is normal and separable, if and only if it is the splitting field of a separable polynomial. A splitting field need not be separable in characteristic \(p\): \(\FF_p(t^{1/p})\) is the splitting field of \(x^p-t\) over \(\FF_p(t)\), and it is not Galois over \(\FF_p(t)\).

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