20. Linear Disjointness（线性无交）
In this section, we study linear disjointness, a technical condition but one with many applications. One way that we use this concept is to extend the definition of separability in a useful way to nonalgebraic extensions. We tacitly assume that all of our field extensions of a given field \(F\) lie in some common extension field \(C\) of \(F\). Problem 6 shows that this is not a crucial assumption. We will also make use of tensor products. By phrasing some results in terms of tensor products, we are able to give cleaner, shorter proofs. However, the basic results on linear disjointness can be proved without using tensor products. Properties of tensor products are given in Appendix D for the benefit of the reader.
Definition 20.1 Let \(K\) and \(L\) be subfields of a field \(C\), each containing a field \(F\). Then \(K\) and \(L\) are linearly disjoint over \(F\) if every \(F\)-linearly independent subset of \(K\) is also linearly independent over \(L\).
Let \(A\) and \(B\) be subrings of a commutative ring \(R\). Then the ring \(A[B]\) is the subring of \(R\) generated by \(A\) and \(B\); that is, \(A[B]\) is the smallest subring of \(R\) containing \(A \cup B\). It is not hard to show that
If \(A\) and \(B\) contain a common field \(F\), then the universal mapping property of tensor products shows that there is a well-defined \(F\)-linear transformation \(\varphi : A \otimes_F B \to A[B]\) given on generators by \(\varphi(a\otimes b) = ab\). We refer to the map \(\varphi\) as the natural map from \(A\otimes_F B\) to \(A[B]\). We give a criterion in terms of tensor products for two fields to be linearly disjoint over a common subfield.
Proposition 20.2 Let \(K\) and \(L\) be field extensions of a field \(F\). Then \(K\) and \(L\) are linearly disjoint over \(F\) if and only if the map \(\varphi : K \otimes_F L \to K[L]\) given on generators by \(a \otimes b \mapsto ab\) is an isomorphism of \(F\)-vector spaces.
Proof. The natural map \(\varphi : K\otimes_F L \to K[L]\) is surjective by the description of \(K[L]\) given above. So, we need to show that \(K\) and \(L\) are linearly disjoint over \(F\) if and only if \(\varphi\) is injective. Suppose first that \(K\) and \(L\) are linearly disjoint over \(F\). Let \(\{k_i\}_{i\in I}\) be a basis for \(K\) as an \(F\)-vector space. Each element of \(K \otimes_F L\) has a unique representation in the form \(\sum k_i \otimes l_i\), with the \(l_i \in L\). Suppose that \(\sum k_i \otimes l_i \in \ker(\varphi)\), so \(\sum k_i l_i = 0\). Then each \(l_i = 0\), since \(K\) and \(L\) are linearly disjoint over \(F\) and \(\{k_i\}\) is \(F\)-linearly independent. Thus, \(\varphi\) is injective, and so \(\varphi\) is an isomorphism.
Conversely, suppose that the map \(\varphi\) is an isomorphism. Let \(\{a_j\}_{j\in J}\) be an \(F\)-linearly independent subset of \(K\). By enlarging \(J\), we may assume that the set \(\{a_j\}\) is a basis for \(K\). If \(\{a_j\}\) is not \(L\)-linearly independent, then there are \(l_j \in L\) with \(\sum a_j l_j = 0\), a finite sum. Then \(\sum a_j \otimes l_j \in \ker(\varphi)\), so \(\sum a_j \otimes l_j = 0\) by the injectivity of \(\varphi\). However, elements of \(K \otimes_F L\) can be represented uniquely in the form \(\sum a_j \otimes m_j\) with \(m_j \in L\). Therefore, each \(l_j = 0\), which forces the set \(\{a_j\}\) to be \(L\)-linearly independent. Thus, \(K\) and \(L\) are linearly disjoint over \(F\). □
Corollary 20.3 The definition of linear disjointness is symmetric; that is, \(K\) and \(L\) are linearly disjoint over \(F\) if and only if \(L\) and \(K\) are linearly disjoint over \(F\).
Proof. This follows from Proposition 20.2. The map \(\varphi : K \otimes_F L \to K[L]\) is an isomorphism if and only if \(\tau : L \otimes_F K \to L[K] = K[L]\) is an isomorphism, since \(\tau = i \circ \varphi\), where \(i\) is the canonical isomorphism \(K \otimes_F L \to L \otimes_F K\) that sends \(a \otimes b\) to \(b \otimes a\). □
Lemma 20.4 Suppose that \(K\) and \(L\) are finite extensions of \(F\). Then \(K\) and \(L\) are linearly disjoint over \(F\) if and only if \([KL : F] = [K : F] \cdot [L : F]\).
Proof. The natural map \(\varphi : K \otimes_F L \to K[L]\) that sends \(k \otimes l\) to \(kl\) is surjective and
Thus, \(\varphi\) is an isomorphism if and only if \([KL : F] = [K : F] \cdot [L : F]\). The lemma then follows from Proposition 20.2. □
Example 20.5 Suppose that \(K\) and \(L\) are extensions of \(F\) with \([K:F]\) and \([L:F]\) relatively prime. Then \(K\) and \(L\) are linearly disjoint over \(F\). To see this, note that both \([K:F]\) and \([L:F]\) divide \([KL:F]\), so their product divides \([KL:F]\) since these degrees are relatively prime. The linear disjointness of \(K\) and \(L\) over \(F\) follows from the lemma.
Example 20.6 Let \(K\) be a finite Galois extension of \(F\). If \(L\) is any extension of \(F\), then \(K\) and \(L\) are linearly disjoint over \(F\) if and only if \(K \cap L = F\). This follows from the previous example and the theorem of natural irrationalities, since
so \([KL:F] = [K:F][L:F]\) if and only if \(K \cap L = F\).
The tensor product characterization of linear disjointness leads us to believe that there is a reasonable notion of linear disjointness for rings, not just fields. Being able to discuss linear disjointness in the case of integral domains will make it easier to work with fields, as we will see in Section 22 and later in this section.
Definition 20.7 Let \(A\) and \(B\) be subrings of a field \(C\), each containing a field \(F\). Then \(A\) and \(B\) are linearly disjoint over \(F\) if the natural map \(A \otimes_F B \to C\) given by \(a \otimes b \mapsto ab\) is injective.
Lemma 20.8 Suppose that \(F\) is a field, and \(F \subseteq A \subseteq A'\) and \(F \subseteq B \subseteq B'\) are all subrings of a field \(C\). If \(A'\) and \(B'\) are linearly disjoint over \(F\), then \(A\) and \(B\) are linearly disjoint over \(F\).
Proof. This follows immediately from properties of tensor products. There is a natural injective homomorphism \(i : A \otimes_F B \to A' \otimes_F B'\) sending \(a \otimes b\) to \(a \otimes b\) for \(a \in A\) and \(b \in B\). If the natural map \(\varphi' : A' \otimes_F B' \to A'[B']\) is injective, then restricting \(\varphi\) to the image of \(i\) shows that the map \(\varphi : A \otimes_F B \to A[B]\) is also injective. □
Example 20.9 Let \(K\) and \(L\) be extensions of a field \(F\). If \(K \cap L\) is larger than \(F\), then \(K\) and \(L\) are not linearly disjoint over \(F\) by the preceding lemma since \(K \cap L\) is not linearly disjoint to itself over \(F\). However, \(K\) and \(L\) may not be linearly disjoint over \(F\) even if \(K \cap L = F\). As an example, let \(F = \mathbb{Q}\), \(K = F(\sqrt[3]{2})\), and \(L = F(\omega\sqrt[3]{2})\), where \(\omega\) is a primitive third root of unity. Then \(K \cap L = F\), but \(KL = F(\sqrt[3]{2},\omega)\) has dimension \(6\) over \(F\), whereas \(K \otimes_F L\) has dimension \(9\), so the map \(K \otimes_F L \to KL\) is not injective.
Lemma 20.10 Suppose that \(A\) and \(B\) are subrings of a field \(C\), each containing a field \(F\), with quotient fields \(K\) and \(L\), respectively. Then \(A\) and \(B\) are linearly disjoint over \(F\) if and only if \(K\) and \(L\) are linearly disjoint over \(F\).
Proof. If \(K\) and \(L\) are linearly disjoint over \(F\), then \(A\) and \(B\) are also linearly disjoint over \(F\) by the previous lemma. Conversely, suppose that \(A\) and \(B\) are linearly disjoint over \(F\). Let \(\{k_1,\dots,k_n\} \subseteq K\) be an \(F\)-linearly independent set, and suppose that there are \(l_i \in L\) with \(\sum k_i l_i = 0\). There are nonzero \(s \in A\) and \(t \in B\) with \(s k_i \in A\) and \(t l_i \in B\) for each \(i\). The set \(\{a_1,\dots,a_n\}\) is also \(F\)-linearly independent; consequently, \(\sum a_i \otimes b_i \neq 0\), since it maps to the nonzero element \(\sum a_i \otimes b_i \in K \otimes_F L\) under the natural map \(A \otimes_F B \to K \otimes_F B\). However, \(\sum a_i \otimes b_i\) is in the kernel of the map \(A \otimes_F B \to A[B]\); hence, it is zero by the assumption that \(A\) and \(B\) are linearly disjoint over \(F\). This shows that \(\{k_i\}\) is \(L\)-linearly independent, so \(K\) and \(L\) are linearly disjoint over \(F\). □
Example 20.11 Suppose that \(K/F\) is an algebraic extension and that \(L/F\) is a purely transcendental extension. Then \(K\) and \(L\) are linearly disjoint over \(F\); to see this, let \(X\) be an algebraically independent set over \(F\) with \(L = F(X)\). From the previous lemma, it suffices to show that \(K\) and \(F[X]\) are linearly disjoint over \(F\). We can view \(F[X]\) as a polynomial ring in the variables \(x \in X\). The ring generated by \(K\) and \(F[X]\) is the polynomial ring \(K[X]\). The standard homomorphism \(K \otimes_F F[X] \to K[X]\) is an isomorphism because there is a ring homomorphism \(\tau : K[X] \to K \otimes_F F[X]\) induced by \(x \mapsto 1 \otimes x\) for each \(x \in X\), and this is the inverse of \(\varphi\). Thus, \(K\) and \(F[X]\) are linearly disjoint over \(F\), so \(K\) and \(L\) are linearly disjoint over \(F\).
The following theorem is a transitivity property for linear disjointness.
Theorem 20.12 Let \(K\) and \(L\) be extension fields of \(F\), and let \(E\) be a field with \(F \subseteq E \subseteq K\). Then \(K\) and \(L\) are linearly disjoint over \(F\) if and only if \(E\) and \(L\) are linearly disjoint over \(F\) and \(K\) and \(EL\) are linearly disjoint over \(E\).
Proof. We have the following tower of fields.
Consider the sequence of homomorphisms
where the maps \(f\), \(\varphi_1\), and \(\varphi_2\) are given on generators by
respectively. Each can be seen to be well defined by the universal mapping property of tensor products. The map \(f\) is an isomorphism by counting dimensions. Moreover, \(\varphi_1\) and \(\varphi_2\) are surjective. The composition of these three maps is the standard map \(\varphi : K \otimes_F L \to K[L]\). First, suppose that \(K\) and \(L\) are linearly disjoint over \(F\). Then \(\varphi\) is an isomorphism by Proposition 20.2. This forces both \(\varphi_1\) and \(\varphi_2\) to be isomorphisms, since all maps in question are surjective. The injectivity of \(\varphi_2\) implies that \(K\) and \(EL\) are linearly disjoint over \(E\). If \(\sigma : E \otimes_F L \to E[L]\) is the standard map, then \(\varphi_1\) is given on generators by \(\varphi_1(k \otimes (e \otimes l)) = k \otimes \sigma(e \otimes l)\); hence, \(\sigma\) is also injective. This shows that \(E\) and \(L\) are linearly disjoint over \(F\).
Conversely, suppose that \(E\) and \(L\) are linearly disjoint over \(F\) and that \(K\) and \(EL\) are linearly disjoint over \(E\). Then \(\varphi_2\) and \(\sigma\) are isomorphisms by Proposition 20.2. The map \(\varphi_1\) is also an isomorphism; this follows from the relation between \(\varphi_1\) and \(\sigma\) above. Then \(\varphi\) is a composition of three isomorphisms; hence, \(\varphi\) is an isomorphism. Using Proposition 20.2 again, we see that \(K\) and \(L\) are linearly disjoint over \(F\). □
Separability of field extensions
One of the benefits of discussing linear disjointness is that it allows us to give a meaningful notion of separability for arbitrary field extensions. In Section 22, we shall see some geometric consequences of this more general notion of separability. We first give an example that will help to motivate the definition of separability for nonalgebraic extensions.
Example 20.13 Let \(K/F\) be a separable extension, and let \(L/F\) be a purely inseparable extension. Then \(K\) and \(L\) are linearly disjoint over \(F\). To prove this, note that if \(\operatorname{char}(F) = 0\), then \(L = F\), and the result is trivial. So, suppose that \(\operatorname{char}(F) = p > 0\). We first consider the case where \(K/F\) is a finite extension. By the primitive element theorem, we may write \(K = F(\alpha)\) for some \(\alpha \in K\). Let \(f(x) = \min(F,\alpha)\) and \(g(x) = \min(L,\alpha)\). Then \(g\) divides \(f\) in \(L[x]\). If \(g(x) = a_0 + \cdots + a_{n-1}x^{n-1} + x^n\), then for each \(i\) there is a positive integer \(r_i\) with \(a_i^{p^{r_i}} \in F\). If \(r\) is the maximum of the \(r_i\), then \(a_i^{p^r} \in F\) for each \(i\), so \(g(x)^{p^r} \in F[x]\). Consequently, \(g(x)^{p^r}\) is a polynomial over \(F\) for which \(\alpha\) is a root. Thus, \(f\) divides \(g^{p^r}\) in \(F[x]\). Viewing these two divisibilities in \(L[x]\), we see that the only irreducible factor of \(f\) in \(L[x]\) is \(g\), so \(f\) is a power of \(g\). The field extension \(K/F\) is separable; hence, \(f\) has no repeated irreducible factors in any extension field of \(F\). This forces \(f = g\), so
From this, we obtain \([KL : F] = [K : F] \cdot [L : F]\), so \(K\) and \(L\) are linearly disjoint over \(F\) by Lemma 20.4.
If \(K/F\) is not necessarily finite, suppose that \(\varphi : K \otimes_F L \to KL\) is not injective. Then there are \(k_1,\dots,k_n \in K\) and \(l_1,\dots,l_n \in L\) with \(\varphi(\sum k_i \otimes l_i) = 0\). If \(K_0\) is the field generated over \(F\) by the \(k_i\), then the restriction of \(\varphi\) to \(K_0 \otimes_F L\) is not injective, which is false by the finite dimensional case. Thus, \(\varphi\) is injective, so \(K\) and \(L\) are linearly disjoint over \(F\).
Definition 20.14 Let \(F\) be a field of characteristic \(p > 0\), and let \(F_{\mathrm{ac}}\) be an algebraic closure of \(F\). Let
and
The field \(F^{1/p^\infty}\) is the composite of all purely inseparable extensions of \(F\) in \(F_{\mathrm{ac}}\). It is, therefore, the maximal purely inseparable extension of \(F\) in \(F_{\mathrm{ac}}\), so \(F^{1/p^\infty}\) is the purely inseparable closure of \(F\) in \(F_{\mathrm{ac}}\).
Definition 20.15 A transcendence basis \(X\) for a field extension \(K/F\) is said to be a separating transcendence basis for \(K/F\) if \(K\) is separable algebraic over \(F(X)\). If \(K\) has a separating transcendence basis over \(F\), then \(K\) is said to be separably generated over \(F\).
Example 20.16 Let \(K = F(x)\) be the rational function field in one variable over a field \(F\) of characteristic \(p\). Then \(\{x\}\) is a separating transcendence basis for \(K/F\). However, \(\{x^p\}\) is also a transcendence basis, but \(K/F(x^p)\) is not separable. This example shows that even if \(K/F\) is separably generated, not all transcendence bases of \(K/F\) are separating transcendence bases.
Example 20.17 If \(K/F\) is algebraic, then \(K\) is separable over \(F\) if and only if \(K/F\) is separably generated, so the definition of separably generated agrees with the definition of separable for algebraic extensions.
We now prove the result that characterizes separability of arbitrary extensions.
Theorem 20.18 Let \(K\) be a field extension of \(F\). Then the following statements are equivalent:
- Every finitely generated subextension of \(K/F\) is separably generated.
- The fields \(K\) and \(F^{1/p^\infty}\) are linearly disjoint over \(F\).
- The fields \(K\) and \(F^{1/p}\) are linearly disjoint over \(F\).
Proof. (1) \(\Rightarrow\) (2): To show that \(K\) and \(F^{1/p^\infty}\) are linearly disjoint over \(F\), it suffices to assume that \(K\) is a finitely generated extension of \(F\). By statement 1, we know that \(K\) is separably generated over \(F\), so there is a transcendence basis \(\{t_1,\dots,t_n\}\) of \(K/F\) for which \(K\) is separable over \(F(t_1,\dots,t_n)\). By Example 20.11, the fields \(F(t_1,\dots,t_n)\) and \(F^{1/p^\infty}\) are linearly disjoint over \(F\). Also, \(K\) and \(F^{1/p^\infty}(t_1,\dots,t_n)\) are linearly disjoint over \(F(t_1,\dots,t_n)\) by Example 20.13, since \(F^{1/p^\infty}(t_1,\dots,t_n)\) is purely inseparable over \(F(t_1,\dots,t_n)\) and \(K\) is separable over \(F(t_1,\dots,t_n)\). Therefore, by Theorem 20.12, the fields \(K\) and \(F^{1/p^\infty}\) are linearly disjoint over \(F\).
(2) \(\Rightarrow\) (3): This is clear since \(F^{1/p}\) is a subfield of \(F^{1/p^\infty}\).
(3) \(\Rightarrow\) (1): Suppose that \(K\) and \(F^{1/p}\) are linearly disjoint over \(F\). Let \(L = F(a_1,\dots,a_n)\) be a finitely generated subextension of \(K\). We use induction on \(n\) to show that \(\{a_1,\dots,a_n\}\) contains a separating transcendence basis for \(L/F\). The case \(n = 0\) is clear, as is the case where \(\{a_1,\dots,a_n\}\) is algebraically independent, since then \(\{a_1,\dots,a_n\}\) is a separating transcendence basis for \(L/F\). We may then assume that \(n > 0\) and that \(\{a_1,\dots,a_m\}\) is a transcendence basis for \(L/F\), with \(m < n\). The elements \(a_1,\dots,a_{m+1}\) are algebraically dependent over \(F\), so there is a nonzero polynomial \(f \in F[x_1,\dots,x_{m+1}]\) of least total degree with \(f(a_1,\dots,a_{m+1}) = 0\). The assumption that \(f\) is chosen of least degree forces \(f\) to be irreducible. We first claim that \(f\) is not a polynomial in \(x_1^p,\dots,x_{m+1}^p\). If \(f(x_1,\dots,x_{m+1}) = g(x_1^p,\dots,x_{m+1}^p)\) for some \(g \in F[x_1,\dots,x_{m+1}]\), then there is an \(h \in F^{1/p}[x_1,\dots,x_{m+1}]\) with \(f = h(x_1,\dots,x_{m+1})^p\), since we are assuming that \(\operatorname{char}(F) = p\) and every coefficient of \(g\) is a \(p\)th power in \(F^{1/p}\). But this implies that \(h(a_1,\dots,a_{m+1}) = 0\). Write \(h(z_1,\dots,z_{m+1}) = \sum_j \alpha_j m_j\), where the \(m_j\) are the monomials occurring in \(h\) and the \(\alpha_j \in F^{1/p}\). Then \(\sum_j \alpha_j m_j(a_1,\dots,a_{m+1}) = 0\), so the \(m_j(a_1,\dots,a_{m+1})\) are linearly dependent over \(F^{1/p}\). However, since each \(m_j\) is a monomial in the \(x_k\), each \(m_j(a_1,\dots,a_{m+1}) \in L \subseteq K\). The assumption that \(K\) and \(F^{1/p}\) are linearly disjoint over \(F\) then forces the \(m_j(a_1,\dots,a_{m+1})\) to be linearly dependent over \(F\). If \(\sum_j \beta_j m_j(a_1,\dots,a_{m+1}) = 0\) with \(\beta_j \in F\), then \(h' = \sum_j \beta_j m_j\) is a polynomial with \(h'(a_1,\dots,a_{m+1}) = 0\) and \(\deg(h') < \deg(f)\). This contradiction verifies our claim that \(f\) is not a polynomial in \(x_1^p,\dots,x_{m+1}^p\). Therefore, for some \(i\) the polynomial \(f\) is not a polynomial in \(x_i^p\). Let
Then \(q(a_i) = 0\), and \(q\) is not a polynomial in \(t^p\). If we can show that \(q\) is irreducible over \(M\), then we will have proved that \(a_i\) is separable over \(M\). To see this, the set \(\{a_1,\dots,a_{i-1},a_{i+1},\dots,a_{m+1}\}\) is a transcendence basis for \(L/F\), so
as rings. Under the map that sends \(x_j\) to \(a_j\) and \(t\) to \(x_i\), the polynomial \(q\) is mapped to \(f\). But \(f\) is irreducible over \(F\), so \(q\) is irreducible in \(F[a_1,\dots,a_{i-1},a_{i+1},\dots,a_{m+1}][t]\). By Gauss' lemma, this means that \(q\) is irreducible over \(M\), the quotient field of \(F[a_1,\dots,a_{i-1},a_{i+1},\dots,a_{m+1}]\). Thus, we have shown that \(a_i\) is separable over \(M\), so \(a_i\) is separable over \(L' = F(a_1,\dots,a_{i-1},a_{i+1},\dots,a_n)\). The induction hypothesis applied to \(L'\) gives us a subset of \(\{a_1,\dots,a_{i-1},a_{i+1},\dots,a_n\}\) that is a separating transcendence basis for \(L'/F\). Since \(a_i\) is separable over \(L'\), this is also a separating transcendence basis for \(L/F\). □
Definition 20.19 A field extension \(K/F\) is separable if \(\operatorname{char}(F) = 0\) or if \(\operatorname{char}(F) = p > 0\) and the conditions in Theorem 20.18 are satisfied; that is, \(K/F\) is separable if every finitely generated subextension of \(K/F\) is separably generated.
We now give some immediate consequences of Theorem 20.18.
Corollary 20.20 If \(K/F\) is separably generated, then \(K/F\) is separable. Conversely, if \(K/F\) is separable and finitely generated, then \(K/F\) is separably generated.
Corollary 20.21 Suppose that \(K = F(a_1,\dots,a_n)\) is finitely generated and separable over \(F\). Then there is a subset \(Y\) of \(\{a_1,\dots,a_n\}\) that is a separating transcendence basis of \(K/F\).
Proof. This corollary is more accurately a consequence of the proof of (3) \(\Rightarrow\) (1) in Theorem 20.18, since the argument of that step is to show that if \(K\) is finitely generated over \(F\), then any finite generating set contains a separating transcendence basis. □
Corollary 20.22 Let \(F\) be a perfect field. Then any finitely generated extension of \(F\) is separably generated.
Proof. This follows immediately from part 3 of Theorem 20.18, since \(F^{1/p^\infty} = F\) if \(F\) is perfect. □
Corollary 20.23 Let \(F \subseteq E \subseteq K\) be fields.
- If \(K/F\) is separable, then \(E/F\) is separable.
- If \(E/F\) and \(K/E\) are separable, then \(K/F\) is separable.
- If \(K/F\) is separable and \(E/F\) is algebraic, then \(K/E\) is separable.
Proof. Part 1 is an immediate consequence of condition 2 of Theorem 20.18. For part 2 we use Theorems 20.18 and 20.12. If \(E/F\) and \(K/E\) are separable, then \(E\) and \(F^{1/p}\) are linearly disjoint over \(F\), and \(K\) and \(E^{1/p}\) are linearly disjoint over \(E\). However, it follows from the definition that \(F^{1/p} \subseteq E^{1/p}\), so \(E F^{1/p} \subseteq E^{1/p}\). Thus, \(K\) and \(E F^{1/p}\) are linearly disjoint over \(E\). Theorem 20.12 then shows that \(K\) and \(F^{1/p}\) are linearly disjoint over \(F\), so \(K\) is separable over \(F\).
To prove part 3, suppose that \(K/F\) is separable and \(E/F\) is algebraic. We know that \(E/F\) is separable by part 1. Let \(L = E(a_1,\dots,a_n)\) be a finitely generated subextension of \(K/E\). If \(L' = F(a_1,\dots,a_n)\), then by the separability of \(K/F\) there is a separating transcendence basis \(\{t_1,\dots,t_m\}\) for \(L'/F\). Because \(E/F\) is separable algebraic, \(EL' = L\) is separable over \(L'\), so by transitivity, \(L\) is separable over \(F(t_1,\dots,t_m)\). Thus, \(L\) is separable over \(E(t_1,\dots,t_m)\), so \(\{t_1,\dots,t_m\}\) is a separating transcendence basis for \(L/E\). We have shown that \(L/E\) is separably generated for every finitely generated subextension of \(K/E\), which proves that \(K/E\) is separable. □
Example 20.24 Let \(F\) be a field of characteristic \(p\), let \(K = F(x)\), the rational function field in one variable over \(F\), and let \(E = F(x^p)\). Then \(K/F\) is separable, but \(K/E\) is not separable. This example shows the necessity for the assumption that \(E/F\) be algebraic in the previous corollary.
Example 20.25 Here is an example of a separable extension that is not separably generated. Let \(F\) be a field of characteristic \(p\), let \(x\) be transcendental over \(F\), and let \(K = F(x)(\{x^{1/p^n} : n \geq 1\})\). Then \(K\) is the union of the fields \(F(x^{1/p^n})\), each of which is purely transcendental over \(F\), and hence is separably generated. Any finitely generated subextension \(E\) is a subfield of \(F(x^{1/p^n})\) for some \(n\) and hence is separably generated over \(F\) by the previous corollary. Therefore, \(K/F\) is separable. But \(K\) is not separably generated over \(F\), since given any \(f \in K\), there is an \(n\) with \(f \in F(x^{1/p^n})\), so \(K/F(f)\) is not separable, since \(K/F(x^{1/p^n})\) is a nontrivial purely inseparable extension.