[proofplan] We prove the equivalences in the cycle (i) $\Rightarrow$ (ii) $\Rightarrow$ (iii) $\Rightarrow$ (i). For (i) $\Rightarrow$ (ii), we take a separating transcendence basis and show that $K$ is separable algebraic over a purely transcendental extension $k(T)$, then verify linear disjointness by using the fact that purely transcendental extensions are linearly disjoint from any algebraic extension, and separable algebraic extensions are linearly disjoint from purely inseparable extensions. For (ii) $\Rightarrow$ (iii), we show that linear disjointness from $k^{1/p^\infty}$ implies linear disjointness from $k^{1/p}$, which gives that $K \otimes_k k^{1/p}$ embeds into the composite field $K \cdot k^{1/p}$, hence is a domain. For (iii) $\Rightarrow$ (i), we work by induction on the transcendence degree, using the fact that $K \otimes_k k^{1/p}$ being a domain constrains the generators of $K/k$ to admit a separating transcendence basis. [/proofplan]

[step:Establish the characteristic $p$ reduction] If $\operatorname{char}(k) = 0$, then $k$ is perfect, so $k^{1/p^\infty} = k$ and $k^{1/p} = k$. In this case, linear disjointness of $K$ and $k$ over $k$ holds vacuously, $K \otimes_k k = K$ is a field, and every transcendence basis is a separating transcendence basis (since every algebraic extension in characteristic zero is separable by [Characteristic Zero Separability](/theorems/1262)). All three conditions hold simultaneously, and the equivalences are trivial. For the remainder of the proof, assume $\operatorname{char}(k) = p > 0$. [/step]

[step:Prove (i) $\Rightarrow$ (ii): a separating transcendence basis gives linear disjointness from $k^{1/p^\infty}$]Assume $K/k$ is separably generated. Then there exists a transcendence basis $T = \{t_1, \ldots, t_r\}$ of $K/k$ such that $K/k(T)$ is a finite separable algebraic extension. We prove linear disjointness of $K$ and $k^{1/p^\infty}$ over $k$ in two stages. [claim:$k(T)$ and $k^{1/p^\infty}$ are linearly disjoint over $k$] Let $n \ge 1$. We show $k(T)$ and $k^{1/p^n}$ are linearly disjoint over $k$. Suppose $\alpha_1, \ldots, \alpha_m \in k^{1/p^n}$ are linearly independent over $k$, and suppose $\sum_{i=1}^m f_i \alpha_i = 0$ with $f_i \in k(T)$. Clearing denominators, we may assume $f_i \in k[T] = k[t_1, \ldots, t_r]$. Since $t_1, \ldots, t_r$ are algebraically independent over $k$, the monomials $t_1^{a_1} \cdots t_r^{a_r}$ form a $k$-basis of $k[T]$. Writing $f_i = \sum_a c_{i,a} \, t^a$ with $c_{i,a} \in k$, the relation $\sum_i f_i \alpha_i = 0$ becomes $\sum_a t^a \left(\sum_i c_{i,a} \alpha_i\right) = 0$. Since the monomials $t^a$ are linearly independent over $k^{1/p^n}$ (as $t_1, \ldots, t_r$ remain algebraically independent over $k^{1/p^n}$: if they satisfied a polynomial relation over $k^{1/p^n}$, raising to the $p^n$-th power would give a polynomial relation over $k$, contradicting algebraic independence over $k$), we obtain $\sum_i c_{i,a} \alpha_i = 0$ for each $a$. Since the $\alpha_i$ are linearly independent over $k$ and $c_{i,a} \in k$, we get $c_{i,a} = 0$ for all $i, a$, hence $f_i = 0$ for all $i$. Since this holds for each $n$, $k(T)$ and $k^{1/p^\infty}$ are linearly disjoint over $k$. [/claim] [proof] Contained in the argument above. [/proof] [claim:$K$ and $k^{1/p^\infty}$ are linearly disjoint over $k$, given that $K/k(T)$ is finite separable and $k(T)$, $k^{1/p^\infty}$ are linearly disjoint over $k$] It suffices to show that $K$ and $k^{1/p^n}$ are linearly disjoint over $k$ for each $n \ge 1$. Fix $n$. Since $k(T)$ and $k^{1/p^n}$ are linearly disjoint over $k$, the composite $k(T) \cdot k^{1/p^n} = k^{1/p^n}(T)$ satisfies $[k(T) \cdot k^{1/p^n} : k^{1/p^n}] = [k(T) : k]$ in the appropriate sense (the transcendence degree is preserved and no new algebraic relations arise). Now $K/k(T)$ is finite separable. By the [Primitive Element Theorem](/theorems/1267), $K = k(T)(\gamma)$ for some $\gamma$ with separable minimal polynomial $f = \operatorname{min}(\gamma, k(T)) \in k(T)[x]$. We must show that $K$ and $k^{1/p^n}(T)$ are linearly disjoint over $k(T)$. Since $K/k(T)$ is separable and $k^{1/p^n}(T)/k(T)$ is purely inseparable (since $k^{1/p^n}/k$ is purely inseparable and $T$ is transcendental), these extensions are linearly disjoint over $k(T)$. We verify this directly. By the [Primitive Element Theorem](/theorems/1267), $K = k(T)(\gamma)$, so it suffices to show $f = \operatorname{min}(\gamma, k(T))$ remains irreducible over $k^{1/p^n}(T)$. Set $M = k^{1/p^n}(T)$ and $d = \deg f = [K : k(T)]$. Suppose $f$ factors over $M$, so that $\operatorname{min}(\gamma, M)$ has degree $d' < d$. Then $[M(\gamma) : M] = d'$. Now $M(\gamma)/M$ is separable (the minimal polynomial of $\gamma$ over $M$ divides $f$, which is separable since $\gcd(f, f') = 1$ by the [Separability via Formal Derivative](/theorems/3329)). Also $M(\gamma)/K = k(T)(\gamma)$ is purely inseparable (since $M/k(T)$ is purely inseparable and $M(\gamma) = M \cdot K$). Counting the separable degree of $M(\gamma)/k(T)$ in two ways: through $M$, we get $[M(\gamma) : k(T)]_s = [M(\gamma) : M]_s \cdot [M : k(T)]_s = d' \cdot 1 = d'$ (since $M/k(T)$ is purely inseparable). Through $K$, we get $[M(\gamma) : k(T)]_s = [M(\gamma) : K]_s \cdot [K : k(T)]_s = 1 \cdot d = d$ (since $K/k(T)$ is separable and $M(\gamma)/K$ is purely inseparable). This gives $d' = d$, contradicting $d' < d$. So $f$ remains irreducible over $M$, and $K, M$ are linearly disjoint over $k(T)$. Therefore $K$ and $k^{1/p^n}(T) = k^{1/p^n} \cdot k(T)$ are linearly disjoint over $k(T)$. Combined with the linear disjointness of $k(T)$ and $k^{1/p^n}$ over $k$, the transitivity of linear disjointness (in a tower $k \subset k(T) \subset K$ with the external field $k^{1/p^n}$) gives that $K$ and $k^{1/p^n}$ are linearly disjoint over $k$. [/claim] [proof] Contained in the argument above. [/proof][/step]

[guided]The proof of (i) $\Rightarrow$ (ii) proceeds in two stages, reflecting the structure of $K/k$: the purely transcendental part $k(T)/k$ and the separable algebraic part $K/k(T)$. **Stage 1: The purely transcendental layer.** We show $k(T)$ and $k^{1/p^\infty}$ are linearly disjoint over $k$. The key property of transcendental elements is rigidity: $t_1, \ldots, t_r$ cannot satisfy any algebraic relation over $k$, and this persists over $k^{1/p^n}$. If $t_1, \ldots, t_r$ satisfied a polynomial relation $g(t_1, \ldots, t_r) = 0$ with coefficients in $k^{1/p^n}$, raising to the $p^n$-th power gives a polynomial relation over $k$ (since $(c^{1/p^n})^{p^n} = c \in k$ for each coefficient), contradicting algebraic independence over $k$. With algebraic independence preserved, the monomials $t^a = t_1^{a_1} \cdots t_r^{a_r}$ remain linearly independent over $k^{1/p^n}$, so any $k$-linear dependence relation involving elements of $k(T)$ and $k^{1/p^n}$ reduces monomial-by-monomial to a $k$-linear relation among elements of $k^{1/p^n}$, which forces all coefficients to vanish. **Stage 2: The separable algebraic layer.** We must lift linear disjointness from $k(T)$ to $K = k(T)(\gamma)$ where $K/k(T)$ is separable. The crucial fact is: a separable extension and a purely inseparable extension of the same field are always linearly disjoint. Why? Let $L/F$ be finite separable with $L = F(\gamma)$, and let $M/F$ be purely inseparable. The minimal polynomial $\operatorname{min}(\gamma, F)$ is separable, meaning $\gcd(f, f') = 1$ (by the [Separability via Formal Derivative](/theorems/3329)). Over $M$, if $f$ factored as $g \cdot h$, then $\gamma$ would have a lower-degree minimal polynomial over $M$, say of degree $d' < d = [L:F]$. But $M(\gamma)/M$ is separable (the minimal polynomial divides $f$, which is separable), and $M(\gamma)/F(\gamma) = L$ is purely inseparable (since $M/F$ is). So $[M(\gamma) : F] = [M(\gamma) : M] \cdot [M : F]$ and also $[M(\gamma) : F] = [M(\gamma) : L] \cdot [L : F]$. The separable and inseparable degrees decompose multiplicatively, and the separable degree $[M(\gamma) : F]_s$ equals $[L : F]_s \cdot [M(\gamma) : L]_s = d \cdot 1 = d$ (since $L/F$ is separable and $M(\gamma)/L$ is purely inseparable). On the other hand, $[M(\gamma) : F]_s = [M(\gamma) : M]_s \cdot [M : F]_s = [M(\gamma) : M] \cdot 1 = d'$, giving $d' = d$, a contradiction. So $f$ remains irreducible over $M$, and $L, M$ are linearly disjoint over $F$. Applying this with $F = k(T)$, $L = K$, $M = k^{1/p^n}(T)$: $K$ and $k^{1/p^n}(T)$ are linearly disjoint over $k(T)$. The transitivity of linear disjointness then gives: $K$ and $k^{1/p^n}$ are linearly disjoint over $k$.[/guided]

[step:Prove (ii) $\Rightarrow$ (iii): linear disjointness from $k^{1/p^\infty}$ implies $K \otimes_k k^{1/p}$ is a domain]Assume $K$ and $k^{1/p^\infty}$ are linearly disjoint over $k$. In particular, $K$ and $k^{1/p}$ are linearly disjoint over $k$ (since $k^{1/p} \subset k^{1/p^\infty}$, linear disjointness from $k^{1/p^\infty}$ implies linear disjointness from $k^{1/p}$). Linear disjointness of $K$ and $k^{1/p}$ over $k$ means the natural map \begin{align*} \mu: K \otimes_k k^{1/p} &\to K \cdot k^{1/p} \subset \overline{k} \\ \alpha \otimes \beta &\mapsto \alpha \beta \end{align*} is injective, where $K \cdot k^{1/p}$ denotes the composite field inside $\overline{k}$. Since $K \cdot k^{1/p}$ is a field (being a subfield of $\overline{k}$), and $\mu$ is an injective ring homomorphism from $K \otimes_k k^{1/p}$ into a field, the ring $K \otimes_k k^{1/p}$ is a domain. Moreover, $K \otimes_k k^{1/p}$ is a finitely generated $k^{1/p}$-algebra that is a domain. Since $K/k$ is finitely generated (as a field extension), $K \otimes_k k^{1/p}$ is a finitely generated $k^{1/p}$-algebra. As an integral domain that is a finitely generated algebra over a field with the same Krull dimension as $K/k$, it is in fact a field (its fraction field equals $K \cdot k^{1/p}$, and the tensor product, being a domain of the correct dimension, equals this fraction field).[/step]

[guided]The key concept here is the relationship between linear disjointness and tensor products. Two field extensions $K/k$ and $L/k$ (inside a common overfield) are linearly disjoint over $k$ if and only if the natural multiplication map $K \otimes_k L \to K \cdot L$ is injective. This is a standard equivalence that connects the algebraic notion of linear disjointness to the ring-theoretic property of the tensor product. Since $K$ and $k^{1/p^\infty}$ are linearly disjoint over $k$, and $k^{1/p} \subset k^{1/p^\infty}$, the fields $K$ and $k^{1/p}$ are also linearly disjoint over $k$ (linear disjointness from a larger extension implies linear disjointness from any subextension). Therefore the multiplication map $\mu: K \otimes_k k^{1/p} \to K \cdot k^{1/p}$ is injective. The composite $K \cdot k^{1/p}$ is a subfield of $\overline{k}$, so it is a domain (in fact, a field). An injective ring homomorphism into a domain has a domain as its source, so $K \otimes_k k^{1/p}$ is a domain. To see that $K \otimes_k k^{1/p}$ is actually a field (not just a domain): $K/k$ is finitely generated, so write $K = k(t_1, \ldots, t_r, \gamma)$ where $t_1, \ldots, t_r$ is a transcendence basis and $\gamma$ is algebraic over $k(t_1, \ldots, t_r)$. Then $K \otimes_k k^{1/p} \cong k^{1/p}(t_1, \ldots, t_r)[x] / (f(x))$ where $f = \operatorname{min}(\gamma, k(t_1, \ldots, t_r))$. If $K \otimes_k k^{1/p}$ is a domain, then $f$ is irreducible over $k^{1/p}(t_1, \ldots, t_r)$, so the quotient is a field.[/guided]

[step:Prove (iii) $\Rightarrow$ (i): $K \otimes_k k^{1/p}$ being a domain implies separable generation]Assume $K \otimes_k k^{1/p}$ is a domain. We must find a transcendence basis $T$ of $K/k$ such that $K/k(T)$ is separable. Let $r = \operatorname{trdeg}_k K$. We proceed by induction on $r$. **Base case $r = 0$.** Then $K/k$ is a finite algebraic extension. We show $K/k$ is separable. Suppose not: then there exists $\alpha \in K$ with inseparable minimal polynomial over $k$. By the [Criterion for Inseparability of Irreducible Polynomials](/theorems/3330), $\operatorname{min}(\alpha, k) = g(x^p)$ for some $g \in k[x]$. Let $d = \deg g$, so $\operatorname{min}(\alpha, k)$ has degree $dp$. Write $f(x) = \operatorname{min}(\alpha, k) = \sum_{i=0}^{d} c_i x^{ip}$ where $c_d = 1$. In $K \otimes_k k^{1/p}$, consider the element $\alpha \otimes 1$. Its minimal polynomial over $1 \otimes k^{1/p} \cong k^{1/p}$ divides $f(x) = \sum_i c_i x^{ip} = \sum_i (c_i^{1/p})^p x^{ip} = \left(\sum_i c_i^{1/p} x^i\right)^p$ in $k^{1/p}[x]$ (writing each $c_i = (c_i^{1/p})^p$ with $c_i^{1/p} \in k^{1/p}$). Set $h(x) = \sum_i c_i^{1/p} x^i \in k^{1/p}[x]$. Then $f(x) = h(x)^p$. Now $K \otimes_k k^{1/p}$ contains a copy of $k(\alpha) \otimes_k k^{1/p} \cong k^{1/p}[x]/(f(x)) = k^{1/p}[x]/(h(x)^p)$. If $h$ has a non-trivial factor, or even if $h$ is irreducible, the ring $k^{1/p}[x]/(h(x)^p)$ contains the nilpotent element $h(\bar{x})$ (where $\bar{x}$ is the image of $x$) satisfying $h(\bar{x})^p = f(\bar{x}) = 0$ but $h(\bar{x}) \neq 0$ (since $\deg h = d < dp = \deg f$). A ring with a non-zero nilpotent element is not a domain, contradicting the assumption. Therefore $K/k$ has no inseparable elements, i.e., $K/k$ is separable. The empty set is a separating transcendence basis. **Inductive step.** Assume $r \ge 1$ and the result holds for extensions of transcendence degree $< r$. Write $K = k(x_1, \ldots, x_n)$ for some finite generating set. Among $x_1, \ldots, x_n$, at least one is transcendental over $k$ (since $\operatorname{trdeg}_k K = r \ge 1$). WLOG $x_1$ is transcendental over $k$. [claim:Setting $t = x_1$ (transcendental over $k$) and $k_1 = k(t)$, the tensor product $K \otimes_{k_1} k_1^{1/p}$ is a domain] We must show the natural multiplication map $K \otimes_{k_1} k_1^{1/p} \to K \cdot k_1^{1/p}$ is injective, where the composite is taken inside $\overline{k}$. [/claim] [proof] Set $k_1 = k(t)$ with $t = x_1$, so $\operatorname{trdeg}_{k_1} K = r - 1$. We have $k_1^{1/p} = k^{1/p}(t^{1/p})$. Since $K \otimes_k k^{1/p}$ is a domain, the multiplication map $K \otimes_k k^{1/p} \to K \cdot k^{1/p}$ is injective, so $K$ and $k^{1/p}$ are linearly disjoint over $k$. Denote the composite field $L = K \cdot k^{1/p}$ (a subfield of $\overline{k}$). We show the multiplication map $\mu: K \otimes_{k_1} k_1^{1/p} \to K \cdot k_1^{1/p}$ is injective by verifying that $K$ and $k_1^{1/p}$ are linearly disjoint over $k_1$. Since $k_1^{1/p} = k^{1/p}(t^{1/p})$, it suffices to show that $K$ and $k^{1/p}(t^{1/p})$ are linearly disjoint over $k(t)$. We verify this in two stages using transitivity of linear disjointness. **Stage 1.** $K$ and $k^{1/p}$ are linearly disjoint over $k$ (established above). Since $t \in K$ is transcendental over $k$, and $k(t) \subset K$, the tower $k \subset k(t) \subset K$ together with the external field $k^{1/p}$ satisfies the hypotheses for transitivity: $k(t)$ and $k^{1/p}$ are linearly disjoint over $k$ (by the argument from the previous step — $t$ remains algebraically independent over $k^{1/p}$), and $K$ and $k^{1/p} \cdot k(t) = k^{1/p}(t)$ are linearly disjoint over $k(t)$ (this follows from $K$ and $k^{1/p}$ being linearly disjoint over $k$ by transitivity of linear disjointness applied to $k \subset k(t) \subset K$). **Stage 2.** We must extend from $k^{1/p}(t)$ to $k^{1/p}(t^{1/p}) = k_1^{1/p}$. The element $t^{1/p}$ has minimal polynomial $x^p - t$ over $k^{1/p}(t)$, so $[k^{1/p}(t^{1/p}) : k^{1/p}(t)] = p$ (since $x^p - t$ is irreducible over $k^{1/p}(t)$ by Eisenstein's criterion at the prime $t$ in $k^{1/p}[t]$). The extension $k^{1/p}(t^{1/p}) / k^{1/p}(t)$ is purely inseparable of degree $p$. From Stage 1, $K$ and $k^{1/p}(t)$ are linearly disjoint over $k(t)$. The extension $K/k(t)$ is finitely generated, so write $K = k(t)(x_2, \ldots, x_n)$. We need to show $K$ and $k^{1/p}(t^{1/p})$ are linearly disjoint over $k(t)$. Consider the composite $K \cdot k^{1/p}(t^{1/p})$ inside $\overline{k}$. We have $K \cdot k^{1/p}(t^{1/p}) = L(t^{1/p})$ where $L = K \cdot k^{1/p}$. Now $t^{1/p}$ satisfies $x^p - t = 0$ over $L$, and either $t^{1/p} \in L$ (in which case $L(t^{1/p}) = L$) or $[L(t^{1/p}) : L] = p$. In either case, $L(t^{1/p})$ is a field. The multiplication map $K \otimes_{k_1} k_1^{1/p} \to L(t^{1/p})$ factors as $K \otimes_{k_1} k_1^{1/p} \to K \cdot k_1^{1/p} \hookrightarrow L(t^{1/p})$. To show injectivity of $\mu$, suppose $\sum_{i=1}^m \alpha_i \otimes \beta_i = 0$ maps to $\sum_i \alpha_i \beta_i = 0$ in $L(t^{1/p})$, where $\alpha_i \in K$ and $\beta_i \in k_1^{1/p} = k^{1/p}(t^{1/p})$. Write $\beta_i = \sum_{j=0}^{p-1} b_{ij} (t^{1/p})^j$ with $b_{ij} \in k^{1/p}(t)$. Then $\sum_i \alpha_i \beta_i = \sum_{j=0}^{p-1} (t^{1/p})^j \sum_i \alpha_i b_{ij} = 0$ in $L(t^{1/p})$. Since $\{1, t^{1/p}, \ldots, (t^{1/p})^{p-1}\}$ are linearly independent over $L$ (either $t^{1/p} \in L$ and $p = 1$, or $[L(t^{1/p}):L] = p$ and they form a basis), we get $\sum_i \alpha_i b_{ij} = 0$ in $L$ for each $j$. Since $K$ and $k^{1/p}(t)$ are linearly disjoint over $k(t)$ (Stage 1), and $\alpha_i \in K$, $b_{ij} \in k^{1/p}(t)$, these relations force each $\sum_i \alpha_i \otimes b_{ij} = 0$ in $K \otimes_{k(t)} k^{1/p}(t)$. Reassembling, $\sum_i \alpha_i \otimes \beta_i = 0$ in $K \otimes_{k_1} k_1^{1/p}$. So $\mu$ is injective and $K \otimes_{k_1} k_1^{1/p}$ is a domain. [/proof] By the inductive hypothesis applied to $K/k_1$ (which has transcendence degree $r - 1$ and satisfies $K \otimes_{k_1} k_1^{1/p}$ is a domain), there exists a transcendence basis $\{t_2, \ldots, t_r\}$ of $K/k_1$ such that $K/k_1(t_2, \ldots, t_r)$ is separable. Then $T = \{t, t_2, \ldots, t_r\}$ is a transcendence basis of $K/k$ and $K/k(T) = K/k_1(t_2, \ldots, t_r)$ is separable, so $T$ is a separating transcendence basis.[/step]

[guided]The implication (iii) $\Rightarrow$ (i) is the heart of Mac Lane's theorem. The idea is that the condition "$K \otimes_k k^{1/p}$ is a domain" prevents any purely inseparable behaviour in the algebraic part of the extension, forcing the existence of a separating transcendence basis. **Base case ($r = 0$).** When $K/k$ is algebraic and finitely generated, hence finite, we need to show $K/k$ is separable. The tensor product $K \otimes_k k^{1/p}$ encodes separability information. If $\alpha \in K$ had inseparable minimal polynomial $f(x) = g(x^p)$ over $k$, then by the [Criterion for Inseparability of Irreducible Polynomials](/theorems/3330), we can write $f(x) = g(x^p)$ with $g \in k[x]$, $\deg g = d$. Over $k^{1/p}$, the polynomial $f(x)$ becomes a perfect $p$-th power: \begin{align*} f(x) = \sum_{i=0}^d c_i x^{ip} = \left(\sum_{i=0}^d c_i^{1/p} x^i\right)^p = h(x)^p \end{align*} where $h(x) = \sum_i c_i^{1/p} x^i \in k^{1/p}[x]$. This uses the Frobenius identity in characteristic $p$: $(a + b)^p = a^p + b^p$. The subring $k(\alpha) \otimes_k k^{1/p} \cong k^{1/p}[x]/(f(x)) = k^{1/p}[x]/(h(x)^p)$ contains the non-zero element $\bar{h} = h(\bar{x})$, where $\bar{x}$ is the residue of $x$. Since $\deg h = d < dp = \deg f$, we have $\bar{h} \neq 0$ in $k^{1/p}[x]/(h^p)$. But $\bar{h}^p = h(\bar{x})^p = f(\bar{x}) = 0$. So $\bar{h}$ is a non-zero nilpotent, and $k^{1/p}[x]/(h^p)$ is not a domain. Since $k(\alpha) \otimes_k k^{1/p}$ is a quotient ring of $K \otimes_k k^{1/p}$ (via the surjection induced by $K \twoheadrightarrow k(\alpha)$... actually $k(\alpha) \subset K$, so $k(\alpha) \otimes_k k^{1/p}$ is a subring of $K \otimes_k k^{1/p}$). A subring of a domain is a domain, contradicting the nilpotent. So no inseparable element exists. **Inductive step ($r \ge 1$).** We choose a transcendental element $t = x_1 \in K$ over $k$ and set $k_1 = k(t)$. The extension $K/k_1$ has transcendence degree $r - 1$. The critical step is to verify that $K \otimes_{k_1} k_1^{1/p}$ remains a domain, so the inductive hypothesis applies. Since $K \otimes_k k^{1/p}$ is a domain, $K$ and $k^{1/p}$ are linearly disjoint over $k$. We need to promote this to: $K$ and $k_1^{1/p} = k^{1/p}(t^{1/p})$ are linearly disjoint over $k_1 = k(t)$. We do this in two stages. *Stage 1: $K$ and $k^{1/p}(t)$ are linearly disjoint over $k(t)$.* Since $t$ is transcendental over $k$, it remains transcendental over $k^{1/p}$ (if $t$ satisfied a polynomial relation over $k^{1/p}$, raising to the $p$-th power gives a relation over $k$, contradicting transcendence). So $k(t)$ and $k^{1/p}$ are linearly disjoint over $k$. By transitivity of linear disjointness applied to the tower $k \subset k(t) \subset K$ with external field $k^{1/p}$: we have $K$ and $k^{1/p}$ linearly disjoint over $k$ (given) and $k(t)$ and $k^{1/p}$ linearly disjoint over $k$ (just shown), which gives $K$ and $k^{1/p}(t)$ linearly disjoint over $k(t)$. *Stage 2: Extend from $k^{1/p}(t)$ to $k^{1/p}(t^{1/p})$.* The element $t^{1/p}$ satisfies $x^p - t = 0$ over $k^{1/p}(t)$, which is irreducible by Eisenstein's criterion at the prime $t$ in $k^{1/p}[t]$, so $[k^{1/p}(t^{1/p}) : k^{1/p}(t)] = p$. Set $L = K \cdot k^{1/p}$ (a field, since $K$ and $k^{1/p}$ are linearly disjoint). Then $K \cdot k_1^{1/p} = L(t^{1/p})$, and $t^{1/p}$ either lies in $L$ (so $L(t^{1/p}) = L$) or satisfies $[L(t^{1/p}) : L] = p$. In either case $L(t^{1/p})$ is a field, and $\{1, t^{1/p}, \ldots, (t^{1/p})^{p-1}\}$ are linearly independent over $L$. Now suppose $\sum_i \alpha_i \beta_i = 0$ in $L(t^{1/p})$ with $\alpha_i \in K$, $\beta_i \in k_1^{1/p} = k^{1/p}(t^{1/p})$. Write $\beta_i = \sum_{j=0}^{p-1} b_{ij}(t^{1/p})^j$ with $b_{ij} \in k^{1/p}(t)$. Then $\sum_j (t^{1/p})^j \sum_i \alpha_i b_{ij} = 0$. Linear independence of the powers of $t^{1/p}$ over $L$ gives $\sum_i \alpha_i b_{ij} = 0$ for each $j$. By Stage 1 ($K$ and $k^{1/p}(t)$ linearly disjoint over $k(t)$), each relation forces the corresponding tensor $\sum_i \alpha_i \otimes b_{ij} = 0$ in $K \otimes_{k(t)} k^{1/p}(t)$. Reassembling, $\sum_i \alpha_i \otimes \beta_i = 0$ in $K \otimes_{k_1} k_1^{1/p}$, proving injectivity. So $K \otimes_{k_1} k_1^{1/p}$ is a domain. By the inductive hypothesis, there exists a separating transcendence basis $\{t_2, \ldots, t_r\}$ for $K/k_1$ (i.e., $K/k_1(t_2, \ldots, t_r)$ is separable). Then $\{t, t_2, \ldots, t_r\}$ is a transcendence basis for $K/k$ and $K/k(t, t_2, \ldots, t_r) = K/k_1(t_2, \ldots, t_r)$ is separable, completing the induction.[/guided]