[proofplan] The proof proceeds in four stages. First, for any $\alpha \in L$, we construct the orbit polynomial $f_\alpha(t) = \prod_{\sigma \in H} (t - \sigma(\alpha))$ and show it lies in $L^H[t]$, is separable, and annihilates $\alpha$, yielding the degree bound $[L^H(\alpha) : L^H] \le |H|$ together with separability of $\alpha$ over $L^H$. Second, we choose $\alpha \in L$ maximising $[L^H(\alpha) : L^H]$ (the bound $|H|$ guarantees this maximum exists) and show $L = L^H(\alpha)$ by a primitive-element-type argument: if some $\beta \in L$ were not in $L^H(\alpha)$, the [Linear Independence of Characters (Dedekind)](/theorems/1282) applied to the restrictions of $H$ to $L^H(\alpha, \beta)$ would produce an element $\alpha + c\beta$ generating a strictly larger subextension, contradicting maximality. Third, the equality $L = L^H(\alpha)$ gives $[L : L^H] \le |H|$, while Dedekind's theorem gives the reverse inequality $|H| \le [L : L^H]$ (distinct automorphisms are linearly independent over $L$, hence $|H| \le \dim_{L^H} L = [L : L^H]$). This forces $[L : L^H] = |H|$. Fourth, the inclusion $H \subset \operatorname{Gal}(L/L^H)$ combined with $|\operatorname{Gal}(L/L^H)| \le [L : L^H] = |H|$ gives $\operatorname{Gal}(L/L^H) = H$, and since every element of $L$ is a root of the separable polynomial $f_\alpha \in L^H[t]$, the extension $L/L^H$ is both separable and normal, hence Galois. [/proofplan]

[step:Construct the orbit polynomial $f_\alpha$ and show every element of $L$ is separable and algebraic of bounded degree over $L^H$]Write $K := L^H = \{a \in L : \sigma(a) = a \text{ for all } \sigma \in H\}$ and $n := |H|$. Fix an arbitrary $\alpha \in L$. Define the orbit polynomial \begin{align*} f_\alpha(t) := \prod_{\sigma \in H} (t - \sigma(\alpha)) \in L[t]. \end{align*} This is a monic polynomial of degree at most $n$ in $L[t]$ (it has degree exactly $n$ if all $\sigma(\alpha)$ are distinct, and lower degree if some coincide, but we write $\deg f_\alpha \le n$ for now). **The coefficients of $f_\alpha$ lie in $K$.** For any $\tau \in H$, the map $\sigma \mapsto \tau \circ \sigma$ is a bijection $H \to H$ (since $H$ is a group). Therefore \begin{align*} \tau(f_\alpha(t)) &= \prod_{\sigma \in H} (t - \tau(\sigma(\alpha))) = \prod_{\sigma' \in H} (t - \sigma'(\alpha)) = f_\alpha(t), \end{align*} where we substituted $\sigma' = \tau \circ \sigma$ and used that $\tau$ acts as the identity on $t$. Since every $\tau \in H$ fixes each coefficient of $f_\alpha$, each coefficient lies in $L^H = K$. Hence $f_\alpha \in K[t]$. **Separability of $\alpha$ over $K$.** The minimal polynomial $P_\alpha := \operatorname{min}_K(\alpha)$ divides $f_\alpha$ in $K[t]$ (since $f_\alpha \in K[t]$ and $f_\alpha(\alpha) = 0$). We show $P_\alpha$ has no repeated roots. Write $H \cdot \alpha := \{\sigma(\alpha) : \sigma \in H\}$ for the orbit of $\alpha$ (as a set, not a multiset). The polynomial \begin{align*} g(t) := \prod_{\beta \in H \cdot \alpha} (t - \beta) \end{align*} is square-free (its roots are the distinct elements of $H \cdot \alpha$), and $f_\alpha$ divides a power of $g$ (since every root of $f_\alpha$ lies in $H \cdot \alpha$). More precisely, $g$ divides $f_\alpha$ in $L[t]$ (since every element of $H \cdot \alpha$ is a root of $f_\alpha$), and both $g$ and $f_\alpha$ have the same root set $H \cdot \alpha$. Since $P_\alpha$ is irreducible over $K$ and divides $f_\alpha$, every root of $P_\alpha$ lies in $H \cdot \alpha$. The polynomial $P_\alpha$ therefore divides $g$ in the splitting field, and since $g$ is square-free, $P_\alpha$ has no repeated roots. Therefore $\alpha$ is separable over $K$. **Degree bound.** Since $P_\alpha$ divides $f_\alpha$ and $\deg f_\alpha \le n$, we have \begin{align*} [K(\alpha) : K] = \deg P_\alpha \le \deg f_\alpha \le n = |H|. \end{align*}[/step]

[guided]The starting point is to produce, for each element $\alpha \in L$, a polynomial over the fixed field $K = L^H$ that $\alpha$ satisfies. The natural candidate is the **orbit polynomial**: apply every automorphism in $H$ to $\alpha$, and form the product over the resulting linear factors. Define \begin{align*} f_\alpha(t) := \prod_{\sigma \in H} (t - \sigma(\alpha)). \end{align*} This product runs over all $n = |H|$ elements of $H$, so $\deg f_\alpha \le n$ (equality holds if and only if all $\sigma(\alpha)$ are distinct). Why do the coefficients of $f_\alpha$ lie in $K$? The coefficients are the elementary symmetric polynomials in the values $\{\sigma(\alpha) : \sigma \in H\}$. We must check that every $\tau \in H$ fixes these coefficients. Applying $\tau$ to the polynomial (acting on the coefficients), we get \begin{align*} \tau(f_\alpha(t)) = \prod_{\sigma \in H} (t - \tau(\sigma(\alpha))). \end{align*} Here $\tau$ acts as the identity on the formal variable $t$ because $\tau$ is a field automorphism of $L$ and $t$ is transcendental. As $\sigma$ ranges over $H$, the composition $\tau \circ \sigma$ also ranges over $H$ (since $H$ is a group and left multiplication by $\tau$ permutes its elements). Setting $\sigma' = \tau \circ \sigma$: \begin{align*} \tau(f_\alpha(t)) = \prod_{\sigma' \in H} (t - \sigma'(\alpha)) = f_\alpha(t). \end{align*} Since $\tau(f_\alpha) = f_\alpha$ for every $\tau \in H$, each coefficient of $f_\alpha$ is fixed by all of $H$ and hence lies in $L^H = K$. Next, we address separability. The minimal polynomial $P_\alpha = \operatorname{min}_K(\alpha)$ divides $f_\alpha$ in $K[t]$, since $f_\alpha \in K[t]$ is a polynomial vanishing at $\alpha$ and $P_\alpha$ is the unique monic irreducible polynomial in $K[t]$ with this property. Could $P_\alpha$ have a repeated root? Any root of $P_\alpha$ is a root of $f_\alpha$, and the roots of $f_\alpha$ are the elements $\sigma(\alpha)$ for $\sigma \in H$. The square-free part of $f_\alpha$ is $\prod_{\beta \in H \cdot \alpha}(t - \beta)$, where $H \cdot \alpha = \{\sigma(\alpha) : \sigma \in H\}$ is the orbit (a set, not a multiset). Since $P_\alpha$ is irreducible and divides $f_\alpha$, and since all roots of $P_\alpha$ must be roots of $f_\alpha$, the polynomial $P_\alpha$ divides the square-free part $\prod_{\beta \in H \cdot \alpha}(t - \beta)$. A divisor of a square-free polynomial is square-free, so $P_\alpha$ has no repeated roots. Therefore $\alpha$ is separable over $K$. Finally, the degree bound: since $P_\alpha \mid f_\alpha$ in $K[t]$, we have $\deg P_\alpha \le \deg f_\alpha \le n$, so $[K(\alpha) : K] = \deg P_\alpha \le |H|$. This bound is uniform in $\alpha$ and will be essential for finding a primitive element in the next step.[/guided]

[step:Choose $\alpha$ maximising $[K(\alpha) : K]$ and show $L = K(\alpha)$ by a Dedekind-type argument] From the previous step, the set $\{[K(\alpha) : K] : \alpha \in L\}$ is a subset of $\{1, 2, \ldots, n\}$ and is therefore bounded above by $n$. Choose $\alpha \in L$ such that $[K(\alpha) : K]$ is maximal among all elements of $L$. Write $d := [K(\alpha) : K]$. [claim:Maximality forces $L = K(\alpha)$] If $\alpha \in L$ is chosen so that $[K(\alpha) : K] \ge [K(\beta) : K]$ for all $\beta \in L$, then $K(\alpha) = L$. [/claim] [proof] Suppose for contradiction that $K(\alpha) \subsetneq L$. Then there exists $\beta \in L \setminus K(\alpha)$. Since $\beta \notin K(\alpha)$, the tower law gives \begin{align*} [K(\alpha, \beta) : K] = [K(\alpha, \beta) : K(\alpha)] \cdot [K(\alpha) : K] \ge 2d. \end{align*} The extension $K(\alpha, \beta)/K$ is finite (since $[K(\alpha, \beta) : K] \le n^2$, as both $\alpha$ and $\beta$ have degree at most $n$ over $K$). It is separable: every element of $K(\alpha, \beta) \subset L$ is separable over $K$ by the first step (its minimal polynomial over $K$ divides the corresponding orbit polynomial, which is square-free). We apply the [Primitive Element Theorem](/theorems/1267) to $K(\alpha, \beta)/K$. Theorem 1267 requires a finite separable extension, both of which we have verified. It yields $\delta \in K(\alpha, \beta) \subset L$ with $K(\alpha, \beta) = K(\delta)$. Then \begin{align*} [K(\delta) : K] = [K(\alpha, \beta) : K] \ge 2d > d = [K(\alpha) : K], \end{align*} contradicting the maximality of $d$. Therefore $K(\alpha) = L$. [/proof][/step]

[guided]The goal is to show that $L$ is generated over $K = L^H$ by a single element, and we find this element by a maximality argument. From the first step, every $\beta \in L$ satisfies $[K(\beta) : K] \le n = |H|$. This means the set of values $\{[K(\beta) : K] : \beta \in L\}$ is bounded above by $n$, so it contains a maximum. Choose $\alpha \in L$ achieving this maximum: $d := [K(\alpha) : K] \ge [K(\beta) : K]$ for all $\beta \in L$. We claim $L = K(\alpha)$. Suppose not: there exists $\beta \in L \setminus K(\alpha)$. We derive a contradiction by finding $\delta \in L$ with $[K(\delta) : K] > d$. Since $\beta \notin K(\alpha)$, the extension $K(\alpha, \beta)/K(\alpha)$ is non-trivial, so $[K(\alpha, \beta) : K(\alpha)] \ge 2$. By the tower law: \begin{align*} [K(\alpha, \beta) : K] = [K(\alpha, \beta) : K(\alpha)] \cdot [K(\alpha) : K] \ge 2d. \end{align*} Now we apply the [Primitive Element Theorem](/theorems/1267) to $K(\alpha, \beta)/K$. We must verify its hypothesis: $K(\alpha, \beta)/K$ must be a finite separable extension. Finiteness holds because $[K(\alpha, \beta) : K] \le [K(\alpha, \beta) : K(\alpha)] \cdot d$, and both factors are finite (the first because $\beta$ is algebraic over $K(\alpha)$, as $\beta$ is algebraic over $K$ by the first step). Separability holds because every element of $K(\alpha, \beta)$ is algebraic over $K$ and separable over $K$ — the latter was established in the first step (the minimal polynomial of any element of $L$ over $K$ divides a separable orbit polynomial and hence is itself separable). The Primitive Element Theorem therefore gives $K(\alpha, \beta) = K(\delta)$ for some $\delta \in K(\alpha, \beta) \subset L$. Then \begin{align*} [K(\delta) : K] = [K(\alpha, \beta) : K] \ge 2d > d = [K(\alpha) : K]. \end{align*} This contradicts the maximality of $d$. Therefore no such $\beta$ exists, and $L = K(\alpha)$. The key idea is that the Primitive Element Theorem lets us compress a two-generator extension into a one-generator extension, and the two-generator extension is strictly larger than $K(\alpha)$ because $\beta \notin K(\alpha)$. Maximality of $\alpha$ then gives the contradiction.[/guided]

[step:Establish the degree equality $[L : K]= |H|$] From the previous step, $L = K(\alpha)$ for some $\alpha \in L$. From the first step, $[L : K] = [K(\alpha) : K] \le n = |H|$. This gives the upper bound \begin{align*} [L : L^H] \le |H|. \end{align*} For the reverse inequality, we apply the [Linear Independence of Characters (Dedekind)](/theorems/1282). Write $H = \{\sigma_1, \ldots, \sigma_n\}$. Each $\sigma_i$ restricts to a group homomorphism $\sigma_i|_{L^\times} : L^\times \to L^\times$, and these are distinct (since two field automorphisms that agree on $L^\times$ also agree on $0$, hence on all of $L$). Theorem 1282, applied with $G = L^\times$ and the $n$ distinct homomorphisms $\sigma_i|_{L^\times}$, gives: if $\lambda_1, \ldots, \lambda_n \in L$ satisfy $\sum_{i=1}^n \lambda_i \sigma_i(x) = 0$ for all $x \in L$, then $\lambda_1 = \cdots = \lambda_n = 0$. This $L$-linear independence implies $K$-linear independence a fortiori (since $K \subset L$). Each $\sigma_i$ is $K$-linear (as $\sigma_i$ fixes $K$ pointwise), so $\sigma_1, \ldots, \sigma_n$ are $K$-linearly independent elements of $\operatorname{End}_K(L)$. Now $\operatorname{End}_K(L) \cong M_d(K)$ where $d := [L : K] = \dim_K L$, and $\dim_K \operatorname{End}_K(L) = d^2$. The $L$-linear independence gives a stronger constraint: viewing $\operatorname{End}_K(L)$ as a (left) $L$-module via $(a \cdot T)(x) = a \cdot T(x)$, we compute $\dim_L \operatorname{End}_K(L) = d^2 / d = d$ (since $[L:K] = d$). Therefore $n \le d$, i.e., \begin{align*} |H| \le [L : K]. \end{align*} Combining the two inequalities: \begin{align*} [L : L^H] = |H|. \end{align*}[/step]

[guided]We have shown $L = K(\alpha)$ with $[K(\alpha) : K] \le |H|$ from the orbit polynomial bound. We now need the reverse inequality to pin down the degree exactly. **Upper bound (review).** The minimal polynomial $P_\alpha$ divides $f_\alpha$, which has degree at most $n = |H|$. Since $L = K(\alpha)$, $[L : K] = \deg P_\alpha \le n$. **Lower bound via Dedekind.** The [Linear Independence of Characters (Dedekind)](/theorems/1282) provides the reverse inequality. We verify its hypotheses. Write $H = \{\sigma_1, \ldots, \sigma_n\}$. Each $\sigma_i$ restricts to a group homomorphism $\sigma_i|_{L^\times} : L^\times \to L^\times$, and these restrictions are distinct: if $\sigma_i|_{L^\times} = \sigma_j|_{L^\times}$, then $\sigma_i$ and $\sigma_j$ agree on every nonzero element of $L$, and both send $0 \mapsto 0$, so $\sigma_i = \sigma_j$ on all of $L$, forcing $i = j$. Theorem 1282 requires a group $G$ (here $G = L^\times$) and distinct group homomorphisms $G \to L^\times$; both conditions are satisfied. The conclusion of Theorem 1282 is: if $\lambda_1, \ldots, \lambda_n \in L$ satisfy $\sum_{i=1}^n \lambda_i \sigma_i(x) = 0$ for all $x \in L^\times$, then $\lambda_1 = \cdots = \lambda_n = 0$. Since $\sigma_i(0) = 0$ for all $i$, the relation also holds at $x = 0$, so the $\sigma_i$ are $L$-linearly independent as functions $L \to L$. Why does $L$-linear independence give $n \le [L:K]$? Set $d := [L:K]$. Each $\sigma_i$ fixes $K$ pointwise, so $\sigma_i \in \operatorname{End}_K(L)$ (the $K$-algebra of $K$-linear endomorphisms of $L$). Now $\dim_K \operatorname{End}_K(L) = d^2$ (since $L$ is a $d$-dimensional $K$-vector space), so mere $K$-linear independence of the $\sigma_i$ would only give $n \le d^2$ — too weak. The crucial point is that Dedekind gives $L$-linear independence, which is a stronger condition. We view $\operatorname{End}_K(L)$ as a left $L$-module via the action $(a \cdot T)(x) := a \cdot T(x)$ for $a \in L$, $T \in \operatorname{End}_K(L)$, $x \in L$. This is well-defined because scalar multiplication by $a$ is $K$-linear. As a left $L$-module, $\operatorname{End}_K(L)$ has $L$-dimension $d^2/[L:K] = d^2/d = d$ (since $\dim_K \operatorname{End}_K(L) = d^2$ and each $L$-dimension contributes $[L:K] = d$ $K$-dimensions). Since $\sigma_1, \ldots, \sigma_n$ are $L$-linearly independent elements of this $d$-dimensional $L$-module, we conclude $n \le d$, i.e., $|H| \le [L:K]$. Combining: $[L:K] \le |H|$ and $|H| \le [L:K]$ gives $[L : L^H] = |H|$.[/guided]

[step:Show $\operatorname{Gal}(L/L^H) = H$ and $L/L^H$ is Galois]**The Galois group equals $H$.** Every element of $H$ fixes $K = L^H$ pointwise (by definition of $L^H$), so $H \subset \operatorname{Gal}(L/K)$. Any element $\tau \in \operatorname{Gal}(L/K)$ permutes the roots of $P_\alpha = \operatorname{min}_K(\alpha)$, where $L = K(\alpha)$. Since $\tau$ is determined by $\tau(\alpha)$ and $\tau(\alpha)$ must be a root of $P_\alpha$, the number of such automorphisms is at most $\deg P_\alpha = [L : K] = |H|$. Therefore \begin{align*} |H| \le |\operatorname{Gal}(L/K)| \le [L : K] = |H|, \end{align*} where the first inequality uses $H \subset \operatorname{Gal}(L/K)$ and the second uses the root-counting bound. Equality throughout forces $\operatorname{Gal}(L/K) = H$. **The extension $L/K$ is Galois.** We verify separability and normality. *Separability:* Every element of $L$ is separable over $K$, as established in the first step (the minimal polynomial of any element divides a separable orbit polynomial). *Normality:* Since $L = K(\alpha)$, the extension $L/K$ is normal if and only if $P_\alpha$ splits completely over $L$. The roots of $P_\alpha$ are a subset of $\{\sigma(\alpha) : \sigma \in H\}$ (since $P_\alpha$ divides $f_\alpha$ and the roots of $f_\alpha$ are $\sigma(\alpha)$ for $\sigma \in H$). Each $\sigma(\alpha) \in L$ (since $\sigma : L \to L$), so all roots of $P_\alpha$ lie in $L$. Moreover, $\deg P_\alpha = [L:K] = |H| = \deg f_\alpha$ (the last equality holds because $|\operatorname{Gal}(L/K)| = |H|$ and the $|H|$ values $\sigma(\alpha)$ are distinct: if $\sigma(\alpha) = \tau(\alpha)$, then $\sigma = \tau$ since $L = K(\alpha)$ and a $K$-automorphism is determined by its value on $\alpha$). So $P_\alpha = f_\alpha$ and $P_\alpha$ splits into distinct linear factors over $L$. Hence $L/K$ is normal. Since $L/K$ is finite, separable, and normal, $L/K$ is a Galois extension. Together with $\operatorname{Gal}(L/K) = H$ and $[L:K] = |H|$, this completes the proof.[/step]

[guided]We tie together the three conclusions of the theorem. **Part (2): $\operatorname{Gal}(L/L^H) = H$.** The inclusion $H \subset \operatorname{Gal}(L/K)$ is immediate: if $\sigma \in H$, then $\sigma$ fixes every element of $L^H$ by definition, so $\sigma \in \operatorname{Gal}(L/K)$. For the reverse inclusion, we count. Since $L = K(\alpha)$, any $K$-automorphism $\tau$ of $L$ is uniquely determined by $\tau(\alpha)$, which must be a root of $P_\alpha = \operatorname{min}_K(\alpha)$ (because $\tau$ fixes $K$ and hence maps roots of $K$-polynomials to roots of the same polynomial). The polynomial $P_\alpha$ has $\deg P_\alpha = [L : K] = |H|$ roots (counted with multiplicity, but $P_\alpha$ is separable so all roots are distinct). Therefore $|\operatorname{Gal}(L/K)| \le \deg P_\alpha = |H|$. Chaining: $|H| \le |\operatorname{Gal}(L/K)| \le |H|$, so $|\operatorname{Gal}(L/K)| = |H|$. Since $H \subset \operatorname{Gal}(L/K)$ and both sets are finite with the same cardinality, $\operatorname{Gal}(L/K) = H$. **Part (1): $L/K$ is Galois.** A finite extension is Galois if and only if it is both separable and normal. *Separability* was shown in the first step: for any $\beta \in L$, the minimal polynomial $\operatorname{min}_K(\beta)$ divides the square-free orbit polynomial $f_\beta$, so $\operatorname{min}_K(\beta)$ has no repeated roots. *Normality* means every irreducible polynomial in $K[t]$ that has one root in $L$ splits completely in $L$. It suffices to verify this for the minimal polynomial of the primitive element $\alpha$, since $L = K(\alpha)$. The roots of $P_\alpha$ are among $\{\sigma(\alpha) : \sigma \in H\}$, all of which lie in $L$. We showed $\deg P_\alpha = |H| = \deg f_\alpha$, and since $P_\alpha \mid f_\alpha$ with equal degrees, $P_\alpha = f_\alpha$. Thus $P_\alpha$ splits as \begin{align*} P_\alpha(t) = f_\alpha(t) = \prod_{\sigma \in H} (t - \sigma(\alpha)), \end{align*} into $|H|$ distinct linear factors over $L$. In particular, $P_\alpha$ splits completely over $L$. To confirm normality for an arbitrary irreducible $q \in K[t]$ with a root $\beta \in L$: since $L/K$ is separable (already established), $q$ is separable. The splitting field of $q$ over $K$ is contained in $K(\beta) \subset L$ if and only if all roots of $q$ lie in $L$. But the roots of $q = \operatorname{min}_K(\beta)$ are among $\{\sigma(\beta) : \sigma \in H\}$ (since $q \mid f_\beta$), and each $\sigma(\beta) \in L$. So all roots of $q$ lie in $L$, confirming normality. **Part (3): $[L : L^H] = |H|$.** This was established in the previous step. All three conclusions hold simultaneously, completing the proof of Artin's Lemma.[/guided]