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  "body": "\n## Proof\n\nWe treat finite and infinite base fields separately. For infinite $K$ we first handle two generators, then argue by induction.\n\n1. Let $L/K$ be a [[finite|0NRYPM9Y27]] [[separable extension|0N019F6RE9]], and let $n = [L : K]$.\n2. Suppose first that $K$ is a [[finite field|08RTDHX1SK]] with $q$ elements. As a vector space, $L$ is isomorphic to $K^n$, so $L$ has $q^n$ elements and is a finite field.\n3. By [[the cyclicity of the multiplicative group of a finite field|0EBZ3G455G]], there is $\\theta \\in L$ such that every nonzero element of $L$ is a power $\\theta^k$.\n4. Each power $\\theta^k$ lies in the [[field|0MA5B744K3]] $K(\\theta)$, and so does $0$. Hence $L = K(\\theta)$, and $L/K$ is [[simple|0P6SBR7YM2]].\n5. From now on, suppose that $K$ is infinite. We first prove a lemma: for $\\alpha, \\beta \\in L$, there is $\\theta \\in L$ with $K(\\alpha, \\beta) = K(\\theta)$.\n6. Since [[finite extensions are algebraic|0DFYTNZ6AM]], $\\alpha$ and $\\beta$ have [[minimal polynomials|0GK1WB9R98]] $f$ and $g$ over $K$. Both are monic and nonzero.\n7. Since $L/K$ is separable, $g$ is a [[separable polynomial|03J8BS7FMV]]. So there are $a, b \\in K[x]$ with $a g + b g' = 1$, where $g'$ is the formal derivative.\n8. Let $M$ be a [[splitting field|02NZZ1J0FH]] of $fg$ over $L$, which [[exists|01441KDGBY]]. Then $f$ and $g$ [[split|0SHMC00RJV]] over $M$.\n9. Write $f = (x - \\alpha_1) \\cdots (x - \\alpha_s)$ and $g = (x - \\beta_1) \\cdots (x - \\beta_t)$ in $M[x]$. Since $f(\\alpha) = 0$ and $g(\\beta) = 0$, we may number the roots so that $\\alpha_1 = \\alpha$ and $\\beta_1 = \\beta$.\n10. The roots $\\beta_1, \\dots, \\beta_t$ are distinct. Indeed, if $\\beta_j = \\beta_k$ with $j \\ne k$, then $(x - \\beta_j)^2$ divides $g$. So $\\beta_j$ has [[multiplicity|03XVP8Q4GY]] at least $2$.\n11. Then $g(\\beta_j) = 0$ and $g'(\\beta_j) = 0$, by [[the derivative criterion for repeated roots|067F8YVYMH]]. Evaluating $a g + b g' = 1$ at $\\beta_j$ gives $0 = 1$, a contradiction.\n12. For $1 \\le i \\le s$ and $2 \\le j \\le t$, step 10 gives $\\beta \\ne \\beta_j$. So the equation $\\alpha_i + y \\beta_j = \\alpha + y \\beta$ has exactly one solution $y = (\\alpha_i - \\alpha)/(\\beta - \\beta_j)$ in $M$.\n13. These are finitely many elements of $M$, and $K$ is infinite. So we can choose $c \\in K$ different from all of them. Then $\\alpha_i + c \\beta_j \\ne \\alpha + c \\beta$ for all $i$ and all $j \\ge 2$.\n14. Let $\\theta = \\alpha + c \\beta$ and $F = K(\\theta)$. Since $\\theta \\in K(\\alpha, \\beta)$, we have $F \\subseteq K(\\alpha, \\beta)$.\n15. Let $h(x) = f(\\theta - c x)$. Its coefficients lie in $F$, because $f \\in K[x]$, $c \\in K$ and $\\theta \\in F$. Also $h(\\beta) = f(\\alpha) = 0$.\n16. The ring $F[x]$ is a [[Euclidean domain|0DQG39K2NH]]. By [[Bézout's identity|0VY3WPP9Y3]], $g$ and $h$ have a [[greatest common divisor|99J1WXG3F7]] $d \\in F[x]$ with $d = u g + v h$ for some $u, v \\in F[x]$.\n17. Since $d$ divides $g \\ne 0$, $d \\ne 0$. Dividing by its leading coefficient, we may assume that $d$ is monic. This keeps $d$ a combination of $g$ and $h$.\n18. Evaluating at $\\beta$ gives $d(\\beta) = u(\\beta) g(\\beta) + v(\\beta) h(\\beta) = 0$. So $d$ is not constant.\n19. Let $2 \\le j \\le t$. If $d(\\beta_j) = 0$, then $h(\\beta_j) = 0$, since $d$ divides $h$. By step 9, $f(\\theta - c\\beta_j) = 0$ forces $\\theta - c \\beta_j = \\alpha_i$ for some $i$.\n20. That would give $\\alpha_i + c \\beta_j = \\alpha + c \\beta$, against step 13. So $d(\\beta_j) \\ne 0$ for all $j \\ge 2$.\n21. Write $g = d e$ with $e \\in F[x]$; then $e \\ne 0$. For $j \\ge 2$, $0 = g(\\beta_j) = d(\\beta_j) e(\\beta_j)$ in the field $M$, so $e(\\beta_j) = 0$.\n22. So $e$ has the $t - 1$ distinct roots $\\beta_2, \\dots, \\beta_t$ in $M$. By [[the bound on the number of roots|0KJVQV3E0F]], $\\deg e \\ge t - 1$.\n23. Hence $\\deg d = \\deg g - \\deg e \\le 1$. With steps 17 and 18, $d$ is monic of degree $1$ with root $\\beta$. So $d = x - \\beta$.\n24. Since $d \\in F[x]$, we get $\\beta \\in F$. Then $\\alpha = \\theta - c \\beta \\in F$.\n25. The field $K(\\alpha, \\beta)$ is the smallest subfield of $L$ containing $K$, $\\alpha$ and $\\beta$. Hence $K(\\alpha, \\beta) \\subseteq F$, and with step 14, $K(\\alpha, \\beta) = K(\\theta)$. This proves the lemma.\n26. Let $w_1, \\dots, w_n$ be a basis of $L$ over $K$. Every subfield of $L$ containing $K$ and all $w_k$ contains all their $K$-linear combinations. So $L = K(w_1, \\dots, w_n)$.\n27. We show by induction on $k$ that $K(w_1, \\dots, w_k) = K(\\theta_k)$ for some $\\theta_k \\in L$. For $k = 1$, take $\\theta_1 = w_1$.\n28. Suppose $K(w_1, \\dots, w_k) = K(\\theta_k)$ with $k \u003c n$. A subfield of $L$ containing $K$ contains $w_1, \\dots, w_{k+1}$ if and only if it contains $\\theta_k$ and $w_{k+1}$. So $K(w_1, \\dots, w_{k+1}) = K(\\theta_k, w_{k+1})$.\n29. By the lemma, $K(\\theta_k, w_{k+1}) = K(\\theta_{k+1})$ for some $\\theta_{k+1} \\in L$. This completes the induction.\n30. For $k = n$, step 26 gives $L = K(\\theta_n)$. So $L/K$ is a simple extension.\n\n## Idea\n\nFor a finite field, a generator of the multiplicative group already generates everything. For an infinite field, we try $\\theta = \\alpha + c\\beta$ and choose $c$ to avoid finitely many bad values. Then $\\beta$ is the only common root of $g(x)$ and $f(\\theta - cx)$, so their gcd over $K(\\theta)$ is $x - \\beta$. This puts $\\beta$, and then $\\alpha$, inside $K(\\theta)$.\n",
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