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  "body": "\n## Proof\n\n1. Let $K \\subseteq L \\subseteq M$ be fields with $M/L$ and $L/K$ [[algebraic|09EYMKMSB9]]. Let $a \\in M$. We show that $a$ is [[algebraic|0NPW1DWECM]] over $K$.\n2. Since $a$ is algebraic over $L$, it has a [[minimal polynomial|0GK1WB9R98]] $m = x^n + c_{n-1} x^{n-1} + \\dots + c_0$ over $L$, with $c_i \\in L$.\n3. Let $E_0 = K$, and for $1 \\le i \\le n$ let $E_i = E_{i-1}(c_{i-1})$, the [[subfield of $M$ generated|0MA5B744K3]] by $E_{i-1}$ and $c_{i-1}$. Then $K = E_0 \\subseteq E_1 \\subseteq \\dots \\subseteq E_n \\subseteq M$.\n4. Each $c_{i-1}$ is algebraic over $K$, since $L/K$ is algebraic. A nonzero polynomial over $K$ with root $c_{i-1}$ is also one over $E_{i-1}$. So $c_{i-1}$ is algebraic over $E_{i-1}$.\n5. By the [[degree of a simple extension|0N5QW9K7MH]], $[E_i : E_{i-1}]$ equals the degree of the minimal polynomial of $c_{i-1}$ over $E_{i-1}$. So it is finite.\n6. All coefficients of $m$ lie in $E_n$, so $m \\in E_n[x]$ is nonzero with $m(a) = 0$. So $a$ is algebraic over $E_n$, and $[E_n(a) : E_n]$ is finite by the same result.\n7. By the [[tower law|0A7KJ68XHT]], applied $n$ times, $[E_n(a) : K] = [E_n(a) : E_n] \\cdot [E_n : E_{n-1}] \\cdots [E_1 : E_0]$. This is a product of finitely many finite numbers, so it is finite.\n8. So $E_n(a)/K$ is a [[finite extension|0NRYPM9Y27]]. Since [[finite extensions are algebraic|0DFYTNZ6AM]], the element $a \\in E_n(a)$ is algebraic over $K$.\n9. Since $a \\in M$ was arbitrary, $M/K$ is algebraic.\n\n## Idea\n\nThe element $a$ satisfies a polynomial with only finitely many coefficients from $L$. Adjoining these coefficients to $K$ gives a finite extension, and $a$ is algebraic over it. The tower law then makes the whole extension containing $a$ finite over $K$, and finite extensions are algebraic.\n",
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