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  "body": "\n## Proof\n\n1. Let $L/K$ be a [[field extension|0WQ1A8SHTZ]], let $a \\in L$ be algebraic over $K$, and let $m_a$ be its [[minimal polynomial|0GK1WB9R98]], of degree $n$.\n2. By the [[structure of simple extensions|0ZN39QDTP9]], $K(a) = \\set{ f(a) : f \\in K[x] }$.\n3. Spanning. Let $z \\in K(a)$, so $z = f(a)$ for some $f \\in K[x]$ by Step 2.\n4. Since $m_a$ is monic, [[polynomial division|0MES3KS0ZC]] gives $f = q m_a + r$ with $\\deg r \u003c n$. Write $r = \\sum_{i=0}^{n-1} c_i x^i$ with $c_i \\in K$.\n5. Then $z = f(a) = q(a) m_a(a) + r(a) = r(a) = \\sum_{i=0}^{n-1} c_i a^i$. So $1, a, \\dots, a^{n-1}$ [[span|088RYKAWK9]] $K(a)$ over $K$.\n6. [[Linear independence|0EVHJFTG9Z]]. Let $c_0, \\dots, c_{n-1} \\in K$ with $\\sum_{i=0}^{n-1} c_i a^i = 0$, and let $r = \\sum_{i=0}^{n-1} c_i x^i$. Then $r(a) = 0$.\n7. By [[divisibility|06K7SARZKR]], $r = m_a s$ for some $s \\in K[x]$. If $s \\ne 0$, then $\\deg r = n + \\deg s \\ge n$ by the [[degree of a product|0T233AZ1GE]], which is false.\n8. So $s = 0$, hence $r = 0$, and all $c_i = 0$.\n9. By Steps 5 and 8, $1, a, \\dots, a^{n-1}$ is a [[basis|0VTQTPQGAV]] of $K(a)$ over $K$.\n10. This basis has $n$ elements, so by the definition of [[dimension|05G2G0P224]], $[K(a) : K] = n$, as a [[degree|0GS74FGFEK]].\n\n## Idea\n\nEvery element of $K(a)$ is a polynomial in $a$, and dividing by $m_a$ cuts its degree below $n$. So the powers $1, a, \\dots, a^{n-1}$ span. A linear relation among them would be a nonzero polynomial of degree less than $n$ with root $a$, which the minimality of $m_a$ forbids.\n",
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