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  "body": "\n## Statement\n\nLet $L/K$ be a [[field extension|0WQ1A8SHTZ]] and let $a \\in L$ be algebraic over $K$ with [[minimal polynomial|0GK1WB9R98]] $m_a$.\nThen the [[evaluation map|0125QVNCFQ]] $K[x] \\to L$, $f \\mapsto f(a)$, induces a ring isomorphism\n$$K[x]/(m_a) \\cong K(a).$$\nIn particular, $K(a) = K[a] = \\{ f(a) : f \\in K[x] \\}$.\n\n## Remarks\n\nThe proof: the kernel of the evaluation map is $(m_a)$, by [[the divisibility property|06K7SARZKR]]. The [[first isomorphism theorem|0ZMYX7GMH0]] gives $K[x]/(m_a) \\cong K[a]$.\nSince $m_a$ is irreducible, $K[x]/(m_a)$ is a [[field|0AWMPX35CW]]. So $K[a]$ is a field, and $K[a] = K(a)$.\nIf $a$ is transcendental over $K$, then the evaluation map is injective, and it extends to an isomorphism $K(x) \\cong K(a)$.\n\n## Explanations\n\n### L1 human\n\nAn algebraic element adds nothing to the field beyond the relation given by its minimal polynomial.\nSo $K(a)$ is the same as polynomials in a variable, computed modulo the minimal polynomial.\nFor example, $\\mathbb{Q}(\\sqrt{2}) \\cong \\mathbb{Q}[x]/(x^2 - 2)$.\n",
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