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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]].\n2. Let $\\operatorname{ev} \\colon K[x] \\to L$, $f \\mapsto f(a)$. By the [[universal property of the polynomial ring|0DQMDSVSXJ]], applied to the inclusion $K \\to L$ and $x \\mapsto a$, it is a ring homomorphism. It is the restriction of the [[evaluation homomorphism|0125QVNCFQ]] of $L[x]$ at $a$.\n3. Let $K[a] = \\set{ f(a) : f \\in K[x] }$, the image of $\\operatorname{ev}$.\n4. The [[kernel|0AAHE7GEW6]] of $\\operatorname{ev}$ is $(m_a)$. Indeed, if $f = q m_a$, then $f(a) = q(a) m_a(a) = 0$.\n5. Conversely, if $f(a) = 0$, then [[$m_a$ divides $f$|06K7SARZKR]], so $f \\in (m_a)$.\n6. By the [[first isomorphism theorem for rings|0ZMYX7GMH0]], the map $f + (m_a) \\mapsto f(a)$ is a [[ring isomorphism|0B6QC0C0G1]] from the [[quotient ring|0VFW4Y69BT]] $K[x]/(m_a)$ onto $K[a]$.\n7. The polynomial $m_a$ is [[irreducible|0NFEG48RDR]]. So [[$K[x]/(m_a)$ is a field|0AWMPX35CW]].\n8. By Steps 6 and 7, $K[a]$ is a subring of $L$ that is a field. So every nonzero element of $K[a]$ has its inverse in $K[a]$, and $K[a]$ is a [[subfield|07QY5RPFBR]] of $L$.\n9. The subfield $K[a]$ contains $K$, as the values of the constant polynomials, and contains $a = \\operatorname{ev}(x)$.\n10. The field [[$K(a)$|0MA5B744K3]] is the intersection of all subfields of $L$ that contain $K$ and $a$. By Step 9, $K(a) \\subseteq K[a]$.\n11. Conversely, $K(a)$ is a subfield containing $K$ and $a$. It is closed under sums and products, so it contains every $f(a) = \\sum_i c_i a^i$ with $c_i \\in K$. So $K[a] \\subseteq K(a)$.\n12. Hence $K(a) = K[a] = \\set{ f(a) : f \\in K[x] }$. With Step 6, the map induced by $\\operatorname{ev}$ is a ring isomorphism $K[x]/(m_a) \\cong K(a)$.\n\n## Idea\n\nEvaluation at $a$ is a ring homomorphism whose kernel is generated by the minimal polynomial. The first isomorphism theorem identifies its image $K[a]$ with $K[x]/(m_a)$. Since $m_a$ is irreducible, this quotient is a field, so the ring $K[a]$ is already the field $K(a)$.\n",
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