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  "body": "\n## Proof\n\n1. Let $L/K$ be a [[field extension|0WQ1A8SHTZ]], let $a \\in L$ be [[algebraic|0NPW1DWECM]] over $K$, and let $m_a$ be its [[minimal polynomial|0GK1WB9R98]]. Let $f \\in K[x]$ with $f(a) = 0$.\n2. The polynomial $m_a$ is [[monic|0BDX4MBKT4]]. By [[polynomial division|0MES3KS0ZC]], there are $q, r \\in K[x]$ with $f = q m_a + r$ and $\\deg r \u003c \\deg m_a$.\n3. [[Evaluation at $a$|0125QVNCFQ]] is a ring homomorphism. So $f(a) = q(a) m_a(a) + r(a)$, and $0 = q(a) \\cdot 0 + r(a)$. Hence $r(a) = 0$.\n4. Suppose that $r \\ne 0$, and let $c \\ne 0$ be its leading coefficient. Then $c^{-1} r$ is monic, vanishes at $a$, and has degree less than $\\deg m_a$.\n5. This contradicts the minimality of $\\deg m_a$. So $r = 0$.\n6. Then $f = q m_a$, so $m_a$ [[divides|04GNK7K4VP]] $f$ in $K[x]$.\n\n## Idea\n\nDivide $f$ by $m_a$. The remainder also vanishes at $a$ and has smaller degree than $m_a$. Since $m_a$ has the least degree among monic polynomials with root $a$, the remainder must be zero.\n",
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        "url": "https://en.wikipedia.org/wiki/Minimal_polynomial_(field_theory)"
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  "name": "Proof that the minimal polynomial divides every polynomial vanishing at the element",
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