{
  "aliases": null,
  "backlinks": null,
  "body": "\n## Proof\n\n1. Let $K$ be a [[field|0QA8CHH8MG]] and $f \\in K[x]$ with $\\deg f \\ge 1$. The [[units|019NFGFPR5]] of $K[x]$ are the nonzero constants, by the [[degree of a product|0T233AZ1GE]].\n2. Among the [[monic|0BDX4MBKT4]] divisors of $f$ of degree at least $1$, choose one, $g$, of least degree. One exists: $f$ times the inverse of its leading coefficient.\n3. The polynomial $g$ is [[irreducible|00X2A9G9W3]]. It is nonzero and not a unit, since $\\deg g \\ge 1$.\n4. Suppose $g = a b$ with $a, b \\in K[x]$ both of degree at least $1$. Then $\\deg a \u003c \\deg g$, and $a$ times the inverse of its leading coefficient is a monic divisor of $f$. This contradicts the choice of $g$.\n5. So in every factorisation $g = a b$, one factor has degree $0$, so it is a unit. Write $f = g k$ with $k \\in K[x]$.\n6. Let $L = K[x]/(g)$, the [[quotient ring|0VFW4Y69BT]]. Since $g$ is irreducible, [[$L$ is a field|0AWMPX35CW]].\n7. The map $K \\to L$, $c \\mapsto c + (g)$, is a ring homomorphism from a field to a nonzero ring. So it is [[injective|09N7DZ7SDB]]. As in [[adjoining a root|0M893X7YY7]], we identify $K$ with its image, so $L/K$ is a [[field extension|0WQ1A8SHTZ]].\n8. Let $\\alpha = x + (g) \\in L$. For every $h \\in K[x]$, the quotient map is a ring homomorphism that fixes $K$, so $h + (g) = h(\\alpha)$.\n9. In particular $g(\\alpha) = g + (g) = 0$. Then $f(\\alpha) = g(\\alpha) k(\\alpha) = 0$, so $\\alpha$ is a [[root|00EHX6FWC0]] of $f$ in $L$.\n10. Since $g$ is monic, irreducible and $g(\\alpha) = 0$, the [[minimal polynomial|0GK1WB9R98]] $m_\\alpha$ of $\\alpha$ [[divides|06K7SARZKR]] $g$. As $\\deg m_\\alpha \\ge 1$, the cofactor is a unit, and since both are monic, $g = m_\\alpha$.\n11. By Step 8, $L = \\set{ h(\\alpha) : h \\in K[x] }$. By the [[structure of simple extensions|0ZN39QDTP9]], this set is $K(\\alpha)$, so $L = K(\\alpha)$.\n12. By the [[degree of a simple extension|0N5QW9K7MH]], $[L : K] = \\deg m_\\alpha = \\deg g$.\n13. Since $f = g k$ with $f \\ne 0$, we have $k \\ne 0$ and $\\deg f = \\deg g + \\deg k \\ge \\deg g$. So $[L : K] \\le \\deg f$, and $f$ has the root $\\alpha$ in $L$.\n\n## Idea\n\nPick an irreducible factor $g$ of $f$ and force a root of $g$ to exist by working modulo $g$. The quotient $K[x]/(g)$ is a field because $g$ is irreducible, and the class of $x$ is a root of $g$, hence of $f$. Its degree over $K$ is $\\deg g$, which is at most $\\deg f$.\n",
  "commit": "4a3290d299c8b2cd6493fce2296920ba7f82d942",
  "front": {
    "kind": "proof",
    "msc": [
      "12F05"
    ],
    "name": "Proof that every polynomial has a root in some extension field",
    "proves": "0VG02M1FEZ",
    "sources": [
      {
        "locator": "Constructing splitting fields",
        "url": "https://en.wikipedia.org/wiki/Splitting_field"
      },
      {
        "locator": "Chapter 1, Stem fields",
        "url": "https://www.jmilne.org/math/CourseNotes/FT.pdf"
      }
    ],
    "tag": "86ZBRVY9WN",
    "uses": [
      "0QA8CHH8MG",
      "0WQ1A8SHTZ",
      "00EHX6FWC0",
      "00X2A9G9W3",
      "0BDX4MBKT4",
      "0T233AZ1GE",
      "019NFGFPR5",
      "0AWMPX35CW",
      "0M893X7YY7",
      "09N7DZ7SDB",
      "0VFW4Y69BT",
      "0GK1WB9R98",
      "06K7SARZKR",
      "0ZN39QDTP9",
      "0N5QW9K7MH"
    ]
  },
  "kind": "proof",
  "link": "https://math2.tamnd.com/node/86ZBRVY9WN/",
  "links": [
    {
      "field": "proves",
      "target": "0VG02M1FEZ"
    },
    {
      "field": "uses",
      "target": "0QA8CHH8MG"
    },
    {
      "field": "uses",
      "target": "0WQ1A8SHTZ"
    },
    {
      "field": "uses",
      "target": "00EHX6FWC0"
    },
    {
      "field": "uses",
      "target": "00X2A9G9W3"
    },
    {
      "field": "uses",
      "target": "0BDX4MBKT4"
    },
    {
      "field": "uses",
      "target": "0T233AZ1GE"
    },
    {
      "field": "uses",
      "target": "019NFGFPR5"
    },
    {
      "field": "uses",
      "target": "0AWMPX35CW"
    },
    {
      "field": "uses",
      "target": "0M893X7YY7"
    },
    {
      "field": "uses",
      "target": "09N7DZ7SDB"
    },
    {
      "field": "uses",
      "target": "0VFW4Y69BT"
    },
    {
      "field": "uses",
      "target": "0GK1WB9R98"
    },
    {
      "field": "uses",
      "target": "06K7SARZKR"
    },
    {
      "field": "uses",
      "target": "0ZN39QDTP9"
    },
    {
      "field": "uses",
      "target": "0N5QW9K7MH"
    }
  ],
  "markdown": "https://math2.tamnd.com/node/86ZBRVY9WN.md",
  "msc": [
    "12F05"
  ],
  "name": "Proof that every polynomial has a root in some extension field",
  "prov": null,
  "role": "",
  "source": "https://github.com/tamnd/math2/blob/main/nodes/86Z/86ZBRVY9WN.md",
  "status": null,
  "tag": "86ZBRVY9WN",
  "uri": "m2://86ZBRVY9WN",
  "url": "https://math2.tamnd.com/wiki/proof-that-every-polynomial-has-a-root-in-some-extension-field/",
  "writing": {
    "tag": "86ZBRVY9WN",
    "kind": "proof",
    "level": "W0",
    "next": [
      {
        "what": "an inventory row",
        "note": "the row of the inventory that the node accounts for",
        "have": 0,
        "want": 1
      }
    ]
  }
}