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  "body": "\n## Definition\n\nLet $\\Omega$ be a field and let $K$, $E$ and $F$ be [[subfields|07QY5RPFBR]] of $\\Omega$ with $K \\subseteq E$ and $K \\subseteq F$.\nThe fields $E$ and $F$ are *linearly disjoint over $K$* if the following holds.\nEvery finite family of elements of $E$ that is [[linearly independent|0EVHJFTG9Z]] over $K$ is linearly independent over $F$.\n\n## Remarks\n\nThe definition is symmetric in $E$ and $F$.\nIt holds if and only if the $K$-linear map $E \\otimes_K F \\to \\Omega$ with $x \\otimes y \\mapsto xy$ is injective, where $\\otimes_K$ is the [[tensor product|0G31VJ5YS5]].\nIf $E/K$ is finite, it holds if and only if $[EF:F] = [E:K]$, where $EF$ is the [[compositum|X4ZZKC319Q]].\nIf $E/K$ is finite Galois, it holds if and only if $E \\cap F = K$.\n\nExample: $E = \\Q(\\sqrt{2})$ and $F = \\Q(\\sqrt{3})$ in $\\C$ are linearly disjoint over $\\Q$.\nCheck: $EF = \\Q(\\sqrt{2}, \\sqrt{3})$ has degree $4$ over $\\Q$, so $[EF:F] = 2 = [E:\\Q]$.\n\nNon-example: let $\\omega = e^{2\\pi i/3}$, $E = \\Q(\\sqrt[3]{2})$ and $F = \\Q(\\omega\\sqrt[3]{2})$.\nThen $E \\cap F = \\Q$, but $EF = \\Q(\\sqrt[3]{2}, \\omega)$ has degree $6$, so $[EF:F] = 2 \u003c 3 = [E:\\Q]$.\nHence the $\\Q$-basis $1, \\sqrt[3]{2}, \\sqrt[3]{4}$ of $E$ is linearly dependent over $F$.\n",
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