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  "body": "\n## Definition\n\nLet $K$ be a field and let $E$ and $F$ be [[field extensions|0WQ1A8SHTZ]] of $K$.\nThe *tensor product* $E \\otimes_K F$ is the [[tensor product|0G31VJ5YS5]] of the $K$-vector spaces $E$ and $F$, with the unique $K$-bilinear multiplication such that\n$$(a \\otimes b)(c \\otimes d) = ac \\otimes bd \\quad \\text{for } a, c \\in E \\text{ and } b, d \\in F.$$\nIt is a [[commutative ring|0GYGHKX834]] with identity $1 \\otimes 1$ and a $K$-algebra.\n\n## Remarks\n\nThe maps $a \\mapsto a \\otimes 1$ and $b \\mapsto 1 \\otimes b$ embed $E$ and $F$ into $E \\otimes_K F$.\nIf $E = K(\\alpha)$ is finite and $f$ is the [[minimal polynomial|0GK1WB9R98]] of $\\alpha$ over $K$, then $E \\otimes_K F \\cong F[x]/(f)$.\nIf moreover $f = g_1 \\cdots g_r$ with distinct monic irreducible $g_i \\in F[x]$, the Chinese remainder theorem gives $E \\otimes_K F \\cong \\prod_{i=1}^r F[x]/(g_i)$, a product of fields.\n\nExample: $\\C \\otimes_\\R \\C \\cong \\C \\times \\C$.\nCheck: $\\C = \\R(i)$ with $f = x^2 + 1$, and over $F = \\C$ one has $x^2 + 1 = (x - i)(x + i)$, so $\\C \\otimes_\\R \\C \\cong \\C[x]/(x - i) \\times \\C[x]/(x + i) \\cong \\C \\times \\C$.\n\nNon-example: $E \\otimes_K F$ need not be a field, even when $E$ and $F$ are; $\\C \\otimes_\\R \\C$ has the zero divisors $(1, 0)$ and $(0, 1)$ in the product above.\n",
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