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  "body": "\n## Definition\n\nLet $R$ be a [[commutative ring|0310REJJY8]] and let $A$, $B$ be [[$R$-algebras|DPBJ04JMH8]].\nThe *tensor product* $A \\otimes_R B$ is the [[tensor product of $R$-modules|F0HQXQ0HCD]] with the multiplication determined by\n$$(a \\otimes b)(a' \\otimes b') = aa' \\otimes bb' \\quad \\text{for } a, a' \\in A,\\ b, b' \\in B,$$\nextended bilinearly to finite sums.\nWith this multiplication and identity $1 \\otimes 1$, it is an $R$-algebra, with structure map $r \\mapsto r \\otimes 1 = 1 \\otimes r$.\n\n## Remarks\n\nThe multiplication is well defined because the formula is $R$-bilinear in each pair of arguments.\nThe maps $a \\mapsto a \\otimes 1$ and $b \\mapsto 1 \\otimes b$ are $R$-algebra [[homomorphisms|0VTH8KPCG2]], and their images commute.\nIf $A$ and $B$ are commutative, then $A \\otimes_R B$ is commutative, and it has a universal property.\nFor every commutative $R$-algebra $C$ and $R$-algebra maps $f \\colon A \\to C$, $g \\colon B \\to C$, there is exactly one $R$-algebra map $h \\colon A \\otimes_R B \\to C$ with $h(a \\otimes b) = f(a) g(b)$.\n\nExample: $R[x] \\otimes_R R[y] \\cong R[x, y]$, with $x^i \\otimes y^j \\mapsto x^i y^j$; both sides are free $R$-modules with these bases, and the map respects products.\nAnother example is $\\C \\otimes_\\R \\C \\cong \\C[t]/(t^2 + 1) \\cong \\C \\times \\C$, by the Chinese remainder theorem, since $t^2 + 1 = (t - i)(t + i)$ over $\\C$.\n\nNon-example: the direct product $A \\times B$ with componentwise operations is in general not $A \\otimes_R B$.\nFor $A = B = \\Q[x]$ over $R = \\Q$, the product $\\Q[x] \\times \\Q[x]$ has zero divisors, such as $(1, 0)(0, 1) = 0$, while $\\Q[x] \\otimes_\\Q \\Q[x] \\cong \\Q[x, y]$ is an integral domain.\n",
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