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  "body": "\n## Definition\n\nLet $F$ be a field and let $V$ be a [[vector space|000AYPA9AJ]] over $F$.\nFor $k \\ge 1$, let $T^k(V) = V \\otimes \\cdots \\otimes V$ be the [[tensor product|0BF2DE2SYK]] of $k$ copies of $V$, and let $T^0(V) = F$.\nThe *tensor algebra* of $V$ is the vector space\n$$T(V) = \\bigoplus_{k \\ge 0} T^k(V),$$\nwith the bilinear multiplication determined by\n$$(v_1 \\otimes \\cdots \\otimes v_k) \\cdot (w_1 \\otimes \\cdots \\otimes w_l) = v_1 \\otimes \\cdots \\otimes v_k \\otimes w_1 \\otimes \\cdots \\otimes w_l$$\nfor $v_i, w_j \\in V$, and with unit $1 \\in T^0(V)$.\n\n## Remarks\n\n$T(V)$ is an associative unital algebra over $F$, graded by the degree $k$, and $V = T^1(V)$ sits inside it.\nUniversal property: for every associative unital $F$-algebra $A$ and every linear map $f \\colon V \\to A$, there is exactly one unital algebra homomorphism $\\bar f \\colon T(V) \\to A$ with $\\bar f(v) = f(v)$ for $v \\in V$.\nExample: if $V = F x$ has dimension $1$, then $T^k(V) = F x^{\\otimes k}$, and $x^{\\otimes k} \\mapsto x^k$ is an isomorphism $T(V) \\cong F[x]$; the check is that both products add degrees.\nNon-example: for $V = F^2$ with basis $x, y$, the polynomial ring $F[x, y]$ is not $T(V)$, because $x \\otimes y \\ne y \\otimes x$ in $T(V)$ while $xy = yx$ in $F[x, y]$.\nIf $V$ has a basis $(e_i)_{i \\in I}$, then $T(V)$ is the free associative algebra on the set $I$.\nThe same construction works for a module over a commutative ring.\n",
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