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  "body": "\n## Definition\n\nLet $R$ be a ring, and let $P$ and $Q$ be left $R$-[[modules|0NZ8JA25WK]] such that $P \\oplus Q \\cong F$ for a [[free module|0DZ9DKAKZR]] $F$.\nLet $F^{(\\infty)} = F \\oplus F \\oplus \\cdots$ be the direct sum of countably many copies of $F$.\nThe *Eilenberg swindle* is the chain of isomorphisms\n$$P \\oplus F^{(\\infty)} \\cong P \\oplus (Q \\oplus P) \\oplus (Q \\oplus P) \\oplus \\cdots \\cong (P \\oplus Q) \\oplus (P \\oplus Q) \\oplus \\cdots \\cong F^{(\\infty)},$$\nobtained by regrouping the summands of a countable direct sum.\nIts conclusion is that $P \\oplus F^{(\\infty)} \\cong F^{(\\infty)}$, where $F^{(\\infty)}$ is free.\n\n## Remarks\n\nExample: every [[projective module|BFFA03GE3V]] $P$ is a direct summand of a free module $F$, so the swindle applies to it.\nTake $R = \\Z/6\\Z$ and $P = 3\\Z/6\\Z$, $Q = 2\\Z/6\\Z$; then $P \\oplus Q \\cong R$, and the swindle gives $P \\oplus R^{(\\infty)} \\cong R^{(\\infty)}$, although $P$ itself is not free.\n\nNon-example: the regrouping needs infinitely many summands.\nWith finitely many summands the argument breaks.\nIndeed, for $R$ commutative and $P \\ne 0$, the isomorphism $P \\oplus R^n \\cong R^n$ fails.\nTo see this, compare dimensions over the residue field at a prime where $P$ is nonzero.\n\nFor this reason the Grothendieck group $K_0(R)$ uses only finitely generated projective modules: with countably generated ones every class would vanish.\n",
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