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  "body": "\n## Definition\n\nLet $R$ be a [[commutative ring|0310REJJY8]] and let $P$ be an [[$R$-module|0NZ8JA25WK]].\nThe module $P$ is *projective* if it satisfies the following lifting property.\nLet $M$, $N$ be $R$-modules, $g \\colon M \\to N$ a surjective $R$-[[linear map|0NRYHJPQG2]], and $f \\colon P \\to N$ an $R$-linear map.\nThen there is an $R$-linear map $h \\colon P \\to M$ with $g \\circ h = f$.\n\n## Remarks\n\nThe following are equivalent for an $R$-module $P$:\n\n1. $P$ is projective.\n2. Every surjective linear map $M \\to P$ has a right inverse, that is, it splits.\n3. $P$ is a direct summand of a [[free module|0DZ9DKAKZR]]: there is a module $Q$ with $P \\oplus Q$ free.\n4. The functor $\\Hom_R(P, -)$ takes [[exact sequences|JB7ZBFBA9C]] to exact sequences.\n\nThe same definition works over any ring, commutative or not.\n\nExample: every free module $F$ with basis $(e_i)$ is projective.\nGiven $g$ and $f$, choose $m_i \\in M$ with $g(m_i) = f(e_i)$ and let $h(e_i) = m_i$; then $g \\circ h$ and $f$ agree on a basis, so they are equal.\nOver $R = \\Z/6\\Z$, the module $\\Z/2\\Z$ is projective but not free, since $R \\cong \\Z/2\\Z \\oplus \\Z/3\\Z$ as $R$-modules.\n\nNon-example: over $R = \\Z$, the module $\\Z/2\\Z$ is not projective.\nTake $g \\colon \\Z \\to \\Z/2\\Z$ the reduction and $f$ the identity of $\\Z/2\\Z$.\nA lift $h \\colon \\Z/2\\Z \\to \\Z$ must be zero, because $2h(1) = h(0) = 0$ and $\\Z$ has no nonzero element of order $2$; then $g \\circ h = 0 \\ne f$.\n",
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