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  "body": "\n## Definition\n\nLet $R$ be a [[commutative ring|0310REJJY8]] and let $E$ be an [[$R$-module|0NZ8JA25WK]].\nThe module $E$ is *injective* if it satisfies the following extension property.\nLet $M$, $N$ be $R$-modules, $i \\colon M \\to N$ an injective $R$-[[linear map|0NRYHJPQG2]], and $f \\colon M \\to E$ an $R$-linear map.\nThen there is an $R$-linear map $g \\colon N \\to E$ with $g \\circ i = f$.\n\n## Remarks\n\nThe following are equivalent for an $R$-module $E$:\n\n1. $E$ is injective.\n2. Every injective linear map $E \\to N$ has a left inverse, that is, it splits.\n3. The functor $\\Hom_R(-, E)$ takes [[exact sequences|JB7ZBFBA9C]] to exact sequences.\n4. ([[Baer's criterion|MR5HNV0Q07]]) Every linear map from an ideal $I$ of $R$ to $E$ extends to $R$.\n\nOver a [[principal ideal domain|0CMD0BCH8T]], a module $E$ is injective exactly when it is divisible: $rE = E$ for every nonzero $r \\in R$.\nThe same definition works over any ring, commutative or not.\n\nExample: $\\Q$ is an injective $\\Z$-module.\nBy Baer's criterion it suffices to extend a map $f \\colon n\\Z \\to \\Q$ with $n \\ne 0$; with $q = f(n)$, the map $g(x) = xq/n$ on $\\Z$ satisfies $g(n) = q$.\n\nNon-example: $\\Z$ is not an injective $\\Z$-module.\nTake $i \\colon \\Z \\to \\Z$, $x \\mapsto 2x$, and $f$ the identity of $\\Z$.\nAn extension $g$ would satisfy $1 = f(1) = g(2) = 2g(1)$, which has no solution in $\\Z$.\n",
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