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  "body": "\n## Statement\n\nLet $R$ be a ring with identity and let $E$ be a left [[$R$-module|0NZ8JA25WK]].\nThen $E$ is an [[injective module|85H7PRHDPZ]] if and only if the following holds.\nFor every left [[ideal|096Z9MP2ME]] $I$ of $R$, every $R$-[[linear map|0NRYHJPQG2]] $f \\colon I \\to E$ extends to an $R$-linear map $g \\colon R \\to E$.\n\n## Remarks\n\nThe condition says that for every such $f$ there is $e \\in E$ with $f(a) = ae$ for all $a \\in I$; take $e = g(1)$.\nOne direction is the definition applied to the inclusion $I \\subseteq R$.\nFor the other, let $M \\subseteq N$ and $f \\colon M \\to E$; by [[Zorn's lemma|04RXTQA5ZA]] there is a maximal extension $(M', f')$ of $f$ to a submodule $M' \\subseteq N$.\nIf $x \\in N$ is not in $M'$, then $I = \\set{a \\in R : ax \\in M'}$ is a left ideal, the map $a \\mapsto f'(ax)$ extends to $R$ by hypothesis, and this extends $f'$ to $M' + Rx$, a contradiction; so $M' = N$.\n\nExample: over a [[principal ideal domain|0CMD0BCH8T]] $R$, a divisible module $E$ is injective.\nA nonzero ideal is $I = (r)$ with $r \\ne 0$, and a map $f \\colon I \\to E$ is fixed by $f(r)$; choose $e$ with $re = f(r)$, then $a \\mapsto ae$ extends $f$.\nFor instance $\\Q$ and $\\Q/\\Z$ are injective $\\Z$-modules.\n\n## Explanations\n\n### L1 human\n\nTo check that a module is injective, one should extend maps from every submodule of every module, which is a lot of work.\nBaer's criterion says that it is enough to extend maps defined on ideals of the ring into the whole ring.\nSo injectivity only has to be tested against one module, the ring itself.\n\n### L2 human\n\nThe criterion is the main tool for proving that a given module is injective.\nOver a principal ideal domain it shows that injective means divisible, which gives the injective $\\Z$-modules $\\Q$ and $\\Q/\\Z$.\nFrom this one shows that every module embeds in an injective one, first for abelian groups and then for modules over any ring via $\\Hom_\\Z(R, -)$.\nThis existence result is what makes injective resolutions, and so right derived functors such as $\\operatorname{Ext}$, available.\n",
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