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  "body": "\n## Statement\n\nLet $R$ be a ring and let $M$ and $N$ be [[simple|G00C22AEDH]] left [[$R$-modules|09VDS2W1E8]].\nLet $f \\colon M \\to N$ be a homomorphism of $R$-modules.\nThen either $f = 0$ or $f$ is an isomorphism.\n\n## Remarks\n\nThe kernel of $f$ is a submodule of $M$, so it is $0$ or $M$.\nThe image of $f$ is a submodule of $N$, so it is $0$ or $N$.\nIf $f \\ne 0$, the kernel is not $M$ and the image is not $0$, so $f$ is injective and surjective.\nExample: a nonzero group homomorphism $\\Z/p\\Z \\to \\Z/p\\Z$, for $p$ prime, is a bijection.\nSimplicity is needed: the map $\\Z/4\\Z \\to \\Z/4\\Z$, $x \\mapsto 2x$, is nonzero and not an isomorphism.\nA common form concerns an algebraically closed field $k$, a $k$-algebra $A$, and a simple $A$-module $M$ of finite dimension over $k$.\nThen every endomorphism of $M$ is multiplication by a scalar.\n\n## Explanations\n\n### L1 human\n\nA simple module has no proper nonzero pieces.\nA map between two simple modules therefore has no room to be \"partly\" zero: it either kills everything or matches the two modules exactly.\n\n### L2 human\n\nSchur's lemma is the first tool of representation theory.\nIt shows that the endomorphism ring of a simple module is a division ring.\nFor representations of a group over $\\C$ it gives that the endomorphisms of an irreducible representation are scalars.\nThis is the key step in the orthogonality relations for characters and in the decomposition of semisimple modules into isotypic parts.\n",
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