----- > [!proposition] Proposition. ([[Schur's Lemma for modules]]) > Let $R$ be a [[ring]], Let $M,N$ be [[simple module|simple]] $R$-[[module|modules]] and $\varphi:M \to N$ an $R$-[[linear map]]. Then either $\varphi=0$ or $\varphi$ is a [[module isomorphism]]. ^proposition > [!proof]- Proof. ([[Schur's Lemma for modules]]) > > Look at the [[kernel of a module homomorphism|kernel]] of $\varphi$; by [[kernel iff submodule]] it is either $\{ 0 \}$ or $N$. If $\ker \varphi=\{ 0 \}$, then the [[first isomorphism theorem for modules]] gives a diagram: > > ```tikz > \usepackage{tikz-cd} > \usepackage{amsmath} > \begin{document} > % https://tikzcd.yichuanshen.de/#N4Igdg9gJgpgziAXAbVABwnAlgFyxMJZABgBpiBdUkANwEMAbAVxiRAFkQBfU9TXfIRQBGclVqMWbAHLdeIDNjwEiZYePrNWiDgAIA9LoA6R4MRNdjRgMYEA5rs5dxMKHfhFQAMwBOEALZIZCA4EEgATNSaUjomaFgg1Ax0AEYwDAAK-MpCID5YdgAWOHLefoGIwaFIohJabCb0PmiFCTxlARHU1Yi10dogJhA0MD4MWGAwwI10za1ciSF0WAxshRAQANaLyWmZ2YJs+UUlzlxAA > \begin{tikzcd} > M \arrow[d, "\pi"'] \arrow[r, "\varphi"] & N \\ > M / \{0\} \cong M \arrow[ru, "\overline{\varphi}"', hook] & > \end{tikzcd} > \end{document} > ``` > > where $\overline{\varphi}$ defined by $\overline{\varphi}(\{ m \})=\varphi(m)$ is an [[injection]] into $N$ (an [[isomorphism]] onto its image). Clearly this implies $\varphi$ is itself an [[injection]]. Its image is a [[submodule]] of $N$; since $N$ is [[simple module|simple]], $\im \varphi$ is either trivial or $M$ itself; but it can't be trivial due to [[injection|injectivity]] considerations, so $\im \varphi=N$ and $\varphi$ is also a [[surjection]]. > > Otherwise suppose $\ker \varphi=N$. Then the [[first isomorphism theorem for modules]] gives a diagram > > ```tikz > \usepackage{tikz-cd} > \usepackage{amsmath} > \begin{document} > % https://tikzcd.yichuanshen.de/#N4Igdg9gJgpgziAXAbVABwnAlgFyxMJZABgBpiBdUkANwEMAbAVxiRAFkQBfU9TXfIRQBGclVqMWbAHLdeIDNjwEiZYePrNWiDgAIA9Lva6AOiYDGBAOamTwYma7dxMKFfhFQAMwBOEALZIZCA4EEgATNSaUjpmaFgg1Ax0AEYwDAAK-MpCID5YVgAWOHLefoGIwaFIohJabGb0PmiFCTxlARHU1Yi10dogZhA0MD4MWGAwwI10za1O3XRYDGyFEBAA1okgyWmZ2YJs+UUlXBRcQA > \begin{tikzcd} > M \arrow[d, "\pi"'] \arrow[r, "\varphi"] & N \\ > M / M \cong \{0\} \arrow[ru, "\overline{\varphi}"', hook] & > \end{tikzcd} > \end{document} > ``` > In this case the map $\overline{\varphi}$ is clearly trivial. Since $\varphi$ factors through a trivial map, $\varphi$ is trivial as well. > ----- #### ---- #### References > [!backlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM [[]] > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ``` > [!frontlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM outgoing([[]]) > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ```