“Theorem of Witt” for modules

For modules $M_1 \oplus N \cong M_2 \oplus N$, why is $M_1 \cong M_2$, if $Hom(M_1,N)=\{0\}=Hom(M_2,N)$?

It seams similar to Witt's cancellation theorem for quadratic forms.

Regards, Khanna

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In a broader question on cancellation I posted the title of a paper by T.Y. Lam on the topic. – rschwieb Jun 20 '12 at 14:28

Let $\phi\colon M_1 \oplus N \to M_2 \oplus N$ an isomorphism. Denote by $i_1\colon M_1 \to M_1 \oplus N$, $i_2\colon N \to M_1 \oplus N$, $j_1\colon M_2 \to M_2 \oplus N$ and $j_2 \colon N \to M_2 \oplus N$ the injections and by $p_1\colon M_1 \oplus N \to M_1$, $p_2 \colon M_1 \oplus N \to N$, $\pi_1\colon M_2 \oplus N \to M_2$ and $\pi_2 \colon M_2 \oplus N \to N$ the projections. Define $\psi \colon M_1 \to M_2$ by $\psi := \pi_1\phi i_1$. We will prove that $\psi$ is an isomorphism.

Let $m_1 \in M_1$ such that $\psi(m_1) = 0$, then $\pi_1\phi(m_1,0) = 0$. As $\pi_2 \phi i_1 \colon M_1 \to N$, $\pi_2\phi i_1 = 0$, hence $\pi_2 \phi(m_1,0) = 0$. So $\phi(m_1,0) = 0$, which implies $m_1 = 0$. So $\psi$ is a monomorphism.

Let $m_2 \in M_2$. We have that $p_2\phi^{-1}j_1\colon M_2 \to N$, so $p_2\phi^{-1}j_1 = 0$. Set $m_1 := p_1\phi^{-1}j_1(m_2)$, then $\phi^{-1}(m_2,0) = (m_1,0)$, or $\phi(m_1,0) = (m_2,0)$, which implies $\psi(m_1) =m_2$. So $\psi$ is onto.

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Thanks a lot! Khanna – khanna Jun 20 '12 at 13:38

I want to add a categorical phrasing of this problem.

Consider the quotient category $R\text{-Mod}/\langle N\rangle_\oplus$ of the category $R\text{-Mod}$ at the additive closure $\langle N\rangle_\oplus$ of $N$: Its objects are again the $R$-modules, but for two $R$-modules $A,B$ we define $\text{Hom}_{R\text{-Mod}/\langle N\rangle_\oplus}(A,B)$ to be the quotient of $\text{Hom}_{R\text{-Mod}}(A,B)$ by the subgroup of those morphisms which factor through a (finite) direct sum of copies of $N$.

The canonical functor ${\mathscr N}: R\text{-Mod}\to R\text{-Mod}/\langle N\rangle_\oplus$ is additive, and moreover $N\cong 0$ in $R\text{-Mod}/\langle N\rangle_\oplus$ since the identity on $N$ is annihilated; I will use $\simeq$ for isomorphism in $R\text{-Mod}/\langle N\rangle_\oplus$ for the rest of the answer. Also, ${\mathscr N}$ is fully faithful when restricted to the subcategory $^{\perp}N$ of those $R$-modules $A$ with $\text{Hom}_{R\text{-Mod}}(A,N)=0$.

Under your assumptions, you have $M_1\simeq M_1\oplus N\cong M_2\oplus N\simeq M_2$, so $M_1$ and $M_2$ are identified, up to isomorphism, by the fully faithful functor ${\mathscr M}|_{^{\perp}N}: {^{\perp}N}\to R\text{-Mod}/\langle N\rangle_\oplus$, and are therefore already isomorphic in ${^{\perp}}N$, and in particular in $R\text{-Mod}$.

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