# “Theorem of Witt” for modules

If I have 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.