Endomorphism Ring and Indecomposability
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Endomorphism Ring and Indecomposability
Let \( V \) be a module and let \( End(V) \) be its endomorphism ring. Show that if \( End(V) \) is local, then \( V \) is indecomposable. Is the converse true?
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Re: Endomorphism Ring and Indecomposability
Suppose that the ring \(\displaystyle{\left(\rm{End}_{R}(V),+,\circ\right)}\) is local.
Also, suppose that the \(\displaystyle{R}\) - module \(\displaystyle{V}\) is not indecomposable.
Then, \(\displaystyle{V=V_1\oplus V_2}\), where \(\displaystyle{V_1\,,V_2}\) are
non trivial and proper \(\displaystyle{R}\)-
submodules of \(\displaystyle{V}\) . Consider the endomorphism
\(\displaystyle{p_1:V\longrightarrow V\,,v_1+v_2\mapsto v_1}\) with \(\displaystyle{\rm{Ker}(p_1)=V_2}\)
and \(\displaystyle{\rm{Im}(p_1)=V_1}\) .
Since \(\displaystyle{p_1\in\rm{End}_{R}(V)}\), \(\displaystyle{p_1}\) is not invertble and the ring
\(\displaystyle{\rm{End}_{R}(V)}\) is local, we have that \(\displaystyle{Id_{V}-p_1}\) is
invertible. This is not true, because
\(\displaystyle{\begin{aligned} \rm{Ker}(Id_{V}-p_1)&=\left\{v_1+v_2\in V: (Id_{V}-p_1)(v_1+v_2)=0\right\}\\&=\left\{v_1+v_2\in V: v_1+v_2-v_1=0\right\}\\&=\left\{v_1\in V: v_1\in V_1\right\}\\&=V_1\neq\left\{0\right\}\end{aligned}}\)
so, the \(\displaystyle{R}\) - module \(\displaystyle{V}\) is indecomposable.
To be continued...
Also, suppose that the \(\displaystyle{R}\) - module \(\displaystyle{V}\) is not indecomposable.
Then, \(\displaystyle{V=V_1\oplus V_2}\), where \(\displaystyle{V_1\,,V_2}\) are
non trivial and proper \(\displaystyle{R}\)-
submodules of \(\displaystyle{V}\) . Consider the endomorphism
\(\displaystyle{p_1:V\longrightarrow V\,,v_1+v_2\mapsto v_1}\) with \(\displaystyle{\rm{Ker}(p_1)=V_2}\)
and \(\displaystyle{\rm{Im}(p_1)=V_1}\) .
Since \(\displaystyle{p_1\in\rm{End}_{R}(V)}\), \(\displaystyle{p_1}\) is not invertble and the ring
\(\displaystyle{\rm{End}_{R}(V)}\) is local, we have that \(\displaystyle{Id_{V}-p_1}\) is
invertible. This is not true, because
\(\displaystyle{\begin{aligned} \rm{Ker}(Id_{V}-p_1)&=\left\{v_1+v_2\in V: (Id_{V}-p_1)(v_1+v_2)=0\right\}\\&=\left\{v_1+v_2\in V: v_1+v_2-v_1=0\right\}\\&=\left\{v_1\in V: v_1\in V_1\right\}\\&=V_1\neq\left\{0\right\}\end{aligned}}\)
so, the \(\displaystyle{R}\) - module \(\displaystyle{V}\) is indecomposable.
To be continued...
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Re: Endomorphism Ring and Indecomposability
I give away the answer of the second part of the question: The converse does not hold! Can you give a counter-example?
It might be helpful to follow this approach when looking for one:
Definition 1: Let \( J \) be a set. A \( J \)-module is given by \( \left( V , \phi_{j} \right)_{j \in J} \), where
Definition 2: Let \( \left( V , \phi_{j} \right)_{j \in J} \; , \; \left( W , \psi_{j} \right)_{j \in J} \) be \( J \)-modules. A \( J \)-module homomorphism is a \( \mathbb{K} \)-linear map \( f \ \colon V \longrightarrow W \) such that for each \( j \in J \; , \; f \circ \phi_{j} = \psi_{j} \circ f \).
It might be helpful to follow this approach when looking for one:
Definition 1: Let \( J \) be a set. A \( J \)-module is given by \( \left( V , \phi_{j} \right)_{j \in J} \), where
- \( V \) is a \( \mathbb{K} \)-vector space
- \( \phi_{j} \ \colon V \longrightarrow V \) is a \( \mathbb{K} \)-linear map
Definition 2: Let \( \left( V , \phi_{j} \right)_{j \in J} \; , \; \left( W , \psi_{j} \right)_{j \in J} \) be \( J \)-modules. A \( J \)-module homomorphism is a \( \mathbb{K} \)-linear map \( f \ \colon V \longrightarrow W \) such that for each \( j \in J \; , \; f \circ \phi_{j} = \psi_{j} \circ f \).
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