- Let \( \displaystyle D \) be a division ring and let \( \displaystyle R = \mathbb{M}_{n}(D) \) be the ring of \( \displaystyle n \times n \) matrices over \( \displaystyle D \). Show that \( \displaystyle R \) has a unique (up to isomorphism) simple left \( \displaystyle R \)-module.
- Let \( \displaystyle \left\{ D_{j} \right\} _{j=1}^{k} \) be division rings and let \( \displaystyle \left\{ \mathbb{M}_{n_{j}}(D_{j}) \right\} _{j=1}^{k} \) be the respective rings of \( \displaystyle n_{j} \times n_{j} \) matrices over \( \displaystyle D_{j} \). Furthermore, let \( \displaystyle S = \prod_{j=1}^{k} \mathbb{M}_{n_{j}}(D_{j}) \) be their direct product. Can the above be generalised for \( \displaystyle S \)? And if so, how? (In other words, how many isomorphism classes of simple left \( \displaystyle S \)-modules exist?)
Isomorphic Simple Modules
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- Community Team
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Isomorphic Simple Modules
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- Community Team
- Posts: 426
- Joined: Mon Nov 09, 2015 1:52 pm
Re: Isomorphic Simple Modules
It's known from \(\displaystyle{\rm{Wedderburn}}\) theorem, that if an associative ring with unity
is semesimple, then \(\displaystyle{R\cong \prod_{i=1}^{s}\mathbb{M}_{n_{i}}(D_{i})}\) as rings,
where \(\displaystyle{n_i\in\mathbb{N}}\) and \(\displaystyle{D_{i}}\) are division rings. The number
\(\displaystyle{s}\) measures the simple \(\displaystyle{R}\) - modules, which, are not isomorphic.
So, the answer in our case is \(\displaystyle{1}\) and \(\displaystyle{k}\), respectively.
is semesimple, then \(\displaystyle{R\cong \prod_{i=1}^{s}\mathbb{M}_{n_{i}}(D_{i})}\) as rings,
where \(\displaystyle{n_i\in\mathbb{N}}\) and \(\displaystyle{D_{i}}\) are division rings. The number
\(\displaystyle{s}\) measures the simple \(\displaystyle{R}\) - modules, which, are not isomorphic.
So, the answer in our case is \(\displaystyle{1}\) and \(\displaystyle{k}\), respectively.
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