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I'm working on an expository paper and need a little clarification.

Let $R = \mathbb{M}_n(D)$ where $D$ is a division ring. Let $E_{ij}$ be the matix whose $(i,j)$-entry is 1 and all other are zeros. Then, $R = {\bf a_1} \oplus \cdots \oplus {\bf a_n}$, where $\bf a_j$ is the left $D$-subspace of $R$ spanned by $E_{1j}, \dots E_{nj}$. The paper I'm working on this from has a proof for $End_R({\bf a_j}) \cong D^\circ$. Define the map $\phi: D \to End_R({\bf a_j})$ by $\phi(\lambda) = {\lambda}E_{jj}$. This map is an antihomomorphism, and is proven to be an isomorphism. My question is, why does this prove that $End_R({\bf a_j}) \cong D^\circ$? This would imply that a division ring is isomorphic to its opposite ring, which doesn't always hold. So, I feel I'm missing something here. Thanks in advance.

Travis62
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    Antihomomorphism reverses the order of multiplication: we have $\phi(\lambda\mu) =\phi(\mu) \phi(\lambda) $, , that is, it's a homomorphism to $D^\circ$. – Berci Jun 19 '18 at 21:06
  • So we have a composition of maps, which is an isomorphism? That is, we have $D^\circ \to D \to End_R({\bf a_j})$, whose composition is an isomorphism? – Travis62 Jun 20 '18 at 16:28
  • No. There's no any map $D^\circ\to D$ involved. For rings (semigroups) an antihomomorphism $R\to S$ is the same as a homomorphism $R^\circ\to S$ (or even $R\to S^\circ$). – Berci Jun 20 '18 at 18:33
  • Thank you, appreciate the response. It's clear now. – Travis62 Jun 20 '18 at 19:24

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