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Let $V$ be a finite-dimensional left vector space over a division ring $K$, and let $V^*$ the dual right vector space (consisting of all linear functions from $V$ to $K$). We can (and will) treat $V^*$ as a left vector space over the division ring $K^*$ (obtained from $K$ by reversing the arguments of the multiplication).

If $K$ is a field, then we have $V \cong V^*$. If $K$ is only a division ring, we can of course not have this as $V^*$ is not a left vector space over $K$ but over $K^*$.

However: Does ist hold that $P(V)$ and $P(V^*)$ are still isomorphic? That is, are the projective spaces consisting of the one- and two-dimensional vector spaces still isomorphic? And if so, could you please give me the collineation and explain why it works?

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    I believe exactly that. But here is my problem: $P(V^*)$ is isomorphic to the dual of $P(V)$, so $P(V)$ being self-dual is equivalent to $P(V) \cong P(V^*)$. But now, if this is only true for $K \cong K^*$, then not every projective plane of the form $P(V)$ would be self-dual. However, this is something that Wikipedia explicitly states: http://en.wikipedia.org/wiki/Duality_%28projective_geometry%292012-11-24

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