I stumbled over a statement on Wikipedia http://en.wikipedia.org/wiki/Duality_%28projective_geometry%29 and would like to ask how this could possibly be true. It states the following

The projective planes $PG(2,K)$ for any division ring $K$ are self-dual.

Is this really true? Or rather: What is wrong with the following argument that (if correct) would contradict the statement:

For a finite dimensional left vector space $V$ over a division ring $K$, we have the dual left vector space $V^{\star}$ over $K^{op}$ (the opposite division ring of $K$). Now, it is certainly true that $P(V^{\star})$ is isomorphic to the dual of $P(V)$. Thus, $P(V)$ is isomorphic to its dual if and only if $P(V) \cong P(V^{\star})$. Since $V$ and $V^{\star}$ have the same dimension, does that not necessarily imply that $K$ and $K^{op}$ are isomorphic by the Second Fundamental Theorem of projective Geometry?

Since the other direction is almost trivial ($K \cong K^{op}$ certainly implies $P(V) \cong P(V^\star)$): Isn't $PG(2,K)$ (or $PG(n,K)$ for any $n \geq 2$) being self-dual simply eqvuialent to $K \cong K^{op}$ (which is of course is not always the case). I also recall that I have read this somewhere.

So, to put it in one line: What is wrong? The statement on Wikipedia or my reasoning?

EDIT: I found some course notes in which it is also claimed that $PG(n,K)$ is self-dual iff $K$ is self-dual (see 6.1 in http://www.maths.qmul.ac.uk/~pjc/pps/pps6.pdf). This does contradict what Wikipedia says, doesn't it?