The Wikipedia page on Pearson's chi-squared test states that a difference to Fisher's exact test is that the latter makes the "assumption of fixed marginal distributions". I assume that applies to the use of chi-squared as a test of independence, because that is what Fisher's exact test does.
However, the description of the chi-squared test for independence clearly shows that it is a goodness-of-fit test which compares the observed joint distribution $O_{i,j}$ to the one expected under independence from the marginal distributions, $$ E_{i,j} = N p_{i\cdot} p_{\cdot j}, $$ which in turn are estimated from the observed marginals, $$ p_{i\cdot} = \frac{O_{i\cdot}}{N} = \frac1N \sum_j O_{ij}, $$ and correspondingly for $p_{\cdot j}$. As far as I can see that means that the test is conditional on given $N$, $p_{i\cdot}$, and $p_{\cdot j}$.
Can someone clear this up? Maybe Wikipedia is simply wrong?