This question was actually asked by John Stillwell in a comment to an answer to this [question](https://mathoverflow.net/questions/20882/most-unintuitive-application-of-the-axiom-of-choice). I thought I would advertise it as a separate question since no one has yet answered and I am also curious about it.


**Question:** Is the existence of a non-principal ultra-filter on $\omega$ a weaker assumption than the existence of a well-ordering of $\mathbb{R}$?



Ultrafilters vs Well-orderings
It is consistent that there exists a non-principal ultrafilter over $\omega$ while $\mathbb{R}$ is not well-ordered. To see this, suppose that the
partition relation $\omega \to (\omega)^{\omega}$ holds in $L(\mathbb{R})$.
Then forcing with $\mathbb{P}= [\omega]^{\omega}$ adjoins a selective ultrafilter $\mathcal{U}$ over $\omega$ and $\mathcal{P}(\omega)$ cannot be well-ordered in 
$L(\mathbb{R})[\mathcal{U}]$. (See Eisworth's paper: Selective ultrafilters and
$\omega \to (\omega)^{\omega}$.) Thus $L(\mathbb{R})[\mathcal{U}]$ is a model of $ZF$ which contains the nonprincipal ultrafilter $\mathcal{U}$ and yet $\mathbb{R}$ cannot be well-ordered in 
$L(\mathbb{R})[\mathcal{U}]$.


An update: it is perhaps also interesting to note that in $L(\mathbb{R})[\mathcal{U}]$, the ultraproduct $\prod\_{\mathcal{U}} \bar{\mathbb{F}}\_{p}$ of the algebraic closures of the fields of prime order $p$ is not isomorphic to $\mathbb{C}$.


