$0^\sharp$ and the regularity of $\aleph_\omega$ I'm sure I'm missing something trivial, and the most likely of it is that I'm simply wrong on my understanding of the constructible universe $L$, or maybe one of the Wikipedia entries I'm about to cite is simply not precise, but I'm curious what part exactly it is.
First, on $0^\sharp$ on wikipedia, we find the following statement:
"It follows from Jensen's covering theorem that the existence of $0^\sharp$ is equivalent to $\omega_\omega$ being a regular cardinal in the constructible universe $L$."
On the page about regular cardinals we find the following quote:
"$\aleph_\omega$['s] initial ordinal  is the limit of the sequence $\omega, \omega_1, \omega_2 ,...$ and so on, which has order type $\omega$, so $\omega_\omega$ is singular, and so is $\aleph_\omega$."
So, that leaves a couple of possible conclusions: 


*

*Regularity is not absolute, and regularity in $L$ is different from general regularity. Perhaps it's my bad, but I find the argument presented in the second quote above pretty simple and straight-forward, so I'm not clear how this is the explanation - I just don't see how the argument can fail even in $L$.

*One of the wikipedia entries is incorrect. I'd be curious which if this is the case.

*All the math done related to $0^\sharp$ is irrelevant because it's trivially nonexistant. I do think this is the least likely explanation, but hey...


So, which is it?
 A: Your first conclusion is the correct one, the point is that being a cardinal is not absolute. So it may exists $\alpha<\omega_1$, so that $L\models \alpha=(\aleph_1)^L$, simply because there is no bijection between $\alpha$ and $\omega$ in $L$. 
Even worst, it may exist a strictly increasing sequence of countable ordinals $\alpha_n$ such that $L\models" \alpha_n\  \rm{is\ a\ cardinal}".$ It follows that $L\models \omega_1>\aleph_\omega^L$. So, in particular, $\omega_\omega$ is not longer de $\omega$th cardinal in $L$.
A: Different models of set theory have different sets of ordinals. In particular if $0^\#$ exists, the $V$ and $L$ have very different sets of ordinals. For example $\{\omega_n^V\mid n\in\omega\}$ is a set in $V$ but not in $L$.
But you don't need you go as far as $0^\#$ to see that regularity, or even being a cardinal, is not absolute. Simply force with $\omega_1^{<\omega}$ over any model of set theory, the generic filter defines a surjection from $\omega$ onto $\omega_1$. How is this possible? Well, it just means that the ground model and the generic extension disagree about which ordinal is $\omega_1$. But what does that mean? It means that the old $\omega_1$ is now a countable ordinal, so it is not regular, and it is not even a cardinal. But in an inner model it is. 
