Is '$\alpha$ is regular' (downwards) absolute?

Question

In particular, if $\alpha$ is an initial ordinal, and $\alpha$ is regular in $V$, is $\alpha$ regular in $L$?

Answer 1

Yes. If $\alpha$ were singular in $L$, there would be a function $f\in L$ that maps a smaller ordinal onto a cofinal subset of $\alpha$. The same $f$ would witness that $\alpha$ is singular in $V$.

Answer 2

Yes, it is downwards absolute. For a given function $f$, the statement "$f$ is a cofinal function from some ordinal less than $\alpha$ into $\alpha$" is absolute. The non-existence of such a function is downwards absolute.