# Strongest large cardinal axiom compatible with $V = L$?

What is the strongest known natural large cardinal axiom compatible with $V = L$ (strongest in the sense that it implies all known "small" large cardinal axioms, where a large cardinal axiom is said to be "small" if it doesn't imply $V \neq L$)?

One candidate might be "there is an $\alpha$-Erdos cardinal for every countable ordinal $\alpha$."

Using the instrumentalist dodge'' of Steel/Hamkins, it seems that we can "cheat" to obtain stronger large cardinal axioms within the confines of $\operatorname{ZF}+(V = L)$, such as "there is a transitive model of the theory ZFC+"$0^\sharp$ exists."" I will leave whether or not such axioms are "natural" for a matter of debate here.

Disclaimer: I'm an algebraist and number theorist but am currently trying to grasp the big picture in set theory, large cardinals, inner model theory, and the like.

Related MO posts:

Is there a large-cardinal completeness theorem for $L$?

If $0^{\sharp}$ exists, then every uncountable cardinal in $V$ is as large as it can be in $L$.

Erdős cardinals and $0^\sharp$

• Your suggestion is essentially it. The modern presentation of $0^\sharp$ is as what Schimmerling calls a baby mouse, this is a structure $J_\tau$ with an external measure that is iterable (and some technical definability requirements). Asking that for each $\alpha<\omega_1$ there is such a structure, but only $\alpha$-iterable, seems as strong as possible. But this is in essence the same as considering $\alpha$-Erdős cardinals. – Andrés E. Caicedo Dec 13 '14 at 1:23
• Another way to cheat and get slightly stronger large cardinals: it is consistent that there is an ordinal $\alpha$, $V=L$ and in some generic extension (by a set forcing notion) there is an $\alpha$-Erdos cardinal, even when $\alpha$ is not countable in $L$. – Yair Hayut Dec 13 '14 at 15:53

In the paper A large cardinal in the constructible universe'' Silver shows that if $\kappa\to (\alpha)^{<\aleph_0}$ for all countable $\alpha,$ then the same is true for $\kappa$ in $L$.
On the other hand, by a result of Rowbottom, $\kappa \to (\omega_1)^{<\aleph_0}$ contradicts $V=L.$
It does not seem extravagant, then, to assert that, for all practical purposes, $\kappa \to (\omega)^{<\aleph_0}$ is the strongest strong axiom of infinity know to be consistent with $V=L,$ and therein lies its chief interest.