What is the least inaccessible cardinal for Tarski-Grothendieck set theory?

Question

Let ordinal $\alpha$ be the least ordinal such that $V_\alpha\models$ Tarski-Grothendieck set theory.

What position does $\alpha$ have in the hierarchy of inaccessible cardinals?

Answer 1

I missed the inaccessibility requirement initially - fixed!

Since $\mathsf{TG}$ is "just" $\mathsf{ZFC}$ + "There is a proper class of inaccessible cardinals," an inaccessible $\alpha$ satisfies $V_\alpha\models\mathsf{TG}$ iff $\alpha$ is an inaccessible limit of inaccessibles, i.e. a $1$-inaccessible.

That said, if we just ask for $V_\alpha\models\mathsf{TG}$ (that is, we don't demand that $\alpha$ be inaccessible), we can do significantly better: in this case, the relevant $\alpha$ is the smallest worldly limit of inaccessibles. This is bigger than the smallest limit of inaccessibles, but smaller than the smallest $1$-inaccessible.

Recall that $\alpha$ is worldly iff $V_\alpha\models\mathsf{ZFC}$. Generally, the smallest worldly $\alpha$ such that $V_\alpha\models T$ (for any fixed $T$) has countable cofinality by a Lowenheim-Skolem type argument, and this is why the ordinal of the previous paragraph falls short of being $1$-inaccessible (= regular limit of inaccessible cardinals).

In a comment, the OP asks which is a better "gauge" (of strength presumably). I think there is no single answer; which cardinal you look at depends on precisely which sorts of questions you are asking. Sometimes it will make sense to include regularity as a core hypothesis, while other times it won't.