TITLE: Axiom of Choice and Determinacy
QUESTION [13 upvotes]: In my set theory course we have been talking about the axiom of determinacy.  One of the first things we showed was that $AD$ and $AC$ are incompatible.  We later showed that $ZF+AD$ implies the consistency of ZFC by showing (in $ZF+AD$) that $\aleph_1$ is inaccessible in $L$.
This seems odd that $AD$ implies that $AC$ is false, but is consistent with $ZF$.  I don't have really a specific question, but was hoping for someone to explain why this isn't really that odd.  Or that it is odd, but is still ok.  I feel like I understand the proofs of the statements, but it still doesn't seem to fit together quite right for me.
Any thoughts would be appreciated, thanks!

REPLY [4 votes]: Here is a quick analogy, suppose that $\kappa$ is inaccessible. $2^\kappa=\kappa^{++}$ implies that $\sf GCH$ is false, but it also proves the consistency of $\sf GCH$.
Both $\sf AC$ and $\sf AD$ are additional axioms which extend the axioms of $\sf ZF$. Both are very strong and give you a lot of information, but at the same time both are incompatible.
Equally one could replace the above example with "$\mathcal P(\kappa)$ cannot be well-ordered". This would be an axiom that proves that the axiom of choice as false, but $\kappa$ is still inaccessible in $L$ and therefore along with $\sf ZF$, this axiom proves the consistency of $\sf ZFC$.
Relative consistency is not direct implication, it's a "workaround". It means that given one universe of these axioms, we can generate a new universe of those axioms. Having a stronger consistency strength means that you can prove more. Often it means that you can prove the existence of a set which is a universe of a certain theory. $\sf ZFC$ cannot prove the existence of a set model, but $\sf ZF+AD$ can.