So this is about something from Preskill's notes on Quantum Computation and Information, Chapter 4, page 3.
Imagine we have a maximally entangled state (Bell state). We can identify the Bell state by measuring the phase and parity bit with the commuting operators
$$\sigma_1^A \otimes \sigma_1^B \\ \sigma_3^A \otimes \sigma_3^B$$
Now they want to show that you cannot do this locally, so when Alice and Bob both are spacelike separated and can only do measurements on their own qubit A and B.
Following Preskill, they can start by measuring $\sigma_3^A$ and $ \sigma_3^B$ each on their own, and because this commutes with $\sigma_3^A \otimes \sigma_3^B$, this does not disturb the parity bit. So they prepared a simultaneous eigenstate of $\sigma_3^A$ and $ \sigma_3^B$. These eigenstates do not commute with $ \sigma_1^A \otimes \sigma_1^B$ so they disturbed the phase bit so they cannot measure it. Conclusion: they cannot read the information that is in the entanglement.
But my real question now is: why do the first two observables $ \sigma_1^A \otimes \sigma_1^B$ and $ \sigma_3^A \otimes \sigma_3^B$ commute, when $\sigma_3^A$ and $ \sigma_3^B$ will not commute with $ \sigma_1^A \otimes \sigma_1^B$. Because in my opinion, we can write the effect of the measurement of Alice and Bob on the state as;
$$\sigma_3^A \otimes I^B \\ I^A \otimes \sigma_3^B$$
which, when measured after each other will have the effect on $|\psi\rangle$:
$$(\sigma_3^A \otimes I^B) (I^A \otimes \sigma_3^B) |\psi\rangle = \sigma_3^A \otimes \sigma_3^B |\psi\rangle$$
which is the original measurement that can be done when both qubits are at the same place? So what is the difference, why is this not the same eigenstate? I must be missing something, but I don't see where I make an error. Can someone please explain.