PeterDonis said:
No, it doesn't. Superposition is basis dependent. Entanglement is not. I can always find a basis of the joint Hilbert space of the system in which the entangled state is a basis state, not a superposition. But the state will still be entangled.
Entanglement is referring to "subsystems" of a whole system. The most simple example are two qbits, e.g., realized by the spins of two spin-1/2 particles (or polarization states of two photons). The Hilbert space of the whole system ("two spins") is described by the product of the two "single-spin" Hilbert spaces
$$\mathcal{H}=\mathcal{H}_1 \otimes \mathcal{H}_2.$$
Now the total system is by definition in an entangled pure state ##\hat{\rho}=|\Psi \rangle \langle \Psi|## if the corresponding reduced states of the single systems
$$\hat{\rho}_1 = \mathrm{Tr}_2 \hat{\rho}, \quad \hat{\rho}_2 = \mathrm{Tr}_1 \hat{\rho}$$
are not pure states. This is a completely basis-independent definition of entanglement.
It is clear that product states are not entangled. Indeed, if
$$|\Psi \rangle=|\psi_1 \rangle \otimes |\psi_2 \rangle$$
Then
$$\hat{\rho}_1 = |\psi_1 \rangle \langle \psi_1|, \quad \hat{\rho}_2=|\psi_2 \rangle \langle |\psi_2 \rangle,$$
i.e., in this case the two subsystems are both prepared in a pure state.
However, if you have a state like the singlet (total spin 0),
$$|\Psi \rangle=\frac{1}{\sqrt{2}} (|\hbar/2 \rangle \otimes |-\hbar/2 \rangle - |-\hbar/2 \rangle \otimes |\hbar/2 \rangle),$$
then it's easy to show that
$$\hat{\rho}_1=\frac{1}{2} \hat{1}_1, \quad \hat{\rho}_2=\frac{1}{2} \hat{1}_2.$$
So in this case the spins of the two subsystems are no pure state but even "maximum-entropy states", i.e., the single spins are completely indetermined although the whole system is in a pure state with total spin 0. So the two spins are in an entangled state, and because it's even leading to the minimal possible knowledge about each of the subsystems, it's called a "maximally entangled state" or a "Bell state".
So a priori entanglement has nothing to do with superposition (which is always basis dependent) but with being described by a product state or not.
You can of course also use another basis of the whole system. In our case we can use the four Bell states defined by
$$|\Psi^{(\pm)} \rangle = \frac{1}{\sqrt{2}} (|\hbar/2 \rangle \otimes |-\hbar/2 \rangle \pm |-\hbar/2 \rangle \otimes |\hbar/2 \rangle),$$
$$|\Phi^{(\pm)} \rangle = \frac{1}{\sqrt{2}} (|\hbar/2 \rangle \otimes |\hbar/2 \rangle \pm |-\hbar/2 \rangle \otimes |-\hbar/2 \rangle).$$
In this basis our singlet-Bell state is not a superposition of these basis states but simply ##|\Psi^{(-)} \rangle##.
