I'm working through Susskind's "Quantum Mechanics" book (TTM series), which I quite like.
Background
In Lecture 7 (Chapter 7), he studies a 2-spin system. A single spin has eigenvectors:
$$|u\rangle=\begin{pmatrix}1\\0\end{pmatrix},~~ |d\rangle=\begin{pmatrix}0\\1\end{pmatrix}$$
and then a 2-spin state has eigenvectors:
$$|uu\rangle=\begin{pmatrix}1\\0\\0\\0\end{pmatrix},~~ |ud\rangle=\begin{pmatrix}0\\1\\0\\0\end{pmatrix},~~ |du\rangle=\begin{pmatrix}0\\0\\1\\0\end{pmatrix},~~ |dd\rangle=\begin{pmatrix}0\\0\\0\\1\end{pmatrix}$$
Alice studies the first with an operator $\sigma$ and Bob the second with an operator $\tau$ (these are really product operators of single-spin $\sigma_z$ with the identity $I$: $\sigma_z\bigotimes I$ and $I\bigotimes\sigma_z)$:
$$\sigma = \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & -1 \end{pmatrix} ~~~ \tau = \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & -1 \end{pmatrix}$$
Now for the interesting stuff.
We can have a product state where the two spins ("subsystems") are independent (no entanglement):
$$\psi ~=~ (a_1|u\rangle+a_2|d\rangle)\bigotimes(b_1|u\rangle+b_2|d\rangle)$$
$$~~~=~a_1b_1|uu\rangle+a_1b_2|ud\rangle+a_2b_1|du\rangle+a_2b_2|dd\rangle~~~(1)$$
where the $a_i$ and $b_i$ are separately normalized to $1$ so that if we calculate the expectation for either spin the other does not factor in at all. For example $\langle\psi|\sigma|\psi\rangle=a_1^2-a_2^2$ with no appearance of the $b_i$.
Then Susskind says that most randomly chosen coefficients of the $|uu\rangle...$ (normalized) will not factorize as in $(1)$. Then they are entangled. And an example of a maximally entangled state is the singlet state:
$$|S\rangle=\frac{1}{\sqrt 2}(|ud\rangle-|du\rangle)$$
Now $\langle S|\sigma|S\rangle=0$ so you have zero information about the individual spins. However, you have information about correlated measurements, because $\langle S|\tau\sigma|S\rangle=-1$ where by matrix multiplication
$$ \tau_z\sigma_z = \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix} $$
Susskind then discusses how you can test whether a state is entangled or not (and how much entangled) by computing the correlation of operators $A$ and $B$, or checking the eigenvalues of single-state density matricies ($\rho_{2x2})$, which should be $\{1,0,0,0...\}$, or checking if the state coefficients $\{0,\frac{1}{\sqrt 2},-\frac{1}{\sqrt 2},0\}$ can factorize as in $(1)$ (they can't).
Question (rewritten after helpful answers by tparker and Emilio Pisanty)
Aren't these entanglement tests all relative to the chosen 4x4 operators, $\sigma_z$ and $\tau_z$, which reflect a particular choice of dividing the state into subsystems?
Instead of a subdivision based on the two spins, we can subdivide based on $|S\rangle$ and the triplet states $|T_1\rangle=\frac{1}{\sqrt 2}(|ud\rangle+|du\rangle),~~|T_2\rangle=\frac{1}{\sqrt 2}(|uu\rangle+|dd\rangle)$ and $|T_3\rangle=\frac{1}{\sqrt 2}(|uu\rangle-|dd\rangle)$. Let's change basis with a similarity matrix $P=(|T_3\rangle~|T_2\rangle~|T_1\rangle~|S\rangle)$. In this new basis, $|S\rangle...|T_3\rangle$ are basis vectors and
$$ A=\tau_z\sigma_{z,new basis} = \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & -1 \end{pmatrix} =\begin{pmatrix} 1 & 0\\ 0 & -1 \end{pmatrix} \bigotimes \begin{pmatrix} 1 & 0\\ 0 & 1 \end{pmatrix} $$
$$ B=\tau_y\sigma_{y,new basis} = \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & -1 \end{pmatrix} =\begin{pmatrix} 1 & 0\\ 0 & 1 \end{pmatrix} \bigotimes \begin{pmatrix} 1 & 0\\ 0 & -1 \end{pmatrix} $$
We consider the new basis vectors as product vectors isomorphic to single spins which can each be in states labeled $|+\rangle$ and $|-\rangle$ (so as not to confuse with $|u\rangle$ and $|d\rangle$) and we get that
$$ |S\rangle = |{--}\rangle,~~~~ |T_1\rangle = |{-+}\rangle,~~~~ |T_2\rangle = |{+-}\rangle,~~~~ |T_3\rangle = |{++}\rangle $$
Since $A$ and $B$ are of the form of product operators, we can let them define a new subdivision of the full system. Each new subsystem no longer corresponds to an electron at a specific location, as in the original division. A and B can be thought to operate on one label each (A on the first + or -, B on the second).
With this new subdivision, each of $|S\rangle...|T_3\rangle$ are not entangled.
To conclude
Entanglement is in the eye of the beholder (4x4 operator, or subsystem division). Yes?
Answer
I think I understand your question, but I don't understand Aaron Stevens's comments at all, which you claim to be a valid rephrasing, so it's possible that I'm not actually understanding your question correctly. With that caveat:
Your basic idea is right, but your statements aren't quite mathematically precise enough to be completely correct. (For one thing, you're using the words "entangled" and "pure" as if they were mutually exclusive, but they're not - the maximally entangled state that you describe is both entangled and pure.) Yes, whether a state has internal entanglement does indeed depend on how you factor the Hilbert space into subsystems.
But you're missing a key point, which is that the Hilbert spaces for a composite system is a tensor product of the individual systems' Hilbert spaces, not a direct sum. The Hilbert space $\mathcal{H}_{AB} = \{ |uu\rangle, |ud\rangle, |du\rangle, |dd\rangle \}$ for a two-spin system is the tensor product $\mathcal{H}_{AB} = \mathcal{H}_A \otimes \mathcal{H}_B$, where $\mathcal{H}_A$ and $\mathcal{H}_B$ are both isomorphic to the Hilbert space $\{ u, d \}$ for a single spin. So we can meaningfully talk about operator that only act on one subsystem. But the set of linear combinations of the $|S\rangle$ and $|T\rangle$ states forms the direct sum $\{|S\rangle\} \oplus \{|T\rangle\}$, so we can't think of the $|S\rangle$ and $|T\rangle$ states as subsystems that operators can act on independently.
Sometimes, a composite system's Hilbert space can be written as a tensor product in two inequivalent ways. This really does correspond to two different valid ways to divide the complete system into subsystems, and whether or not the subsystems are entangled can indeed depend on that division. (But this is not quite the same thing as basis dependence, because it turns out that the entanglement is independent of the basis one chooses for each subsystem. Once one chooses a division of the complete system into physical subsystems, then any change of basis within one subsystem will not affect the entanglement.)
We can't see this with your two-spin example, but we can see it if we consider a system of three spins $A$, $B$, and $C$, whose Hilbert space is $\mathcal{H}_A \otimes \mathcal{H}_B \otimes \mathcal{H}_C = \{ uuu, uud, udu, udd, duu, dud, ddu, ddd \}$. Consider the state $$\frac{1}{\sqrt{2}} (|u_A d_B\rangle - |d_A u_B\rangle) \otimes |u_C\rangle = \frac{1}{\sqrt{2}}(|udu\rangle - |duu\rangle).$$ In this state the spins A and B are maximally entangled, but the spin C is not entangled with either of them. One person might only have experimental access to operators that act on either (a) the A and B spins or (b) the C spin. This person would naturally consider the A and B spins together as comprising a single subsystem, and the C spin as comprising a separate subsystem. They would therefore naturally factor the Hilbert space as $\mathcal{H} = \mathcal{H}_{AB} \otimes \mathcal{H}_C$, and say that the state is not entangled. They would not observe any unusual correlations between spins in "separate subsystems".
But someone else might have experimental access to a different set of operators, which can only act on either (a) the A spin, or (b) the B and C spins. This second person would naturally consider the A spin as comprising a single subsystem, and the B and C spins together as comprising a separate subsystem. They would therefore naturally factor the Hilbert space as $\mathcal{H} = \mathcal{H}_A \otimes \mathcal{H}_{BC}$, and say that the state is entangled (in fact, maximally entangled). They would observe perfect correlations between (what they describe as) "separate subsystems".
But again, once you specify a particular tensor factorization of your Hilbert space into fixed subsystems, then the entanglement between the subsystems is both basis- and observer-independent.
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