Tensor product of Pseudorandom States

I am reading the paper Pseudorandom Quantum States,where the following candidate was shown to be a Pseudorandom state, called the complex random phase state:
$\mathrm{PRF}_k:X → X$ be a keyed pseudorandom function where $X = \{0, 1, 2, . . . , N − 1\}$ and $N = 2^n$ with $n$ being the security parameter. The family of pseudorandom states is then defined to be $$|\phi_k\rangle =\frac{1}{\sqrt{N}}\sum_{x\in X}\omega^{\mathrm{PRF}_k(x)}|x\rangle,\quad \omega = \exp(2\pi i/N).$$

My question is does the tensor product of two pseudorandom states remain pseudorandom? Given the above example , we can construct another PRS by letting $\omega = \exp(-2\pi i/N)$. If we then take their tensor product we get $$\frac{1}{N}\sum_{x,y\in X}\omega^{\mathrm{PRF}_k(x)-\mathrm{PRF}_k(y)}|x,y\rangle,$$ thus a state whose $N$ out of $N^2$ co-ordinates are fixed to $1/N$. I do not know if this is a PRS or not.

Any help will be really appreciated. Thanks in advance.

This is not a PRS. As you note, the tensor product is superposition of $N^2$ possible states and so if it is a PRS it can only be interpreted be with respect to the set $X'=\{0,\ldots,N^2-1\}$ e.g. by associating the state $x,y$ with $Nx+y$. Likewise, the amplitudes of the states would have to be of the form $\omega'^j$ where $\omega'=\exp(2\pi i/N^2)$, with a pseudorandom function selecting the $j$ quasi-uniformly. Here the amplitudes all lie in a subset of size $N$ (they're all power of $\omega=\omega'^N$), and in particular (as you note) at least $N$ amplitudes that are $\omega'^0$. This is not close to having phases uniform in $X'$ and so is not a PRS.

This is also true even if we tensor $|\phi_k\rangle$ with $|\phi_k'\rangle$ with $k\neq k'$, or even if we tensor PRS drawn from different pseudo-random functions.