How can gravity increase $p$ without increase of $\rho$ or $T$? A point in static fluid has pressure of $\rho$ * g * h, where $\rho$ is the constant density throughout the fluid body, g the gravitational accelaration, and h is the height of the fluid column above that point.
What I can't understand is how this increase in pressure is not accompanied (or caused) by matching increase in density or temperature.
In my understanding, whether it's liquid or gas, pressure basically is the aggregate effect of randomly moving particles exerting force on their surroundings, hence it is connected to the translational kinetic energy of the particles and the frequency of their contact with the surroundings. So it seems any increase in pressure should mean either (1) decrease in volume (or increase in density) or (2) increase in temperature (increased momentum of particles). Apparently this is incorrect as no article on the internet or textbooks I could find suggest this.
What am I getting wrong?
How does weight of a fluid column cause pressure rise in microscopic point of view? By making particles faster? Then how come the temperature remains unaffected?
 A: The density being constant means you assume incompressibility, which is an approximation. In reality the density of the fluid at the bottom will be greater than at the top.
In the case of a liquid this seems confusing, but remember that liquids have a low compressibility (hence the assumption of incompressibility), meaning that $\kappa = \dfrac{\partial p}{\partial \rho}$ is large. So even though the difference in density between the top and bottom of the container containing the liquid is small, the difference in pressure is large.
The reason a small change in density can cause such a huge pressure difference, is ultimately due to Pauli's exclusion principle. Since the electrons in the liquid aren't allowed to occupy the same state, the structure of the states has to be changed in order to compress the liquid. This however requires huge amounts of energy due to the electron-electron repulsion and the repulsion between different nuclei. So we need a huge force (pressure) to compress the liquid.
A: 
What I can't understand is how this increase of pressure is not accompanied (or caused) by matching increase of density or temperature.

There is a small increase in density.  It's just that the pressure increase seems out of proportion when you think about how a gas reacts.
Unlike an ideal gas where we assume the particles have insignificant volume, in a liquid the particles take up most of the volume.  The particles have very little "empty space" they can move around in before colliding.  A fairly tiny density increase can mean a significant reduction in this "empty space".  The less space, the more often collisions happen, and the more force is transmitted.
(Atoms/molecules aren't really hard spheres and there isn't really empty space between them in a liquid, but the analogy isn't bad)  
A: An increase in pressure of a liquid on a surface is not accompanied or caused by an increase in temperature or pressure because, unlike a gas, a liquid is relatively incompressible making the density and temperature of a liquid relatively independent of pressure. For example the density of water at the bottom of the deepest ocean is only about 5% higher than at the surface. That makes the pressure primarily dependent on inter molecular forces as opposed to the velocities of the liquid molecules. Those compressive inter molecular forces increase with the weight of the liquid above the surface. 
So from the microscopic point of view the increase in pressure is due to increased iinter molecular forces which, in turn, is due to the weight of the liquid above the surface.
Hope this helps.
