electric-circuitselectrical-resistanceelectricity
Would current flow through the resistor below?
The power source is an ideal cell without internal resistance, and the wires are also ideal.
I thought that current won't flow through the resistor because there is no potential difference between the points $\text{A}$ and $\text{B}$, but I am not sure.
The voltage across either horizontal resistor is zero so they can be removed from the circuit without changing the solution.
This is most easily seen by simply removing the two horizontal resistors and then it's clear that the nodes the horizontal resistors connect to each have the same voltage. Thus, by Ohm's law, there is no current through either horizontal resistor since there is no voltage across either resistor.
In other words, it doesn't matter if the horizontal resistors are there or not so they can be removed without changing the solution.
You are then left with 3 identical paths with resistance 2R each.
Of course, in a real circuit, the real resistors will never be identical so, this ideal solution is only an approximation.
The word "Earth" is sometimes misleading. I think (if I get the sense of your question right), it is more properly called a "Protective Earth". In home electricity supplies, one side of the supply is "tethered" to the same potential as a protective earth circuit. This latter is simply a system of conductors, going through the third "Earth" pin on the socket outlet, that tethers any conducting surface of an electrical appliance to the one (called the "Neutral") side of the supply. The other side of the supply is called the "active".
If a fault happens in an appliance such that the active touches the conductive housing of an appliance (say of a toaster), we have a dangerous situation, since anyone touching the appliance can then get an electric shock. However, if the housing is connected to the protective earth circuit, there is a redundant path back to the supply's neutral. This leads to a high current in the redundant path and hopefully a blown fuse in the supply live.
A more modern and safer way to achieve this protection is earth leakage protection - a system that detects, through the Ampère-law-begotten "magneto motive force MMF)", when the current through the active is different from that through the neutral. In this system, both active and neutral lines pass through a torus-shapen ferromanetic core. If the current in one doesn't match that in the other (i.e. the current going into the appliance through the active is not the same magnitude but exactly out of phase with the current coming out through the neutral), then there is a nonzero $\oint \vec{H}\cdot{\rm d}\vec{\ell}$ ("magneto-motive force") around the core and thus an AC magnetic field though a sense coil wound threading the torus- by Faraday's law, this begets an EMF in the sense coil which trips a circuit breaker. These devices can be made very cheaply and are extremely effective - shutting off within milliseconds if an imbalance of more than typically $50{\rm \mu A}$ is sensed.
Lastly, electric current can in general flow without a return path through the mechanisms of displacement current and capacitance. See my answer to the question "Does alternating current (AC) require a complete circuit?" for details.
Lastly, there have been some really bizarre truly one-line power transmission systems thought of in the past, where the one line works as a waveguide and does not need a return path. See the Wiki Page for the Goubau Line for details.
Best Answer
If the lower wire from A to B is not assumed to be of zero resistance then definitely current would flow through both of them.
If the upper net resistance is $R_\text{upper}$ and the lower is $R_\text{lower}$ then $$ \frac{I_\text{upper}}{I_\text{lower}}~=~\frac{R_\text{lower}}{R_\text{upper}} \,.$$