Let me try to shed a little light on this topic. W/o a protection diode across the DC relay coil, when the circuit is opened, the coil becomes an energy source due to the energy stored in the electromagnetic field of the coil inductance. That field will begin to collapse in order to maintain the current that was flowing when the circuit became open.. The physics is such that the field will reverse polarity and the voltage generated will spike to a very high level as the circuit tries to maintain the current . You've probably seen an arc occur across a switch or relay contact when switching off an inductive load. That arc is due to the stored energy in the inductance.
A diode across the coil is reverse biased in normal operation and no current flows through it. When the circuit is opened, the field polarity switches and the field current has a path through the now forward biased diode. There will be no voltage spike generated as the current will flow through the diode and back into the coil. The diode's reverse voltage rating only has to be greater than the normal supply voltage. The current that flows in the diode is limited by the coil resistance, so that is the determining factor for the diode's current rating. The field energy is dissipated as heat in the resistance of the diode and relay coil. The greater the inductance of the coil, the longer the current will flow for a given resistance.
Note that the current does NOT flow back into the power source. It can't because the power source is in series with the open switch.
Signal diodes work well for small DC relays, as has been stated. A 1N4001 diode works well with larger DC relays due to its one amp continuous rating.
AC relays aren't as easy to protect. The advent of solid state relays with "zero current crossing" detectors enables them to switch when the current is zero and thus alleviates the problem.
Hope this helps!