You left out one critical piece of information, the amperage draw of the lamp.
You are wanting to drop (lose) some of the voltage (perhaps a volt or two) in the resistor. The voltage drop in the resistor is determined by the formula volts=amps X ohms (E=I*R where E stand for voltage, I for amps and R for ohms).
If, for example, the bulb were to draw 0.06 amps (small indicator lamp) and you want to drop 2 volts, you would divide 2 volts by 0.06 amps to get a required resistance of 33.333 ohms.
Discalimer for the purists; Yes, once you drop some voltage the amount of current that the lamp draws will decrease slightly though not in a linear fashion (characteristic of incandescent lamp) so when you figure the resistance to drop exactly 2 volts it may only end up dropping 1.9 volts. Close enough, we are just dimming a light.
Next, determine how many watts you need to dissipate in the resistor, the power is consumes is converted into heat and they will get warm or hot, need to be rated to handle it.
Watts=volts X amps, or, watts=amps X amps X ohms.
Dropping 2 volts in the resistor at 0.06 amps will cause it to dissipate 0.12 watts. In this case a common 1/4 or 1/2 watt resistor would suffice but if you are using a lamp that draws more current or you want to drop more voltage, it will need to handle a greater wattage.
Don't know how much current the lamp draws?
Here is a trick that I use when I just want to drop a volt or so (especially if the current may vary), use common silicon rectifier diodes. Commonly available in a 1 amp size which would be good for a bulb drawing up to 1 amp (though they will get warm...), as long as you dont exceed the current rating, it doesnt matter how much current the lamp draws, it will always drop about 0.6 volts (affected slightly by the amount of current and the temperature of the diode for you purists). Want to drop more? use 2, 3 or 4 of them in series. You will notice on the diode that there is a band painted on one end or it may be marked with a letter K, this is the cathode. The diode is a "checkvalve" for electricity, current only goes one way through it, if you hook it up backwards it will block the flow but with your 12v supply, no harm is done. With a DC power source, connect from the Positive supply terminal to the Anode of the diode (the end opposite the Cathode), the cathode end goes to the lamp and the other side of the lamp to the neg power supply connection.