I take it you mean 100 ohms, not 100W
Did you, then, try it on four AA cells, and find it was no less bright?
Way it works is:
Assuming it has a resistive ballast.
Voltage drop across LED = 2V (assuming single emitter or parallel emitters) (red LEDs are always 2V). You knock this off the 12V supply voltage to do the calculations, because it is constant and does not vary (significantly) with current.
Voltage drop across ballast resistor = 10V (this is the important bit of the voltage)
Current = 0.2A
=> ballast resistor that it has already = 50 ohms
To halve current, double ballast resistor; this means, add another resistor of the same value (ie. 50 ohms) in series. 47 ohms is the nearest E12 preferred value to 50 ohms. (51 ohms in the E24 series is closer, but E12 values are more common, and it's not that critical.)
If it has a linear regulator ballast, which is a thing that acts like a resistor that gets bigger the more voltage you put on it, then that will compensate for you adding external series resistors by reducing its own resistance, until it gets to zero and runs out of range. So you need to make it turn itself all the way down to zero and supply all the necessary resistance externally yourself. That means adding a 100 ohm external ballast resistor.
If it has a switching ballast, you basically need to do the same thing, but it may not work. A linear regulator is more or less guaranteed to turn itself down to zero if you take away all its headroom, but a switching one may, or may not, or may start doing something weird, depending on its topology. Impossible to predict without seeing its actual circuit.