Reduce Voltage?

Actually you can't do that at all. You have 12V @ 80mA available. That means that you have P = V*I = 960mW available. If you need 9V @ 120mA, then that means that you need 1080mW, which is more power than you have available.

If you can live with 9V @ 80 mA output, or if you can come up with 12V @ 120 mA input, then you can use a three-terminal voltage regulator, for example a

7809:

Hook it up like the datasheet shows in the figure at the bottom left of p.

1. Connect pin 1 to +12V (the positive wire of your +12V supply), connect pin 2 to ground (the negative wire of your +12V supply), and you will get

+9V between pin 3 and pin 2.

The datasheet shows some capacitors at the input and output; it might work without them but it would be better if you put them in. The values are not critical; anything over 0.1uF will do for either cap.

You can buy the 7809 from Digikey

but it is very common so you can probably find it locally too. Get it in the TO-220F package (see the picture on the first page of the datasheet).

A resistor would drop the voltage, but the voltage drop would depend on the current, so the voltage at the load would depend on how much current the load was drawing. That probably isn't what you want, though it might still work.

Jonathan