You need to drop the voltage using a capacitor. I can't remember exactly what value you want, but 330nF sounds familiar. You can work it out easily enough, knowing you want to drop about 230V at 10 to 20mA (depending on your LED) at 50Hz. If you don't know the formulae involved, you probably shouldn't be working with the mains!

You could use an inductor instead, but capacitors are cheaper and more convenient. Don't use a resistor, unless you want to make a heater!

The other thing you need to remember is to put a diode (possibly another LED) back-to-back with your LED, but the opposite way around, so you don't get a big reverse voltage across it in the other half of the cycle. In fact, you can obtain a red LED back-to-back with a green one - these appear yellow with AC current.