How do I reduce the brightness of an automotive lamp?
I don't know how to calculate the proper size resister to safely
  reduce the brightness to say maybe half. There are two lamps, 21w each
  that are fed from one line.


The relevant formulas are \$R = V / I\$ and \$P = V * I\$ (where V = voltage in volts, I = current in amps, R = resistance in ohms, and P = power in watts).  

Calculating the required resistance for an incandescent lamp is a little tricky though, because the filament resistance has a positive temperature coefficient so it reduces at lower voltage (causing higher than expected power draw), while the light output spectrum shifts more towards infrared which is less visible. So you might have to experiment with different resistor values to get the exact effect you want.

Another option might be to simply wire an identical lamp in series. Each lamp should then drop half the total voltage, and you could use the 'ballast' lamp as an indicator that the tail light is on. 

So how much current and power would each lamp draw, and what would be the equivalent resistance to do the same job? I tested a 12 V 15 W festoon lamp on a range of voltages from 0 to 12 V, and this is what I got:-



At 12 V the lamp drew 1.22 A, so its resistance was 12/1.22 = 9.84 &ohm;. At 6V it drew 0.86 A, ~70% of the current at 12 V (a fixed resistor would draw 50%). Its resistance at 6 V was 6/0.86 = 6.98 &ohm;, ~70% of the value at 12 V. 

We can expect that a 12 V 21 W lamp would act similarly. At 12 V it should draw 21/12 = 1.75 A, and the resistance should be 12/1.75 = 6.86 &ohm;. 70% of that is 1.23 A and 4.8 &ohm;. So a 4.7 &ohm; (standard value) power resistor should drop the voltage from 12 V to ~6 V. It would dissipate ~6*1.23 = ~7.4 watts, so should be rated for 15 W or higher (derated 50% to keep surface temperature down), and kept away from anything that doesn't like high temperatures.     

So you could put either a 4.7 &ohm; 15 W resistor or a 12 V 21 W lamp in series to reduce the tail light voltage to ~6 V.