Voltage Divider vs. Resistor in Series
If you draw 1mA from the resistor divider circuit you mentioned, it will output one volt (the upper resistor will have 1.1mA flowing through it, thus dropping 11 volts; of that 1.1mA, 0.1mA will go through the bottom resistor while the remaining 1mA will go into your load). The 6K resistor would drop 6 volts, thus feeding 6 volts into a 100mA load.
If either the load current or the load resistance is a known constant value, one can calculate a series resistance which will convert a known input voltage into any desired known, lower, load voltage. If the load current or resistance isn't known precisely, however, deviations from the ideal will cause the load voltage to vary from what is intended. The greater the difference between the input voltage and the load voltage, the greater the variation in load voltage.
Adding a load resistor will effectively add a known fixed load in addition to the potentially-variable one. Suppose one had a 12-volt source and the intended load were 10uA +/- 5uA at 6 volts. If one just used a series resistor sized for the 10uA case (600K), it would drop only 3V at 5uA (feeding 9 volts to the load) and 9V at 15uA (feeding 3 volts to the load). Adding a 6.06K resistor in parallel with the load would cause the total current draw to be about 1.000mA+/-0.005mA, requiring the upper resistor be changed to 6K; since changes in the load current would only affect the total current by about 0.5%, they would only affect the voltage drop of the upper resistor by about 0.5%.
If the source voltage is stable, and the output current is small, a voltage divider may be a practical means of generating a stable voltage. Unfortunately, for the voltage divider to generate a stable voltage, the amount of current fed through the lower resistor (and thus wasted) must be large relative to the possible absolute variation in load current. This is usually no problem when the output current is on the order of picoamps, is sometimes acceptable when the output current is on the order of microamps, and generally becomes unacceptable when the output current is on the order of amps.
The single resistor doesn't divide the voltage.
For an ideal 12V source with 6k\$\Omega\$ in series, you get 12V with 6k (output) impedance.
The centre of two 10k resistors in series across the same source would provide 6V with an impedance of 5k\$\Omega\$.
So there is no difference between this and a 6V source with 5k in series.
If you really have the 1mA then the single resistor will do. The 1mA will flow into the input of the circuit following the resistor and this will therefore have an input resistance of 6k\$\Omega\$ (6V / 1mA). So you end up with two resistors after all: the one you placed and the input impedance.
In case you're building the divider with the two 10k\$\Omega\$ resistors keep in mind that the input impedance of the following circuit is parallel to the lower resistor. Anything but a high-impedance input (like the input of an opamp) will decrease the 6V at the node.