Variation of peak current and peak voltage with capacitance in an AC circuit The relation of peak current, peak voltage and capacitive reactance in alternating current is given by:
$$i_m=\frac{v_m}{X_c}$$
and $$X_c=\frac{1}{C\omega } \, .$$
So if we have a circuit with a capacitor and a bulb connected in series in an AC circuit, the bulb keeps glowing due the property of AC current. So if we reduce capacitance, $X_c$ increases.
So $i_m$ decreases (and $v_m$ stays constant), but if we rearrange the former equation can't we say that $v_m$ increases (and $i_m$ stays constant).
According to my text $i_m$ decreases (and $v_m$ stays constant). But how do we know that? Why can't it be the other way? Am I missing any concept?
 A: I think that the problem here is that you haven't properly set up the circuit equations.

So if we have a circuit with a capacitor and a bulb connected in
  series in an AC circuit

Since the bulb and capacitor are series connected, the current through each is identical.
Denote the series current phasor as $I_s$.  Assuming the source is an AC voltage source, it is true that the voltage across the source is constant.
But, by Kirchhoff's voltage law, we have
$$V_s = V_c + V_b$$
where the phasor voltages above are the source voltage, capacitor voltage and bulb voltage respectively.
By Ohm's law, we have
$$V_c = I_s \frac{1}{i\omega C}$$
and
$$V_b = I_s R_b $$
Thus, 
$$I_s = \frac{V_s}{R_b + \frac{1}{i\omega C}}$$
Then, holding $V_s$ constant, see that decreasing $C$ decreases $I_s$.
A: 
So $i_m$ decreases (and $v_m$ stays constant), but if we rearrange the former equation can't we say that $v_m$ increases (and $i_m$ stays constant).

Generally when one is dealing with a circuit, the voltage supplied comes from somewhere such as a power supply. Power supplies usually supply a fixed voltage rather than a fixed current, which is why $v_m$ "stays constant". (There are also fixed-current power supplies in which the voltage alters, while the current stays constant. These are not common, though.)

According to my text $i_m$ decreases (and $v_m$ stays constant). But how do we know that? Why can't it be the other way? Am I missing any concept?

It certainly can be the other way around, but it depends on what else is connected to the circuit.
