Does light with a wavelength on the Planck scale become a self-trapping black hole? As the wavelength of a photon shrinks, its energy rises, and so its mass rises (using $E=hc/\lambda$ and $m=E/c^2$). On calculating the Schwarzschild radius for a photon based on its mass derived from those two equations, I found that the Schwarzschild radius of the photon will be equal to $\lambda/2\pi$ in one instance, when the wavelength of the photon equals $2\pi$ times the Planck's length: 
$$\lambda=2\pi \times \mathscr{L}_P\implies r_s=\frac{\lambda}{2\pi}$$  
where $r_s$is the Schwarzschild radius and $\mathscr{L}_P$ is the Planck's length.      
In other words, a photon with a wavelength $\lambda=2\pi \times \mathscr{L}_P$ would gravitationally trap itself in a circular orbit with a radius equal to the plank length.  A photon in a circular path with diameter of $2\pi \mathscr{L}_P$ would have a gravity well that would trap itself at the corresponding radius of the plank length (with an orbital path diameter of $\lambda=2\pi \times \mathscr{L}_P$).  Has this been discussed as a conceptual mechanism as to why the plank length is a lower limit on potential allowed wavelengths, and the resolution of the universe (that a photon with a wavelength of $2\pi \mathscr{L}_P$ in a circular path with a diameter equal to that wavelength 
would in fact be the definition of a black hole?)
 A: I can't improve on Mark's answer, but I would add that while a single ray of light cannot form a black hole, light can form a black hole if multiple rays are focussed onto the same point. This object is known as a Kugelblitz.
The important distinction from the single ray is that in a Kugelblitz the net momentum is zero due to the (approximate) spherical symmetry.
A: A photon of sufficiently small wavelength would not become a black hole. To see this, consider two observers: one who measures the photon as having a Planck-scale wavelength, and another that is traveling at high speed in the same direction as the photon. This second observer will observe the photon to have a wavelength that is longer than that measured by the first observer due to Doppler shifting. The second observer will conclude that the photon does not have enough energy to create a black hole. All observers will agree on whether a black hole exists or not, so the only consistent conclusion is that no black hole forms.
We can also reason in the reverse. A photon of visible light obviously doesn't have enough energy to create a black hole. Otherwise, light bulbs would be dangerous black hole generators. However, due to Doppler shifting, there is a frame of reference traveling at high speed towards the photon in which that photon has a much larger amount of energy--large enough to create a black hole if that was possible.
Be careful when assigning importance to Planck-scale measurements. We do not know if the Planck units have any physical significance. It is not known if spacetime is continuous or not, and if it's not, we have no reason to think that the "resolution" of spacetime is at all related to the Planck length. Just as a counter example, the Planck mass is about 20 micrograms--a small amount, but one that is handled all the time by pharmacists and far larger than any fundamental particle mass.
A: This is no doubt both mathematically and physically correct. When we construct either a proton or neutron we have “cancelled out” charge components. But what is a photon but oscillating opposite charges? Your instinct is probably correct. These charge components become trapped over their own gravity well an their propagation becomes rotation. Forward momentum is now angular momentum. $m=E/c^2$. Mass is energy in orbit which acquires through this angular momentum, inertia, mass.
