Why can you have shiny black objects? Knowing black is supposed to be the "color" (I don't want to get into the color/hue/shade debate, please) that absorbs light. how does one manage to have shiny black surfaces? I know about "gloss black" versus "matte black" finishes, but shouldn't the light passing through the gloss (if they didn't pass through the gloss, you wouldn't see the black, right?) be absorbed by the underlying black object? Then there are black gemstones like jet and opal.
How do black objects shine?
 A: Ptomato's answer is right. I just wanted to add that many glossy black objects in daily life reflect light because they have a thin transparent layer on top. If its index of refraction is high and its surface is smooth, you have a glossy surface. Underneath that, you have your black (and possibly rough) material. This is the case in many cheap plastic objects.
A: At any surface (at least one which has a different index of refraction from air) some light is reflected, depending on the angle at which the light hits the surface and the polarization of the light; the Fresnel equations will tell you what fraction of your light is reflected and what fraction is transmitted. When you see a black object "shine", you are seeing the reflected light. But since the object is black, all the "transmitted" light is simply absorbed.
The difference between a matte black and a gloss black finish is one of index of refraction, I guess, and possibly of rough/smoothness.
A: A perfectly black body is not something you see on your daily life, so when you see a "black object" it's actually a almost black object (black enough, for our perception) but it is reflecting some light, wich we are able to perceive as specular.
A: As I was reading other people's answers I noticed that they were very... complicated. A simpler answer would be that if you have a shiny black object, the shine could be coming from a thin transparent coating, not the black object itself.
A: Reflection of radiowaves is the result of electrical currents or dielectric charge/discharge induced on the surface of conductors, dielectrics and everything in between. For visible light the model is still valid, but may be more affected by atomic/molecular properties of surface.
So black glossy dielectric partially reflects (horizontally) polarized white light and absorbs the rest of energy. The proportion depends on angle. Black matte dielectric or white matte conductor (say PLatinum with porous surface) has more complex path with more than 1 changes of light path, so more energy is absorbed. The purely metal (white) conductor with very complex surface can look like completely matte black. 
A: The light that you see reflected off of a solid surface is a mix of specular reflection and diffuse reflection.  For most surfaces, the apparent color of the object is due almost entirely to the diffuse component.
I can't explain the mechanisms behind the two types of reflection, but when you see a shiny black object, "shiny" means the surface gives a strong specular reflection, and "black" means it only gives a very weak diffuse reflection.
A: Simply, shiny 'black' objects are not black. They may be close to black and will appear black compared to other colours, but, as soon as they reflect some electromagnetic radiation (even a few photons), these surfaces have colour. The difference between a matte black and a shiny black with radiation is that when the photons directly hit a polished object, they are more likely to be absorbed than if the hit is at an angle. The difference between a matte finish and a shiny finish is that, no matter the angle, the light will always hit a part directly as matte is just loads of smooth surfaces facing different directions.
You can get into refraction indices and the Fresnel Equations, but there is really no point as it is just a mathematical way of saying the same thing: matte black is just a polished surface that is always at the correct angle to absorb maximally and, when it does reflect, it can reflect into itself as the photons are obviously small enough to collide with multiple 'bumps'.
Emission of matt surfaces is purely that if you consider a bumpy surface, the surface area is higher (like the villi in the gut), meaning that a larger amount of radiation can be emitted.
A: As ptomato points out, even a "very black" object reflects some light. If the surface is rough (at a small scale), the light is reflected off in all directions, and the object appears, matte. If the surface is smooth, then all of the reflected light will bounce in the same way, so that you can see the reflection of the light source; this reflection will be easier to see against the black surface than it would be on a similar white (and shiny) object.
If my model is correct, then shiny black should look shiny under direct light, but shiny black and matte black should look very similar under indirect light (in monotonous surroundings).
Also, matte black objects are shiny when wet, so if my model is correct then that shine must be from the liquid surface. So I predict that the color of the shine will be the color of the liquid. For example, if a white light shines on a matte maroon object, if you wet it with water the reflections will be white, but if you polish it the reflections will be maroon.
A: Black is all coulers, not a shade, when  learning about the spretrem of light from the sun,
And haw objects only can reflect there own couler of its surface but with a different couler that is not of there the object can not reflect other coulers there for no light bounces of and we do not see a red object but a black object that doesn't repct any light at all. 
But we all know black object reflect of light for why we can see the couler that it is,  witch cpuld mean black absorbs all coulers from the sun rather than one or two but all of them, when looking at black objects we can still see that light is reflected and we can see the object and know the  colour , what we do know is from what we witness, and the only way for black to be able to be seen and have a  reflective surface  is that all  Spectrum  of light is   Absorbed then revealing the object to the naked eye and light can bounce of it becouse it is all  colours from Spectrum of light,rather being just a black blob or object that dose not reflect light.
And when a object becomes black becouse it can only reflect its own   Colour  there for it has its coulée but can not be seen by any other  Spectrum of light so it is not a black object like a b object  that is black and Reflects  light all the other objects.
So yes my  Theory is that black is every  colour . 
