Radio antennas that are much shorter than the wavelength

http://www.antenna-theory.com/antennas/shortdipole.php is a website with useful info., including formulas.
To oversimplify, it seems to say that once the antenna is a tenth or less of the wavelength, the exact ratios don't matter so much. The antenna is inefficient, but it works for both sending and receiving. If you can detect the signal, of course you can amplify it as much as you want.

The key to the efficiency of an antenna (whether for transmitting or receiving - the two processes are essentially reciprocal) is resonance, and impedance matching with the source / receiver. The size also matters in terms of the relationship between power and current.

A nice analysis of the impact of size of an antenna on the power/current relationship is given at this site. Summarizing:

The current in a dipole antenna goes linearly from a maximum at the center to zero at the end. Because the amplitude of the generated E-field from a given point is proportional to the current at that point, the average power dissipated is (equation 3A2 from the above link):

$$\left<P\right>=\frac{\pi^2}{3c}\left(\frac{I_0 \ell}{\lambda}\right)^2$$

(note - this is in cgs units... more about that later). For the same current, as you double the length of your (much shorter than $\lambda/4$) antenna, you quadruple the power.

Directly related to this concept of power is the concept of radiation resistance: if you think of your antenna as a resistor into which you are dissipating power, then you know that

$$\left<P\right> = \frac12 I^2 R$$

and combining that with the above equation for power, we see that we can get an expression for the radiation resistance

$$R = \frac{2\pi^2}{3c}\left(\frac{\ell}{\lambda}\right)^2$$

This is still in cgs, which will drive most electrical engineers nuts. Converting to SI units (so we get resistance in Ohms) we just need a scale factor of $10^{9}/c^2$ (with $c$ in cgs units...); thus we get a simple approximation for radiation resistance in SI units (I now go from $c=2.98\times10^{10}~\rm{cm/s}$ to $c=2.97\times10^8~\rm{m/s}$):

$$R = \frac{2\pi^2 c}{3\times 10^{-7}}\left(\frac{\ell}{\lambda}\right)^2$$

which agrees nicely with the expression given at this calculator for an electrically short dipole (note - their expression is for $\ell_{eff}$ which is $\ell/2$ for a short dipole; and they use slightly rounded numbers which is OK since there are some approximations going on anyway).

But if we are driving with a 50 Ohm cable, and our antenna represents a much smaller "resistance", then most of the power would be reflected and we don't get a good coupling of power into the antenna (remember - because of reciprocity, everything I say about transmitting is true for receiving... but intuitively the transmission case is so much easier to grasp). So to get good efficiency, we need to make sure there is an impedance match between our antenna and the transmitter / receiver.

If you know what frequency you are working at, impedance matching can be done with a simple LC circuit: the series LC represent a low impedance to the antenna, but a high impedance to the receiver. In the process, they convert the large current in the antenna into a large voltage for the receiver (source of image and detailed explanation)

This is an example of resonant matching: it works well at a specific frequency. One can use signal transformers to achieve the same thing over a wider range of frequencies - but this loses you some of the advantages of resonance (all frequencies are amplified equally).

It remains to be shown what the real effect is of reducing antenna size on the received signal. For this, the most extensive reference I could find was this MIT open course lecture. Starting on page 121, this shows that the effective length of a dipole determines how much of the incoming energy can be "harvested", and it again shows that the power is proportional to the square of the size. So an antenna that is twice as short will collect four times less power. But that means it will also collect four times less noise. As long as most of the noise in the system comes from "outside", the ratio (SNR) will be the same, and you don't suffer from the smaller antenna.

This changes once the antenna becomes so small that other sources of noise become significant. It is reasonable to think that this will happen when the conductive (lossy) resistance of the antenna becomes comparable to the reactive (radiation resistance). But since the former scales with the length of the antenna, and the latter with the square of the length, it is obvious there will be a size at which the non-ideal effects will dominate.

The better the conductors, and the better the amplifiers, the smaller the antenna can be.

Summary

So yes, the power transmitted drops with the square of the length, making a short antenna less efficient as a transmitter (and therefore, as a receiver). Much of the time, though, you care about signal to noise ratio - is there more signal than noise coming from your antenna?

For this, we need to look at the Q of the antenna (bandwidth). The higher the Q, the more gain you have at just the frequency of interest (because of resonance); while "noise" is a wide-band phenomenon, "signal" is a narrow-band one, so a high Q amplifies the signal without amplifying (all the) noise. If we can make an antenna with a high Q, then it doesn't matter so much that it is short.

The requirements for transmitting antennae are much higher than for receiving antennas. Transmitting antennas must optimally radiate, so that the signal is not obscured by other stations with better antennas. If a receiver antenna is too short and far away from resonance, all received stations are uniformly weaker. What matters is that the desired signal is not less than the limit of sensitivity of the receiver. This is true for AM and FM.