Interpretation of the mass of the field Before QFT I had never thought if a field should or should not have any mass. Now it turns out that either case is possible.
It might be naive to think this way, but I picture the mass of the field in the same way as that of a particle, but distributed along the field. The problem is how to think of a massless field.

So, my question is: what does it mean to say a field has mass/is massless?

 A: The mass of a field is completely well-defined only for free field theories. There it is the mass of the particle whose 1-particle equation is second quantized. For fields of spin $>1/2$, massless is equivalent to the existence of local gauge transformations. 
Once one adds interactions, the language is kept (based on the noninteracting part) but the meaning becomes somewhat fuzzy due to the need for renormalization. What is meaningful again are only the masses of the asymptotic (scattering) fields - which are the masses of the observable particles.
A: There are a few things to unravel here.
The simplest thing to do would be to say that massless fields have massless excitations. However, this doesn't work because invariant mass doesn't add directly. For example, a system of two gluons can have a mass, as explained here, even though each gluon by itself is massless. This may be detected by the system's inertia.
As a result, you cannot assign an "invariant mass density" $\rho(\mathbf{x})$ to a field state. If you tried this with the two gluon example, it would be zero everywhere, since each gluon is massless, which is wrong. It also doesn't make sense for the classical electromagnetic field, since light in a box has mass.
To get a sensible result we have to restrict to single particle states with definite momentum. In special relativity, we define the invariant mass by
$$E^2 = \mathbf{p}^2 + m^2.$$
Using the de Broglie relations $E = \hbar \omega$ and $p = \hbar k$, and setting $\hbar$ to one, this gives the dispersion relation
$$\omega^2 = k^2 + m^2.$$
That is, the frequency $\omega$ of a single particle state with definite momentum is larger the bigger the mass of the field is, because of the rest mass contribution.
