Gauge a global symmetry of a CFT Suppose we have a CFT with a global symmetry, and let us denote the theory as $T$. Now we gauge the global symmetry, to obtain another theory $T'$. Does $T'$ also have to be a CFT? If not, what is the condition for $T'$ to be a CFT?
If the global symmetry is discrete, I think $T'$ is still a CFT. But if the symmetry is continuous, it looks like not. For example, if we start with a free fermion theory, which is obviously a CFT. Suppose there is a $SU(N)$ flavor symmetry rotating the fermions. Then we can gauge $SU(N)$, to get a $SU(N)$ gauge theory, which is a QCD-like theory. QCD is not conformal, so $T'$ is not conformal.
The difference between the two cases are: for discrete symmetry, one does not introduce a kinetic term for gauge field after gauging, while for continuous symmetry there is a kinetic term which involves a running coupling constant. So I am wondering: if we don't introduce the kinetic term for continuous gauge field, do we still regard the gauged theory $T'$ as a CFT?
 A: I think that you answered your own question quite well:
If your symmetry is a continuous symmetry, the theories $T$ and $T'$ are very different: they don't have the same field content ($T'$ has additional gauge fields), and not even the same symmetries (in $T'$ the global symmetry is gone after gauging: all gauge-invariant observables are scalars). In this case I don't see any reason why $T'$ would be a CFT. $T$ and $T'$ share some similarity at the Lagrangian level, but as quantum field theories they are in fact completely unrelated.
If the symmetry is discrete, then the story is different. The theory $T'$ is closely related to $T$, in the sense that it has exactly the same local observables, except that you choose to look at only a subset of them (those that are uncharged under your discrete symmetry). Then if $T$ is a CFT, $T'$ is certainly a CFT as well. Here is an example: suppose that your theory $T$ is the 2d Ising model, which is a CFT with a discrete $\mathbb{Z}_2$ symmetry; if you gauge this symmetry, then you end up with a theory $T'$ that is the free fermion in 2d, also a CFT.
There can be subtleties regarding non-local observables, but as far as I understand these are beyond the scope of the question.
