From Jeff Schonert:

Q:  First, I’m still pretty confused about the apparent “arbitrariness”
of choosing a renormalization prescriptions and how that squares with
what we measure.  We saw that in writing down counterterms, we could
choose what finite parts to subtract (e.g. minimal subtraction,
etc.).  So say you’re considering the renormalized mass (squared),
which is bare mass squared plus the self-energy.  Do you figure out
what to subtract off in the self-energy (i.e. pick a renormalization
prescription) by making sure your result agrees with the
experimentally measured value?  Or does it not make a difference
because depending on what you subtract off, the other parameters in
the theory change in a way to leave the final result the same, no
matter what prescription you chose?

A:  It is probably best to start by making definitions of the parameters m^2 and g, and the normalization of the field \phi.  Is m the physical mass, or is it something else that might agree with is at tree level (for example, the inverse propagator evaluated at zero momentum) but which be different in general?  For the coupling g there is no obvious choice, since you measure it in a scattering process and the answer depends on energies and angles, so you make some choice.  When you calculate, the definitions that you have made determine the values of the values of the counterterms --- the infinite parts always cancel, the finite parts depend on your definition.  (This should not be a surprise.  Recall that these arose in the first place as a consequence of our definitions: we broke the Lagrangian into two pieces, in a way that was determined by our definitions.)  Thus if m is not chosen to be the physical mass, we have to calculate the relation between these.

Dimensional regularization is a little more abstract because we do not start by defining m and g in a physical way; instead we decide what the finite part of the counterterm will be (zero) and then have to calculate how m and g relate to physical quantities.

I have not yet used the term `bare mass'.  Literally, this would be the mass parameter that appears in the Lagrangian, but this is in general not physically meaningful (especially in dimensional regularization).  When we get to effective field theory (ch. 29) we can make a more physical definition of bare mass.

Q:  Second, you mentioned how dimensional regularization is preferred
because as we will see later in the course, it respects gauge
symmetry.  Does the regularized theory always have to respect the
symmetry of the Lagrangian, or is it more of a convenience to do so?

A:  If you have two different regulators, then one of the principles of renormalization theory is that you get the same theory in the end --- the counterterms are different but the total amplitude is the same.  So if you have a regulator that preserves the symmetry, and one that does not, then you can always choose the counterterms in the latter case to give a symmetric limit.  But you have to work much harder because you don't have the symmetry in the intermediate steps.

Lattice gauge theory is an interesting example: the Lorentz symmetry is broken.  Fortunately, the cubic symmetry of the lattice is a large enough subgroup that the counterterms are strongly restricted.  And it does preserve gauge invariance (for numerical integration of the path integral, there is no known way to implement dimensional regularization).  On the other hand, a momentum cutoff would preserve Lorentz symmetry but break gauge invariance: this would be much more unwieldy.

Q:  For example, in \phi^{4} theory in 4 dimensions, could you ever
generate a \phi^{3} counterterm? (I would think the answer would be
no, because such a counterterm is odd under \phi \mapsto -\phi).

A:  Correct, you don't generate a \phi^3 counterterm because of this symmetry.