Field redefinitions and new counterterms

My question was motivated by my attempt to answer this question. Suppose we are given an action and we make a change of variables such that the theory is non-renormalizable. Does the new theory then require an infinite number of counterterms?

As an explicit example lets consider the situation brought up in the linked question (though I change notation for my convenience). We start with the Lagrangian, $${\cal L}= \frac{1}{2}\partial^\mu\phi_0\partial_\mu\phi_0-\frac{1}{2}m^2\phi_0^2$$

Then we make the substitution, If I make $\phi_0=\phi+\frac{\lambda}{M} \phi^2$ such that $\lambda$ is dimensionless and $M$ is some mass scale. Then the Lagrangian is $${\cal L}= \frac{1}{2}\partial^\mu\phi\partial_\mu\phi-\frac{1}{2}m^2\phi^2+2\frac{\lambda}{M}\phi\partial^\mu\phi\partial_\mu\phi-\frac{\lambda}{M} m^2\phi^3 + 2\frac{\lambda^2}{M^2}\phi^2\partial^\mu\phi\partial_\mu\phi-\frac{1}{2}\frac{\lambda^2 }{M^2}m^2\phi^4$$

Now originally we could have found all the counterterms with calculating a few simple diagrams. On the one hand I'd think that since we still have a single coupling, $\lambda$, we should still have the same number of counterterms in the new theory. On the other hand I've learned that operators get renormalized, and not couplings, so since we have more operators we also need more counterterms. How many counterterms does this new theory actually need?

• Haven't you forget a Jacobian with this change of variable? I think that they usually do the job to make the theory after change of variable ok. – Adam Jun 17 '14 at 13:30