I'm reading the Saclay Lectures on EFT, and I don't understand how it uses the tree-level matching to compute the 1-loop-level matching. To simplify, in this post I'll put its $C_6,\lambda_1=0$ since they're relevant to the full matching but not to my question.
I have these Lagrangians:
$$\mathcal{L}_\text{UV}=\frac{1}{2}[(\partial\phi)^2-m_L^2\phi^2+(\partial H)^2-M^2 H^2]-\frac{\lambda_0}{4!}\phi^4-\frac{\lambda_2}{6!}\phi^2 H^2\tag{2.1}$$ $$\mathcal{L}_\text{EFT}=\frac{1}{2}[(\partial\phi)^2-m^2\phi^2]-\frac{C_4}{4!}\phi^4.\tag{2.2}$$ The tree-level matching is trivial, and states $$m^2=m_L^2, \quad C_4=\lambda_0.\tag{2.10+13}$$ Now onto 1-loop-level matching, focusing on the 2-point function and evaluating it at its pole both in the EFT theory and the UV theory, obtaining two formulas: $$\text{EFT}\rightarrow m_\text{phys}^2=m^2-\frac{C_4 m^2}{32\pi^2}\left[\log\left(\frac{\mu^2}{m^2}\right)+1\right]\tag{2.19}$$ $$\text{UV}\rightarrow m_\text{phys}^2=m_L^2-\frac{\lambda_0 m_L^2}{32\pi^2}\left[\log\left(\frac{\mu^2}{m_L^2}\right)+1\right]-\frac{\lambda_2 M^2}{32\pi^2}\left[\log\left(\frac{\mu^2}{M^2}\right)+1\right].\tag{2.28}$$ The next step is obviously to equate (match) the two quantities at the same scale $\mu$. The book finds $$m^2(\mu)=m_L^2(\mu)-\frac{\lambda_2 M^2}{32\pi^2}\left[\log\left(\frac{\mu^2}{M^2}\right)+1\right]\tag{2.29}$$ where, I presume, the part with $\log(\mu^2/m^2)$ is absent because the tree-level matching results were used. But if they were used to cancel those terms, why can't they also be used to cancel $m$ and $m_L$ in the above equation?
