Why do the counterterms in renormalized $\phi^4$-theory with power two in fields give vertices and not propagators? I am reading Peskin and Schroeder, chapter ten, and my Lagrangian is
$$
\mathcal{L}=\frac{1}{2}(\partial_\mu\phi_r)^2-\frac{1}{2}m^2\phi_r^2-\frac{\lambda}{4!}z^2\phi^4+\frac{1}{2}\delta_Z(\partial_\mu\phi_r)^2-\frac{1}{2}\delta_m\phi^2_r-\frac{\delta_\lambda}{4!}z^2\phi^4.
$$
It was my understanding that terms of power 2 in the fields always give Feynman rules that are propagators. However, it appears that the counterterms with power 2 in the fields give a Feynman rule looking like $$i(p^2\delta_z-\delta_m),$$
instead of something with a denominator that would be more familiar. Like $\frac{i}{p^2-m^2+i\epsilon}$, from the first terms. Why is this the case? Is the idea that any term with power 2 in the fields gives a propagator wrong? 
 A: Consider $\phi^4$ theory:
$$
\mathcal L=\frac12 Z_1(\partial\phi)^2-\frac12 Z_m m^2\phi^2-\frac{1}{4!}\lambda_0\phi^4
$$
There are two approaches to perturbation theory:
First
The propagator is given by
$$
\Delta=\frac{1}{Z_1p^2-Z_m m^2}
$$
and there is one type of vertex, with value
$$
-i\lambda_0
$$
Second
The propagator is given by
$$
\Delta=\frac{1}{p^2-m^2}
$$
and there are two types of vertices, with value
$$
-i((Z_1-1)p^2-(Z_m-1)m^2),\qquad -i\lambda_0
$$
The two approaches are completely equivalent, and they give rise to the exact same expression for a given scattering process.
Note that the coefficients $Z_1,Z_m$ depend on the expansion parameter $\lambda$. This means that the first approach is more cumbersome because it is in general not clear which diagrams contribute to a given order in perturbation theory, inasmuch as both the vertices and the propagators contain powers of $\lambda$. On the other hand, the second approach leads to more diagrams (because there is one more vertex) but it is more convenient (because the propagators are independent of $\lambda$).
A: I want to add to AccidentalFourierTransform's answer:
Assuming the $\delta$'s are small, then we can expand the renormalized term in powers of $(\delta_2p^2-\delta_m)$:
$$\frac{i}{Z_2p^2-Z_mm^2}=\frac{i}{p^2-m^2
}\left(1+\frac{\delta_2p^2-\delta_m}{p^2-m^2}\right)^{-1}=\frac{i}{p^2-m^2
}\left(1-\frac{\delta_2p^2-\delta_m}{p^2-m^2}+\dots\right)=\frac{i}{p^2-m^2
}+\frac{i}{p^2-m^2
}\left(i\delta_2p^2-i\delta_m\right)\frac{i}{p^2-m^2
}+\dots$$
Which is the sum of all the diagrams consisting of the original term + the counter-term, so by identifying $\frac{i}{p^2-m^2}$ as the momentum term, we identify $i(\delta_2p^2-\delta_m)$ as the momentum counter-term.
A: This comes from treating
$$\frac{1}{2}(\partial_\mu\phi_r)^2-\frac{1}{2}m^2\phi_r^2$$
as the free Lagrangian, and treating
$$-\frac{\lambda}{4!}\phi_r^4+\frac{1}{2}\delta_Z(\partial_\mu\phi_r)^2-\frac{1}{2}\delta_m\phi^2_r-\frac{\delta_\lambda}{4!}\phi_r^4$$
as the perturbation. The propagators are then defined as the time-ordered two-point correlation functions of the "free" field theories, just as before. In this case, the propagator of the scalar field is
$$D_F(x-y)\equiv \langle0|T\{\phi(x)\phi(y)\}|0\rangle=\int\frac{d^4p}{(2\pi)^4}\frac{i}{p^2-m^2+i\epsilon}e^{-ip\cdot(x-y)}.$$
The only difference is that we have changed what we are viewing as the "free" theory. This is basically just changing the center point of the perturbation expansion.
The vertices come from the perturbation terms. For example, to the lowest order (one vertex), we expand
$$\exp\left[i\int\mathcal{L}\right]\approx\exp\left[i\int\mathcal{L}_0\right]\Big[1+i\int dx^4 \Big(-\frac{\lambda}{4!}\phi_r^4$$
$$+\frac{1}{2}\delta_Z(\partial_\mu\phi_r)^2-\frac{1}{2}\delta_m\phi^2_r-\frac{\delta_\lambda}{4!}\phi_r^4\Big) + ...\Big].$$
The terms $-(\lambda/4!)\phi_r^4$ and $-(\delta_\lambda/4!)\phi_r^4$ both yield vertices that connect four propagators just as in normal perturbation theory. The term $(\delta_Z/2)(\partial_u\phi_r)^2-(\delta_m/2)\phi_r^2$ yields a vertex that connects two propagators. The derivative terms makes it slightly more complicated to see what the vertex will look like, but by looking at the formula for $D_F(x-y)$ above, we can see that $\partial_\mu$ will just pull down an extra factor of $p_\mu$ from the connected propagators (both propagators will be constrained to have the same momentum by four-momentum conservation).
