Spinor indices in $\not p = m_P$ (mass renormalization) [duplicate]

Question

I'm reading Schwartz QFT, Chapter 18 (mass renormalization) and I'm confused about the equations about on-shell subtraction/pole mass. He writes:

The renormalized propagator should have a single pole at $\not p = m_P$ with residue $i$. The location of the pole is a definition of mass.

But $\not p$ has two spinor indices (i.e. a 4x4 matrix) while $m_P$ is just a number, so how does this make sense? I thought maybe it means $\not p = m_P \mathbf{1}$, but in the Weyl representation for instance, the $\gamma^\mu$ have zeros on the main diagonal so $p_\mu \gamma^\mu$ can't be proportional to $\mathbf{1}$ (unless it's all zeros).

He also goes on to write the equation,

$$i = \lim_{\not{p}\to m_P} (\not p - m_P) \frac{i}{\not p - m_R + \Sigma_R(\not p)} = \lim_{\not{p} \to m_P} \frac{i}{1 + \frac{d}{d\not{p}} \Sigma_R(\not p)} \tag{18.41}$$

I don't understand what $\lim_{\not{p}\to m_P}$ and $\frac{d}{d\not{p}}$ mean if $\not p$ is a 4x4 matrix. Is it a limit/derivative in the 16-dimensional space of the matrix entries?

Edit: It did occur to me that maybe $\not p = m_P$ is a shorthand for $m_P = \sqrt{p^2} = \sqrt{\not p^2}$. However, for there to be a pole at $\not p = m_P$ they would have to coincide fully. $m_P = \sqrt{p^2}$ seems insufficient.

Answer 1

I'm not familiar with Schwartz's book, but I know that in Peskin and Schroesder, they first show that the fourier-space two-point correlation function is given by $$\int d^4x\langle\Omega|T\psi(x)\bar\psi(0)|\Omega\rangle e^{ip\cdot x}=\frac{i(\not p+m_0)}{p^2-m_0^2}+\frac{i(\not p+m_0)}{p^2-m_0^2}(-i\Sigma(p))\frac{i(\not p+m_0)}{p^2-m^2_0}+...$$ Then they argue $\Sigma(p)$ can be viewed as a function of $\not p$ using $p^2=(\not p)^2$, and also rewrote $\frac{\not p+m_0}{p^2-m_0^2}$ as $\frac{1}{\not p-m_0}$. I would assume the meaning is just $\sqrt{(\not p)^2}=m_0$, even though this is a bit of an abuse of notation.