Quantum corrections to massless fermionic field in QED the corrections to electron propagator change the bare electron mass from $m_0$ to $m=m_0+δm=m_0+∑(\not{p}=m)$ (Peskin, formula 7.27). This is the consequence of the fact, that the quantum corrections change the propagator from $i/(\not{p}−m0)$ to  $i/(\not{p}−m_0−∑(\not{p})))$. This part is written very well in Peskin's book, formula 7.23. Then (after fomula 7.30) Peskin goes on to explain that if the fermion was massless, then the quantum corrections would never give a mass to it, in other words, δm is 0 when $m_0$ is 0.
The question is the following. When $m_0$ is 0, the quantum corrections change the massless fermion propagator from $i/\not{p}$ to $i/(\not{p}−∑(\not{p}))$, and to me this 'corrected' propagator looks like massive. Seems like the quantum corrections gave mass to the massless fermion.  To make the point clearer, the quantum corrections to photon propagator leave the denominator as $(1/q^2)$ and don't make it $1/(q^2-A)$, which is what happens to massless fermion propagator. 
My question is only from the point of view of diagrammatic approach (and not from the point of view of left and right spinors and the mass term connecting these two spinors).
Also, when we'll get to renormalization, seems like the fermionic field renormalization via $Z_2^{-1}=1-(d\Sigma/d\not{p})|_{\not{p}=m}$ will not be enough, we also would have to introduce mass renormalization because of that $\Sigma(\not{p})$ term in the 'corrected' propagator. Am I wrong?
 A: We are looking to bare  quantities $\psi_B, m_B$, such as the part of the Lagrangian which does not involve the electromagnetic field may be written : 
$\mathcal L_B = i \bar \psi_B \not \partial\psi_B- m_B \bar \psi_B  \psi_B \tag{1}$
Now, starting from physical quantities $\psi,m$, the inverse electron propagator may be written, in a dimensional regularization procedure (neglecting the finite terms) :
$\not{p}(1+\frac{\lambda ~\alpha}{\epsilon}) - m (1 + \frac{\mu~\alpha}{\epsilon}) \tag{2}$
where $\lambda, \mu$ are some constant, $\alpha = e^2$, and $\epsilon = d-4$. 
So, we have a Lagrangian : 
$\mathcal L_B = i \bar \psi \not \partial (1+\frac{\lambda ~\alpha}{\epsilon})\psi- m(1 + \frac{\mu~\alpha}{\epsilon}) \bar \psi  \psi \tag{3}$
The $\frac{\lambda ~\alpha}{\epsilon}$ factor  may "disappear" in the redefinition of the bare fermionic field : 
$\psi_B = \sqrt{Z_2}\psi= \sqrt{1- \frac{\lambda ~\alpha}{\epsilon}}~\psi = (1- \frac{\lambda ~\alpha}{2\epsilon} + O(\alpha^2))~\psi \tag{4}$ 
If $m=0$, this is enough. 
Now, if $m \neq 0$,  you need also a a redefinition of the bare mass : $m_B=Z_2^{-1}m(1+\frac{\mu~\alpha}{\epsilon}) = m (1+\frac{\lambda ~\alpha}{\epsilon})(1+\frac{\mu~\alpha}{\epsilon}) = m(1+\frac{(\lambda + \mu) ~\alpha}{\epsilon} + O(\alpha^2)) \\
= m +\delta m\tag{5}$.
