As I understand it is still unclear if all neutrinos in the standard model are massive. Neutrino oscillations show that certainly not all neutrinos can be massless, but it might still be the case that one neutrino species is massless.
I found a very simple argument why all neutrinos should be massless, based only on energy-momentum conservation of special relativity and one particular interaction in the standard model. I'm wondering if it is a valid argument, if it is used anywhere, and if it is considered of any value.
Consider the standard model interaction vertex of two neutrinos going into a Z-boson, $\nu + \nu\to Z$. Here $\nu$ can be any neutrino. Suppose that this neutrino were massless. Then, denoting the two neutrino four-momenta by $p^\mu$ and $q^\mu$, and using $E^2 = p^2+m^2$ with $m=0$, we will have $p^0=\pm p^1$. Similarly, $q^0 = \pm q^1$. For simplicity we consider the case where both signs are a $+$, so that we have $p^0 = p^1$ and $q^0 = q^1$. Now, denoting the $Z$-boson four-momentum by $k^\mu$, energy-momentum conservation in the vertex requires that $k^0=p^0 + q^0$, as well as $k^1 = p^1+q^1 = p^0 + q^0$. We see immediately that this implies that $k^1 = k^0$, meaning that the $Z$-boson must be massless.
Of course we know that the $Z$-boson has a mass, so we have arrived at a contradiction. Hence our initial assumption, that the neutrino $\nu$ was massless, must be false. Therefore all neutrinos must be massive.
Any thoughts/comments are appreciated.
EDIT: As Anna points out in the comments, the process should not be $\nu + \nu\to Z$ but $\nu + \bar\nu\to Z$, i.e., incorporating an antineutrino. It does not change the argument, of course.
EDIT: I see I totally forgot to mention that I'm working in $1+1$ dimensions. I also now see, thanks to the answer by Cosmas Zachos, that choosing both $\pm$ signs as a $+$ is questionable, and may not be possible. Perhaps the $\nu + \bar\nu\to Z$ vertex only occurs when the two $\pm$ signs are opposite, in which case my argument above does not work. Or it could be that, as some other people say, the process only occurs in cases where for instance the $Z$ is virtual, so that it can be off-shell.