When considering a general mass term with Majorana masses, how do you exclude the existence of the left-handed neutrino mass? Considering a general mass term describing the neutrino masses:
$$
-\frac{1}{2}
\begin{pmatrix}
\bar\nu_{R} & \bar\nu_L 
\end{pmatrix}
\begin{pmatrix}
M_R & m \\
m & M_L
\end{pmatrix}
\begin{pmatrix}
\nu_{R} \\
\nu_{L} 
\end{pmatrix}
$$
In my coursebook it's written that $m$ is a Dirac mass which can be acquired through Yukawa-coupling, which I understand.
They also say that we need to set $M_L = 0$ because a mass term for $\nu_L$ is forbidden through gauge invariance, while there's no such restriction for $M_R$ because $\nu_R$ is sterile.
This intuitively made sense to me at first but when trying to quantify it I couldn't come to a decent result.
The reason you originally can't use Dirac masses (and why you need Yukawa-coupling) is, in short, because you want the left- and righthanded parts to transform differently under gauge transformations and a Dirac term wouldn't be invariant then because of the differing charges, unless the mass is set to zero. In this case the "problematic" term drops out.
I don't see why this puts a restriction on the Majorana masses. These terms look like mass terms describing the mass of a left-handed neutrino, I don't see how this would not be invariant under gauge transformations.
To put it into formulae (just an example to get the point across):
$$\nu_R(x) \rightarrow \nu_R^\prime(x)=e^{iq_Rf(x)}\nu_R(x)$$
$$\nu_L(x) \rightarrow \nu_L^\prime(x)=e^{iq_Lf(x)}\nu_L(x)$$
LH-Majorana term:
$$M_L\bar\nu_L\nu_L \rightarrow M_L\bar\nu_L^\prime\nu_L^\prime = M_Le^{i(q_L-q_L)f(x)}\bar\nu_L\nu_L = M_L \bar\nu_L\nu_L$$
[Edit]:
Supposedly this expression is correct, while the other one above is wrong:
$$
-\frac{1}{2}
\begin{pmatrix}
\bar\nu_{Rc} & \bar\nu_L 
\end{pmatrix}
\begin{pmatrix}
M_R & m \\
m & M_L
\end{pmatrix}
\begin{pmatrix}
\nu_{R} \\
\nu_{Lc} 
\end{pmatrix}
+ \text{h.c.}
$$
I still don't see how the argument applies though.
 A: When considering appending neutrino masses to the standard model you get terms of the form:
$$ \mathscr{L}_{lept,\phi} = \lambda^{ij} \overline{L}^i \phi R^j + \lambda^{ij}_\nu \overline{L}^i \phi R_\nu^j$$
Where $R^j_\nu$ some RH fermion singlet containing each generations RH lepton neturino. This is the only sensible consturction of a scalar given both $L, \phi$ are fermion and complex scalar doublets respectively. It is this construction that leads to there being no terms diagonal in the matrix. Sub in 
$$ \phi = \frac{1}{\sqrt{2}} (0, v + \eta(x))^T, \\ \overline{L} = (\nu_e(x)_L, e_L(x))^T $$
For a single generation to verify that after SSB your terms are of form $ \overline{\nu}_L \nu_R $
Your transformation listed
However this mechanism is discussed in a common extension to the Standard Model, the see-saw mechanism whereby you can view the $M_R$ as large and retain gauge invariance over a larger gauge group. 
I probably should add that the reason why the term listing isn't gauge invariant is due to in the Dirac Lagrangian, mass terms occur from the mixing of generations, separating a spinor into the 2 chiralities and subbing in should make that apparent.
Hope that has started to clear this up
A: In your LH-Majorana term, you missed the charge conjugation operation of LH neutrino. The charge conjugation will alter the gauge transformation behavior. Since LH neutrino is endowed with electroweak charge, the LH-Majorana mass term would therefore violate gauge invariance.
And for that matter, given the weak SU(2) gauge invariance, LH fermions should always show up in the Lagrangian as isospin doublets. Thus LH-Majorana mass term and Dirac mass term are forbidden, while Yukawa term and RH-Majorana term of neutrinos are allowed. 
