I have a gaussian noise $\nu(t)$ with variance $\sigma^2$. After a FFT I get $X(\omega)$. If now I do the IFFT on the $X(\omega)$ can I say that the result is still a gaussian noise of variance $\sigma^2$? What is the effect of zero padding on FFT? How the zero padding affects the statistics of the noise after FFT and IFFT? Thanks.
Think about both questions separately. First of all, the (I)FFT is just an implementation of the (I)DFT, so I'm going to generalize all this to the DFT.
Does the zero-padded IDFT retain variance?
Parseval's theorem says power out = constant factor · power in, and the power of the zero-padded sequence is the energy of that sequence divided by it's length – and that length is larger than the original length, whereas the energy stayed the same.
Does the zero-padded IDFT retain gaussianness?
Long story short: yes. This is a result from the fact that the DFT is effectively a large sum of sufficiently identically distributed random variables. What you need is independence (which would imply whiteness), but as mentioned below, most people mean "WGN" when they say "GN".
When people say "Gaussian noise" they often mean "white Gaussian noise", but since that would have a constant PSD, and you explicitly made it so that your noise realization's Fourier transform is anything but constant, but comes in a boxcar shape, you obviously lose the whiteness.