Cross correlation and cross power spectrum for signal + noise Quick question about my understanding of cross correlations and cross spectral densities.

Let $C_{ab}$ be the cross correlation between the signals $a$ and $b$, i.e., $C_{ab}(\tau) = \langle a(t)b(t+\tau)\rangle$. Let $S_{ab}$ be the cross spectral density of the signals $a$ and $b$, i.e., $S_{ab}=\mathcal{F}(a)^* \mathcal{F}(b)$ and by the convolution theorem/cross correlation theorem $S_{ab} = \mathcal{F}(C_{ab}(\tau))$.

Say $x(t)$ is some signal, and $\nu(t)$ is some uncorrelated noise signal. Let $s(t)=x(t)+\nu(t)$
Thus, $C_{xs} = C_{xx}+C_{x\nu}$ from the linearity of the expectation value being calculated in the cross correlation. By initial assumption that $x$ and $\nu$ are uncorrelated, we get $C_{xs}=C_{xx}$. (This calculation so far also appears here on page 5, for example so I think it is correct).

Now, if we take the fourier transform of both sides of this, we get $S_{xs}=S_{xx}$. Is this correct so far? But does this not mean that $\mathcal{F}(x)^* \mathcal{F}(s) = \mathcal{F}(x)^*\mathcal{F}(x)$? and hence $\mathcal{F}(s)=\mathcal{F}(x)$, and hence even $s(t)=x(t)$? Clearly there is something wrong here. What mistake am I making?

Thanks!

EDIT: On a similar note, if $C_{x\nu}=0$, does that not mean that $\mathcal{F}(x)^*\mathcal{F}(\nu)=$ and hence (by taking the absolute value of both sides and squaring) that $S_{xx} S_{\nu\nu}=0$? This is patently false. At each frequency this would require either the power in my signal to be zero, or the power in the noise to be zero. Even if my noise is white noise then this is clearly false. I'm doing something fundamentally wrong here. What? Any help?
$S_{xy}=S_{xx}$ is true but this does not mean $F_x*F_y = F_x*F_x$ when $y$ is random. 

The actual definition is $$S_{xy}(f)=\lim_{T\rightarrow \infty}\textbf{E}\{F^*_xF_y\}$$ for random signals. Since $x$ and $v$ are uncorrelated, this would be equal to $$\lim_{T\rightarrow \infty}\textbf{E}\{F^*_xF_x\} = S_{xx}.$$ But your next assumption is not correct.

$F_x$, $F_y$ are only based on single realization of $x$ and $y$ respectively. For random signal, $S_{xy} \ne F_x^*F_y$