Three kinds of spectra In commutative algebra while proving the Nullstellensatz one introduces for a while the Rabinowitsch sprectrum as:
$$\operatorname{Spec}_{\rm Rab}(R)=\{R\cap \mathfrak m: \mathfrak m \text{ is a maximal ideal of } R[X]\}.$$
There is a quite easy to see the inclusion list: $\operatorname{Spec Max}\subseteq\operatorname{Spec}_{\rm Rab}\subseteq \operatorname{Spec}$. I'm asked to prove that the inclusions are strict, by considering $R=K[[Y]]$, $K$ a field, and $S=R[Z]$ by using the ideals $(Y)$ and $(Z)$ in $S$. 
Im kinda stuck, any kind of help would be helpful, and not very familiar with the ring of formar series. Thank you
This is an exercise in the book A Course in Commutative Algebra from George Kemper, page 20 number 1.4.
 A: It is clear that $(y,z)_S$ is a maximal ideal of $S$. 
Now I claim $(z)_S$ is a not maximal prime ideal in $S$, and belongs to the Rabinowitsch Spectrum. Let $\mathbf{m}$ be the ideal $(1-X*y,z)_{S[X]}$ of $S[X]$. It is maximal since the quotient of $S[X]$ by $\mathbf{m}$ is the field $K( (y) )$ of the Laurent series in $y$, and clearly $\mathbf{m}\cap S=(z)_S$.
Moreover $(y)_s$ is clearly prime but not maximal in $S$ and does not belong to the Rabinowitsch Spectrum, since if $\mathbf{m}$ is a maximal ideal of $S[X]$ such that $\mathbf{m}\cap S=(y)_s$, it should contain $(y)_SS[X]$. Then we get a surjective morphism from $S[X]/(y)_SS[X]=K[X,z]$ to the field $S[X]/\mathbf{m}$. If we suppose $K$ algebraically closed, then we have $S[X]/\mathbf{m}=K$, since the kernel of this morphism is a maximal ideal of the polynomial ring $K[X,z]$. This is a contradiction with the fact that $S/\mathbf{m}\cap S = S/(y)_s = K[z]$ injects in $K=S[X]/\mathbf{m}$.
A: For $R=K[[Y]]$, to show that $\text{Spec}_{\text{rab}}(R) \subsetneq \text{Spec}_{\text{max}}(R)$, we consider the ideal $(0)_R$.
It can be shown that $(Y)_R$ is the only maximal ideal of $R$, so $(0)_R$ is not a maximal ideal of $R$.
Now, consider the surjective evaluation homomorphism $\varphi: R[Z]\to R[Y^{-1}], f\mapsto f(Y^{-1})$, where $Z$ is an other intermediate.
It can be shown that $R[Y^{-1}]$ is a field: $R[Y^{-1}]=\text{Quot}(R)=:R((Y))$, the field of formal Laurent series with $Y$ an intermediate.
The kernel $\text{ker}(\varphi)$ of the above homomorphism is a maximal ideal in $R[Z]$, because $R[Y^{-1}]$ is a field (and by using the first isomorphism theorem).
And we have $\text{ker}(\varphi) \cap R = \{0\}$, because for a polynomial $f\in R[Z]$ to satisfy $f(Y^{-1})=0$, its "constant term", which can be seen as an element in $R$, must be zero.
Hence, $(0)_R \in \text{Spec}_{\text{rab}}(R) \setminus \text{Spec}_{\text{max}}(R)$.
