Let $R$ be an integral domain and $\mathfrak{p} \subset R[X]$ a prime ideal of $R[X]$. My question arises from the problem treated in following former thread of mine:Dimension Formula from Lius AG AC
(in schematic language we set $Y:=Spec(R), Spec(R[X]/\mathfrak{p}):=X \subset Z:=Spec(R[X])$; in the linked thread above the problem was to verify the conclusion that $codim(X_{\zeta},Z_{\zeta}) \le 1$ implies $codim(X,Z) \le 1$)
The concern of this thread is that trying to prove it I came to a result that seems quite strange to me so I have keen interest in finding the error in my reasoning.
What I have done:
After reducing our story to commutative algebra we know that $$codim(X,Z)= codim(V(\mathfrak{p}),Z)= height(\mathfrak{p})= dim(R[X]_{\mathfrak{p}})$$
Trying to draw an analogy to $codim(X_{\zeta},Z_{\zeta})$ we encounter the problem that $\mathfrak{p}$ might be not more prime in $R[X] \otimes_R K=K[X]$. Take into account that $X_{\zeta}= Spec(R[X]/\mathfrak{p} \otimes K)= Spec(K[X]/\mathfrak{p})$
Remark: $K:= O_{Y,\zeta}=Frac(R)$.
Geometrically this menas that $X_{\zeta}$ might be not irreducible in $Z_{\zeta}$.
So the first QUESTION is how is in this case $codim(X_{\zeta},Z_{\zeta})$ defined?
My naive idea would be that since the generic point $\zeta_X$ of $X$ is contained in $X_{\zeta}$ (since $f$ is dominant) there exist an irreducible component of $C_X$ of $X_{\zeta}$ (considered as closed subscheme of $Z_{\zeta}$) which contains $\zeta_X$.
Then $C_X= Spec(K[X]/\mathfrak{h})$ where $\mathfrak{h}$ is lying over $\mathfrak{p}$ in the canonic map $R[X] \to K[X]$.
Then
$$codim(X_{\zeta},Z_{\zeta})= codim(C_X, Z_{\zeta}) =codim(V(\mathfrak{h}),Z_{\zeta})=height(\mathfrak{h})$$
Does it make sense? Is the first equality correct?
If yes, then I would argue as follows: As $R \to K=Frac(R)$ is flat (since localization) then via base change $R \to R[X] \to R[X]_{\mathfrak{p}}$ we conclude that
$R[X]_{\mathfrak{p}} \to K[X]_{\mathfrak{p}}$ is flat (flatness stable under base change).
Since $\mathfrak{h}$ is lying over $\mathfrak{p}$ we can use Going Down Theorem for flat morphisms and and conclude $$dim(R[X]_{\mathfrak{p}} \le dim(K[X]_{\mathfrak{p}} = dim(K[X]_{\mathfrak{h}}$$
The last one holds since $codim(X_{\zeta},Z_{\zeta})= codim(C_X, Z_{\zeta}) =dim(K[X]_{\mathfrak{h}})$.
So $codim(X,Z) \le codim(X_{\zeta},Z_{\zeta}) $.
Is this proof correct?
The problem that I see here is that following the same argumentation we would conclude that for every $y \in Y$ the for the fiber $X_y=f^{-1}(y)$ follows
$$codim(X,Z) \le codim(X_{y},Z_{y}) $$
But this unequality seems to be wrong. Where is the error in my reasoning?