I would like to show the following statement:
Let $A$ be a finitely generated $k$-algebra, where $k$ is any field. Let $X=\operatorname{Spec}(A[X_1,\ldots ,X_n])$ and $Y=\operatorname{Spec}(A)$. Let $X'\subseteq X$ and $Y' \subseteq Y$ be irreducible components respectively of $X$, $Y$ such that the image of $X'$ in $Y$ is included in $Y'$. Then $\dim(X')=\dim(Y')+n$.
For the sake of computing dimensions, we may as well assume that $A$ is reduced. We then consider $\mathfrak{q}$ a minimal prime ideal in $A[X_1,\ldots ,X_n]$, $\mathfrak{p}$ a prime ideal in $A$ contained in $\mathfrak{q}\cap A$, so that we get a map $A/\mathfrak{p} \rightarrow A[X_1,\ldots ,X_n]/\mathfrak{q}$ between finitely generated $k$-algebras that are also integral domains.
In order to prove that $\dim(A[X_1,\ldots ,X_n]/\mathfrak{q})=\dim(A/\mathfrak{p})+n$, I thought one should use the fact that $$\dim(A/\mathfrak{p})=\operatorname{Tr.deg}_k(\operatorname{Frac}(A/\mathfrak{p}))$$ and $$\dim(A[X_1,\ldots ,X_n]/\mathfrak{q})=\operatorname{Tr.deg}_k(\operatorname{Frac}(A[X_1,\ldots ,X_n]/\mathfrak{q})).$$
However, I fail to see what the field of fraction looks like in each case. I assume that we can not just replace the brackets with parenthesis, right?
Also, we can write $\operatorname{Frac}(A/\mathfrak{p})=A_{\mathfrak{p}}/\mathfrak{p}A_{\mathfrak{p}}=A_{\mathfrak{p}}$ because $\mathfrak{p}$ is minimal and $A$ is reduced, and similarily for $\operatorname{Frac}(A[X_1,\ldots ,X_n]/\mathfrak{q})$. But then again, I fail to understand how the transcendence degree can be easily computed. The fact that $\mathfrak{p}$ may be strictly included in $\mathfrak{q}\cap A$ upsets me.
Any help with this problem would be gladly appreciated. I thank you in advance.