I have a problem with the following statement:
Let $A \subset B$ be the normalization of a reduced, finite type $\mathbb{Q}$-algebra $A$ (integral closure in total ring of fractions). Then $\mathbb{R} \otimes_\mathbb{Q} A \to \mathbb{R} \otimes_\mathbb{Q} B$ is the normalization of $\mathbb{R} \otimes_\mathbb{Q} A$.
I think this should be true. I know already that $\mathbb{R} \otimes_\mathbb{Q} B$ is normal because of https://stacks.math.columbia.edu/tag/06DF and clearly $\mathbb{R} \otimes_\mathbb{Q} B$ is integral over $\mathbb{R} \otimes_\mathbb{Q} A$ But I wasn't able to embed the total ring of fractions $Q(\mathbb{R} \otimes_\mathbb{Q} A)$ in $Q(\mathbb{R} \otimes_\mathbb{Q} B)$.
The thing missing from my argument for this, is that $\mathbb{R} \otimes_\mathbb{Q} B$ has the same number of minimal primes as $\mathbb{R} \otimes_\mathbb{Q} A$. Can I argue in a different way?
Since $A \to B$ is a finite ring map inducing an isomorphism $Q(A) \to Q(A) \otimes_{A} B$, there is a nonzerodivisor $s \in A$ such that $A[s^{-1}] \to B[s^{-1}]$ is an isomorphism. Consider the base change of these $\mathbb{Q}$-algebras via $\mathbb{Q} \to \mathbb{R}$: $\require{AMScd}$ \begin{CD} A @>{i}>> B @. \phantom{aaa} @. A_{\mathbb{R}} @>{i_{\mathbb{R}}}>> B_{\mathbb{R}} \\ @V{f}VV @VV{g}V @. @V{f_{\mathbb{R}}}VV @VV{g_{\mathbb{R}}}V \\ A[s^{-1}] @>>{\simeq}> B[s^{-1}] @. @. A[s^{-1}]_{\mathbb{R}} @>>{\simeq}> B[s^{-1}]_{\mathbb{R}} \end{CD} The inclusion $i$ is injective by definition, and $f,g$ are injective since $s$ is a nonzerodivisor (here $s$ is also a nonzerodivisor on $B$ since it is a subring of $Q(A)$). Since $\mathbb{Q} \to \mathbb{R}$ is flat, the base changes $i_{\mathbb{R}},f_{\mathbb{R}},g_{\mathbb{R}}$ are injective as well; hence $s$ is a nonzerodivisor on $A_{\mathbb{R}}$ and on $B_{\mathbb{R}}$ (in particular $s$ is not contained in any of the minimal primes of $A_{\mathbb{R}}$ or $B_{\mathbb{R}}$); hence $i_{\mathbb{R}}$ induces an isomorphism $Q(A_{\mathbb{R}}) \simeq Q(B_{\mathbb{R}})$. By 06DF, the base changes $B_{\mathbb{R}},A[s^{-1}]_{\mathbb{R}},B[s^{-1}]_{\mathbb{R}}$ are normal. Since $i_{\mathbb{R}}$ is an integral ring map, $B_{\mathbb{R}}$ is the normalization of $A_{\mathbb{R}}$.