Blow up commute with base change

Question

Let $X$ be an irreducible Noetherian scheme and $X_{red}$ be the reduced subscheme associated to it. This induces a natural morphism from $X_{red}$ to $X$. Fix a point $p$ in $X$ (by a point we mean an irreducible zero dimensional closed subscheme of $X$, so can be non-reduced). Denote by $Bl_pX$ the blow up of $X$ along $p$ in the sense stated in Chapter II.$7$ of "Algebraic Geometry" by Hartshorne. Is it true that the pull-back $Bl_pX \times_X X_{red}$ is the blow up of $X_{red}$ at $p_{red}$?

Answer 1

The answer I think is yes.

Denote by $\pi_1$ the natural projection map from $Bl_pX \times_X X_{red}$ to $X_{red}$. Using Proposition IV-$21$ in "Geometry of Schemes" by Eisenbud and Harris, we conclude that the blow up of $X_{red}$ at $p_{red}$ is the closure of $\pi_1^{-1}(X_{red}\backslash p_{red})$ in $Bl_pX \times_X X_{red}$. Note further that $X\backslash p$ is dense in $Bl_p X$. Since the image of $X_{red}\backslash p_{red}$ is dense in $X$, it should follow that the closure of $\pi^{-1}(X_{red}\backslash p_{red})$ is $Bl_pX \times_X X_{red}$.

EDIT As pointed out below, the argument is true if $p_{red}=p \times_{X} X_{red}$.