Let $V=Z(y^5-x^3(x+1))$ be an irreducible variety in $\mathbb A^2$.
The partial derivatives of the equation $y^5-x^3(x+1)$ at the point $p=(0,0)$ are:
$ \frac{{\partial f}}{{\partial x}}=-3x^2-4x^3$, $ \frac{{\partial f}}{{\partial y}}=5y^4$, then $\frac{{\partial f}}{{\partial x}} | _p=0=\frac{{\partial f}}{{\partial y}} | _p$ i.e. $p$ is a singular point.
1) The equations of the blowing-up at $p$ in $\mathbb A^2\times\mathbb P^1$ are $xu=yt$ together with the morphism restriction $\pi:\Pi\to\mathbb A^2$ where $\Pi$ are given by the equation $xu=yt$ in $\mathbb A^2\times\mathbb P^1$.
2) In the affine chart $t=1$, we have
a) $y^5=x^3(x+1)$
b) $xu=y$.
Thus $Z(xu-y,y^5-x^3(x+1))=Z(y-xu,x)\cup Z(y-xu,x^2u^5-(x+1))$, and the exceptional curve is $Z(y-xu,x)=E$.
Then $E\cap V_1=\emptyset$ with $V_1=Z(y-xu,x^2u^5-(x+1))$
Thus look at the chart affine $u=1$
3) In the affine chart $u=1$, we have
a) $y^5=x^3(x+1)$
b) $x=yt$
In this case, $Z(xu-y,y^5-x^3(x+1))=Z(y,x-yt)\cup Z(y^2-t^3(yt+1),x-yt)$ and the exceptional curve is $E=Z(y,x-yt)$.
Then $E\cap V_2=\{(0,0,0)\}$ with $V_2=Z(y^2-t^3(yt+1),x-yt)$
Now calculate the matrix of partial derivatives of the equations of $V_2$; if $f=x-yt, g=y^2-t^3(yt+1)$
$\begin{bmatrix}{\frac{{\partial f}}{{\partial x}}=1}&{\frac{{\partial f}}{{\partial y}}=-t}&{\frac{{\partial f}}{{\partial t}}=-y}\\{\frac{{\partial f}}{{\partial x}}=0}&{\frac{{\partial f}}{{\partial y}}=2y-t^3}&{\frac{{\partial g}}{{\partial t}}=-4t^3y-3t^2} \end{bmatrix}|_{(0,0,0)}= \begin{bmatrix}{1}&{0}&{0}\\{0}&{0}&{0}\ \end{bmatrix}$
Then the variety $\widetilde{V}$ is singular in $\pi^{-1}(p)$.
Is correct?
I have to do blowing up of the variety $V_2=Z(y^2-t^3(yt+1),x-yt)$ but now on $\mathbb A^3\times\mathbb P^2$?
They could help me do this please.