Let $C = \{(x,y) \in \mathbb{A}^2_k : y^2 = x^3+x^2\}$ as an affine variety over some algebraically closed field $k$. From the real picture we see that this curve is self-intersecting at the origin. I want to show this by blowing up $C$ and determining its fiber at the origin.
Formally, let $C'$ be the closure of $\{ (p,l) \in C \times \mathbb{P}^1 : 0 \neq p \in l \}$ in $\mathbb{A}^2 \times \mathbb{P}^1$, and $\pi : C' \to C$ the canonical projection. I want to show that $\pi^{-1}(\mathbf{0}) = \{(\mathbf{0},(1:1)),(\mathbf{0},(1:-1))\}$.
By covering $C'$ by two affine open subspaces, I was able to show that this fiber consists of at most these two points: Consider for example the affine open subspace $U_0 = \{ ((x,y),(s:t)) \in C' : s \neq 0\}$. It is easy to show that $U_0$ is contained in the common zero set of $Y^2-X^3-X^2$ and $(\frac T S)^2 -X -1$, so the points $(\mathbf{0},(s:t)) \in U_0$ satisfy $s = \pm t$.
But now I have difficulties to show the converse. Why do the points $(\mathbf{0},(1:1))$ and $(\mathbf{0},(1:-1))$ belong to $C'$?
Thank you in advance!
I think the easiest argument is just to show that $C$, together with those two points, is closed and irreducible.
Look at the closed subvariety $V$ of $\mathbb{A}^2\times\mathbb{P}^1$ defined by $y^2=x^3+x^2$ and $uy=vx$. The point is to show that it has two irreducible components: the $\mathbb{P}^1$ lying over the origin, and $C'$.
Take $u\neq 0$; set $u=1$ for simplicity. Then we have $v^2 x^2 = x^3 +x^2$, so we see that there are two components, cut out by $x=0$ and $v^2 = x+1$ respectively. Both points we want already lie in the second curve, so we're done—though in general we should check $v\neq 0$ as well if we think there might be more points there.