# Limit of a Family of Quartics

I've been trying an exercise from a book I'm self-studying, and I can't seem to get the blowup right. I'm not sure where I'm going wrong, so a nudge in the right direction would be helpful.

$\textbf{Summary of problem}$:

Given the following family of parametrized nodal cubics

$$\mu: \mathbb{P}^1 \to \mathbb{P}^2$$ $$[s:t] \to [3b^2st^2: t^3 - 3s^2t: s^3]$$

there is a corresponding family of dual curves of quartics $$[s: t] \to [s^2(s^2- t^2) : 2b^2s^3t: b^2t^2( t^2 + 3s^2)]$$ From here, it is clear that the member corresponding to $b = 0$ has three points of indeterminacy, all when $t=1$ and when $s=0$ or $s= \pm 1$. I need to show that the limit of these quartics should be a three-component curve, two simple lines and double line.

$\textbf{Attempt}$:

The idea is that when I blow up at these points of indeterminacy, I should have that the exceptional divisors get mapped to three lines. The double line comes out cleanly by blowing up at $s=0$, but I'm having issues with the others. I'll do the blowup at $[1:1]$ to illustrate my problem.

We can work with the chart $t= 1$. Here, we have an affine chart $\mathbb{A}^2$ with coordinates $(b,s)$, and we want to blow up at $(0,1)$. This results in the variety $\{ ( (b,s), [b_1: s_1]) | b_1(s-1) = s_1b\} \subset \mathbb{A}^2 \times \mathbb{P}^1$. Again, I restrict to an affine chart, say when $s_1 \neq 0$. In this chart, we have $b = b_1(s-1)$, so we can rewrite our dual map as $$(s, b_1) \to [s^2(s^2- 1) : 2b_1^2(s-1)^2s^3: b_1^2(s-1)^2(1 + 3s^2)]$$ This is the total transform of our map, so to get the proper transform, we divide out be $s-1$ $$(s, b_1) \to [s^2(s + 1) : 2b_1^2(s-1)s^3: b_1^2(s-1)(1 + 3s^2)]$$ Our exceptional divisor is given by $s = 1$, and thus plugging in $s= 1$ in the above map, we get that the image of our exceptional divisor is $[2: 0: 0]$.

From the above, I'm getting that the exceptional divisor gets mapped to a point, not a line. The $b^2$ terms are problematic here, as I'm getting an extra copy of $(s-1)$ in the last two coordinates. Doing the same calculations on the other open affine gives me that it maps to a point as well.

Any help would be appreciated, particularly to whether I'm resolving the map correctly. Thanks!

The quartic has equation $3x^2y^2b^4+4y^4-4x^3zb^6-6xy^2zb^2-y^2z^2$ so when $b=0$, you're left with $4y^4-y^2z^2$ or $y^2(4y^2-z^2)$.
ideal(x-s^4+s^2*t^2,y-2*b^2*s^3*t,z-3*b^2*s^2*t^2-b^2*t^4)