Let $X$ be the affine surface $\{x^2+yt=0\}\subseteq \mathbb C^3$, then $X$ has an $A_1$ singularity at $0$. Consider $X$ as a family of curves via the projection to the last coordiate $$\pi:X\to \mathbb C, (x,y,t)\mapsto t.$$
Let $\sigma:\tilde{X}\to X$ be the blowup of $X$ at $0$ with exceptional divisor $E$, then it is claimed in Harris-Morrison's Moduli of Curves, page 133, that the exceptional divisor $E$ has multiplicity one.
Does the multiplicity means the order of vanishing of $(\sigma\circ\pi)^*(t)$ on the component $E$? How to obtain $E$ has multiplicity one?
Here is another example in Harris-Morrison, where the family is $y^2-x^3-t=0$ and it is shown that exceptional divisor of blowup acquires with multiplicity two. (Note the major difference between two examples is the smoothness of the original total space.)
I was imitating its computation to work in my case: Blowup $0\in \mathbb C^3$ and get $W\subseteq \mathbb C^3\times \mathbb P^2_{[\alpha,\beta,\gamma]}$ with equations
$$x\beta=y\alpha,x\gamma=t\alpha,y\gamma=t\beta.$$
Then $\tilde{X}\subset W$ is the strict transform of this blowup and $E$ is the conic $\{\alpha^2+\beta\gamma=0\}\subseteq \mathbb P^2_{[\alpha,\beta,\gamma]}$. By setting $\gamma=1$, we have affine equation of $\tilde{X}$ $$x=t\alpha, \ y=t\beta, \ t^2(\alpha^2+\beta)=0.$$
To me, the appearance of the $t^2$ means the vanishing order of $t$ on exceptional $E$ has order two.
This contradicts the conclusion in the book. Am I missing something?
Edited: The picture attached below is a piece from page 133 of Moduli of curves, and the last sentence is where I am trying to understand:
Best Answer
Sasha's already covered this in the comments, but I want to make what was said there a little more explicit. The problem here is that the blowup that has been computed (I made this mistake too!) was the blowup of a point inside $\Bbb A^3$, not a point inside $X$.
To be precise, if one has $Z\subset Y\subset X$ a chain of closed subvarieties, we computed $Bl_Z X$ and looked at the total transform of $Y$, that is, the preimage of $Y$ under the map $\pi:Bl_Z X\to X$. Instead, what one should do to find the blowup $Bl_Z Y$ is to consider the strict transform of $Y$ in $Bl_Z X$. See for instance this MSE question, remark 9.11 in Gathmann's notes, Hartshorne examples I.4.9.1 and II.7.15.1, etc.
In our case, the blowup of $X$ at $0$ is the strict transform of $X$ under the blowup of $\Bbb A^3$ at zero. The strict transform is covered by three charts $\alpha=1$, $\beta=1$, and $\gamma=1$ which have coordinate algebras $k[x,\beta,\gamma]/(1+\beta\gamma)$, $k[\alpha,y,\gamma]/(\alpha^2+\gamma)$, and $k[\alpha,\beta,t]/(\alpha^2+\beta)$ respectively. The exceptional divisor in each of these patches is just given by taking the closed subschemes associated to $x=0$, $y=0$, and $t=0$, respectively, and the subschemes we get this way are of multiplicity one.