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Let $\phi: \mathbb{P}^1 \longrightarrow \mathbb{P}^9_{(Z_0,Z_1,Z_2,Z_3,Z_4,Z_5,Z_6,Z_7,Z_8,Z_9)}$ defined by $(t,s) \longmapsto (t^{18},t^{16}s,t^{14}s^2,t^{12}s^3, t^{10}s^4, t^8 s^5, t^6 s^6, t^4 s^7, t^2 s^8, s^9)$.

Then $I(X)=<${$Z_{i-1}Z_{i+1}-2Z_{i}$}$_{i=1,\dotso,8}>$ where $X=Im(\phi)$.

I want to know singular point of $X$ if it exists.

So I considered Jacobian and tried to calculate its rank. But it is too complicate to calculate.

Can you give some tips or another useful way?

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If we replace $t^2$ by $t$, the map $\phi$ becomes the $9$-uple embedding of $\mathbb{P}^1$. In particular, it is a closed immersion and the image is smooth. – Piotr Achinger Dec 29 '12 at 9:24
Use Macaulay2 for this type of calculations. It can tell you the codimension of the singular locus. – Mahdi Majidi-Zolbanin Dec 29 '12 at 15:21
Why do you want to use Macaulay2 ? It is smooth by the argument of Piotr (replacing $t^2$ by $t$ to get really a map from $P^1$ to $P^9$). If you want to do it in coordinates, look at the set where t=1 and the set where s=1. This seems to be a trivial exercise, not really for MO. – Jérémy Blanc Dec 29 '12 at 18:38

This is not a well-defined map from $\mathbb P^1$: if we multiply $s$ and $t$ by the same factor, the right-hand side is going to change.

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It becomes homogeneous if $s$ has degree 2. – rita Dec 29 '12 at 11:04

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