Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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?

share|improve this question
3  
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

1 Answer 1

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.

share|improve this answer
3  
It becomes homogeneous if $s$ has degree 2. –  rita Dec 29 '12 at 11:04

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.