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Let n, m $\in \mathbb{N}$.

I'm trying to show that

$M(n,m) = \{[z_0 : z_1 : … : z_n] \in \mathbb{C}P^n | \sum^{n}_{i=0} z^m_i = 0\}$

is a submanifold and codim(M(n,m))=2.

My idea was to use the regular value theorem.

Thanks in advance.

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    What went wrong with your idea?2012-11-01
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    I guess f should be $f(z) =\sum^{n}_{i=0} z^m_i $, where $z \in \mathbb{C}P^n $. At first: How can I show that f is differentiable?2012-11-01
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    Well, the same $f$ is differentiabe from $\Bbb C^n$, as all its partial derivatives exist and are continuous.2012-11-01
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    The codimension is meant *over $\Bbb R$*, isn't it?2012-11-01
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    Yes, that's right, codimension over $\mathbb{R}$. Ok, so if f is differentiable, I have to show that 0 is a regular value. How do I see that? And if I take $z=0$ then, f(z)=0, so 0 has to be a regular point. But isn't the rank of the Jacobian matrix 0 for z=0?2012-11-01
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    How can I show that 0 is a regular value?2012-11-03

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