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I'm reading some lecture notes (unfortunatey in italian, https://me.unitn.it/system/files/Bernardi%20Alessandra/tesi_teroni_1.pdf , page 5), where there's this statement (without proof): Suppose we have $X\subset\mathbb{P}(\mathbb{C}^n)$ variety which is locally isomorphic to $\mathbb{C}^m$ (obviously $m\leq n$), and a point $[p]\in X$ with local coordinates $[x_0,\ldots,x_m]$. Then the basis of the tangent space $T_{[p]}(X)$ is made of $\{\frac{\partial}{\partial x_0}, \ldots, \frac{\partial}{\partial x_m}\}$, so $$\dim(T_{[p]}(X))=m+1=\dim(X)+1.$$ I can't understand why this should be correct: we have $\dim(X)=\dim(\mathbb{C}^m)=m$, but then why I can pass to a basis with $m+1$ coordinates? I mean, I know that if $X$ is a variety then $\dim(T_p(X))=\dim(X)$ if $p$ is smooth. But I don't know how this changes if I consider a variety locally isomorphic to $\mathbb{C}^m$ embedded in $\mathbb{P}^n$.

Although this statement is unclear to me, I don't thinnk it's an error since other authors use this passage

https://www.mimuw.edu.pl/~jabu/conf/2013/notes_Lukecin_AB.pdf (page 11)

https://arxiv.org/pdf/math/0701409.pdf (beginning of page 5)

Any hint would be much appreciate, I'm struggling with this problem and it's quite important for me to understand this passage. Thanks in advance. (I know this is a duplicate of the samee question on math.stackexchange, but as I said it's quite important for me to find why this works, and I can't do this by myself)

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    $\begingroup$ Looking at all of those examples, it seems that the distinction is just whether the authors are considering the vector space dimension of a finite dimensional vector space, e.g., the Zariski tangent space, or the Krull dimension of the projective space associated to that vector space. Also, the first author seems to be computing the Zariski tangent space of the affine cone of $X$ in $\mathbb{C}^n$, rather than the Zariski tangent space of $X$ in $\mathbb{C}\mathbb{P}^{n-1}$. $\endgroup$
    – Jason Starr
    Commented Jul 14, 2018 at 10:22
  • $\begingroup$ sorry but I can't understand: the dimension of a projective variety $X$ is one less the Krull dimension of the coordinate ring, but also $\dim(T_p(X))=\dim(X))$, so I should have $\dim(T_p(X))=\dim_K(A)-1$. I can't see why it should be $\dim(X)+1=\dim(T_p(X))$. How is related the krull dimension of a variety and the krull dimension of the projective tangent space? $\endgroup$
    – christmas_light
    Commented Jul 14, 2018 at 11:06
  • $\begingroup$ I do not understand what you wrote. For a field $k$, for a $k$-vector space $V$ that has a basis of size $n$, the dimension of the associated $k$-scheme $\mathbb{P}(V)$, i.e., the Krull dimension of the ring of regular functions on any dense open affine, equals $n-1$. Thus, the vector space $\mathbb{C}^n$ has vector space dimension $n$, yet the associated projective space $\mathbb{CP}^{n-1$} has (Krull) dimension $n-1$. $\endgroup$
    – Jason Starr
    Commented Jul 14, 2018 at 11:11
  • $\begingroup$ I agree with this, but my problem is understand why the equality $\dim(X)+1=\dim(T_p(X))$ holds. And I can't see how the krull dimension helps me with this. $\endgroup$
    – christmas_light
    Commented Jul 14, 2018 at 11:15
  • $\begingroup$ Here is another way to say the same thing. A subvariety $X$ of $\mathbb{P}(\mathbb{C}^n)$ admits at each point $p$ a projective tangent space $\mathbb{P}T_p(X)\subset\mathbb{P}(\mathbb{C}^n)$; you can define it for instance as the intersection of the hyperplanes tangent to $X$ at $p$. If $p$ is smooth these projective subspace has dimension $m=\dim(X)$, so it comes from a $(m+1)-$dimensional vector subspace in $\mathbb{C}^n$, which the authors you quote denote by $T_p(X)$. $\endgroup$
    – abx
    Commented Jul 14, 2018 at 12:28

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