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I know this is probably a simple problem but I have got myself stuck trying to prove this fact myself. I'd be very grateful if anyone could clear this up for me.

Let $V$ be an affine variety in $\mathbb{A}^n$ (say, over an algebraically closed field $K$) and suppose $P = (0,\dots, 0)\in V$ is the origin of affine space. I am trying to prove there's an isomorphism between the "abstract" Zariski tangent space $T_P (V) = (\mathfrak{m}_P/\mathfrak{m}_P^2)^*$ and the "naive" vector space

$$W = \left\{(\alpha_1, \dots, \alpha_n): \sum_{i=1}^n \alpha_i \frac{\partial f}{\partial x_i} (P)=0, \quad \forall f\in I(V)\right\}$$

I've shown there's an injection $\phi: W\to T_P (V)$ as follows: for $Q = (\alpha_1, \dots, \alpha_n)\in W$ we get a linear functional $\phi_Q : \mathfrak{m}_P/\mathfrak{m}_P^2 \to K$ via

$$\phi(Q) = \phi_Q = \sum_{i=1}^n \alpha_i \frac{\partial}{\partial x_i}\lvert_P$$

which corresponds to taking a "weighted total derivative" of an equivalence class of a function $r\in \mathfrak{m}_P$ modulo $\mathfrak{m}_P^2$ i.e.

$$\phi_Q (\bar{r}) = \sum_{i=1}^n \alpha_i \frac{\partial r}{\partial x_i} (P)$$

where $\bar{r}$ denotes the image of $r$ in the quotient. This is indeed a linear functional on $\mathfrak{m}_P/\mathfrak{m}_P^2$. Moreover the mapping $\phi$ is an injective $K$-linear map $W\to T_P (V)$. What I'm struggling to do is show surjectivity of this map. Every element $g\in T_P (V)$ can be written as

$$ g = \sum_{i=1}^n \alpha_i g_i$$

where $g_i (\bar{x_j}) = \delta_{i j}$ is one of the dual basis vectors. What isn't clear to me is that the coefficients $\alpha_i$ come from some $Q = (\alpha_1, \dots, \alpha_n)\in W$ i.e. that

$$\sum_{i=1}^n g(\bar{x_i}) \frac{\partial f}{\partial x_i} (P) = 0, \quad \forall f\in I(V)$$

Can anyone explain why this map is surjective?

Edit: I mistakenly forgot to include the condition that the linear combination sums to zero in the definition of the "naive" tangent space in the original post, which caused some confusion. This has been amended above.

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  • $\begingroup$ A very well formulated question: congratulations ! (And +1, of course) $\endgroup$ Mar 16, 2016 at 19:25
  • $\begingroup$ @GeorgesElencwajg thank you for your kind words! A few days ago I read this wonderful answer of yours: math.stackexchange.com/a/1041689/184547 and decided I ought to develop some more "concrete" skills in algebraic geometry in addition to the abstract material. This question could be seen as an attempt to start on that. $\endgroup$
    – Alex Saad
    Mar 16, 2016 at 19:36
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    $\begingroup$ By the way, I find this question not trivial at all and the treatment of the relation between the clasical and the Zariski tangent spaces in the literature is not always satisfactory. So this question might prove useful as a reference in the future. $\endgroup$ Mar 16, 2016 at 21:17

1 Answer 1

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The inverse morphism $\psi=\phi^{-1}:(\mathfrak{m}_P/\mathfrak{m}_P^2 )^*\to W$ is defined as follows.

Given a $K$-linear map $t: \mathfrak{m}_P/\mathfrak{m}_P^2 \to K$ we define $\psi(t)= (t(x_1),...,t(x_n))\in W$
But why is that vector $(t(x_1),...,t(x_n))$ in $W$?
We have to check that for $f\in I(V)$ we have $\sum_{i=1}^n t(x_i) \frac{\partial f}{\partial X_i} (P)=0$
This is true: write $$f= \sum_{i=1}^n \frac{\partial f}{\partial X_i} (P).X_i+q$$ where $q$ is a polynomial with no constant nor linear term.
Since $f$ becomes zero already in $\mathcal O(V)$, we have a fortiori $f=0\in \mathfrak m_P$ and we get by applying the $K$-linear map $t$ to $f$: $$0=t(f)=\sum_{i=1}^n \frac{\partial f}{\partial X_i} (P).t(x_i) +t(q) $$ However $q\in \mathfrak m^2_P$ forces $t(q)=0$ so that we obtain our crucial relation $\sum_{i=1}^n t(x_i) \frac{\partial f}{\partial X_i} (P)=0$.

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  • $\begingroup$ From the set-theoretic/notational point of view, $(\mathfrak{m}_P/\mathfrak{m}_P^2)^*$ is a nightmare: you have to take polynomials $f$, then their classes in $\mathcal O(V)$, then the classes of those in $\mathfrak m_P \subset \mathcal O_{V,P}$, then the classes of those in $\mathfrak{m}_P/\mathfrak{m}_P^2$ ! And finally you have to take the dual vector space... Needless to say, I didn't try to keep up notationally and committed several abus de notation in my answer . $\endgroup$ Mar 16, 2016 at 21:46
  • $\begingroup$ I'm very happy to see such a lucid proof of this result - as you said in your comment above, I've found proofs in the literature to be either somewhat unsatisfactory or nonexistent. I tried to prove it myself but I wasn't sure if I was going down the right path. It's great to see the proof finished in such a clear way. Thank you for a fantastic answer! $\endgroup$
    – Alex Saad
    Mar 16, 2016 at 22:25
  • $\begingroup$ You are welcome, dear Alex: kind comments like yours are a great reward and make me feel I'm not wasting my time answering questions on this site! $\endgroup$ Mar 16, 2016 at 23:15
  • $\begingroup$ @AlexSaad Besides Georges (as per usual) excellent answer, it might behoove you to understand how both relate to derivations. Or, equivalently, how to think about $T_x X$ as being $X_x(k[\varepsilon])$ where these homomorphisms are taken in the category of pointed-schemes. This is nice because it allows one to easily see the geometric content in something like 'formal smoothness', and other basic parts of deformation theory. By the way, a great reference for all of this (both your original question and my comment) is Gortz and Wedhorn's excellent text. In particular, Chapter 6. $\endgroup$ Mar 17, 2016 at 0:47
  • $\begingroup$ @AlexYoucis thanks for this reference! I've done some thinking about tangent spaces for schemes before and it is very helpful to have all these different ways of thinking about the tangent space - in its "naive" form, "functionals" form and "functorial points" manifestations. Also, Gortz and Wedhorn looks like good supplementary reading to my main text (Qing Liu's book). I've just got a copy out from the library and I like how it seems to motivate all the abstract constructions really well. $\endgroup$
    – Alex Saad
    Mar 17, 2016 at 12:47

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