I have recently been learning about sheaves of differentials, and in particular the conormal sheaf. That is, the sheaf of differentials corresponding to the diagonal morphism. I have become rather confused about the difference between normal, conormal, tangent, and cotangent in algebraic geometry.

Let $X$ be a scheme. For the sake of this question, everything will be local so say $X = \text{spec} A$ is affine. If $Y$ is a closed subscheme with closed immersion $i: Y \longrightarrow X$, let $\mathcal{I}$ be the corresponding sheaf of ideals. I recently learned the definition of the conormal sheaf, $\mathcal{I}/\mathcal{I}^{2}$, on $X$, or the pullback of this on $Y$. I wanted to relate this to the idea of a tangent space at a point. Suppose $p\in X$ is a closed point and give it the reduced induced subscheme structure. In other words, we assign the residue field $\kappa(p)$ to the point $p$ making $Z = \{ p \}$ a scheme. Then as usual the closed immersion $j: Z \longrightarrow X$ corresponds to a surjection of sheaves, $$ j^{\#}: \mathcal{O}_{X} \longrightarrow j_{*}\mathcal{O}_{Z}. $$ The sheaf $j_{*}\mathcal{O}_{Z}$ is then the skyscraper sheaf on $X$ assigning the residue field $\kappa(p)$ to the point $p \in X$. We obtain an exact sequence of stalks at $p$, $$ 0 \longrightarrow I_{p} \simeq \mathfrak{m}_{p} \longrightarrow \mathcal{O}_{X, p} \longrightarrow \kappa(p) \longrightarrow 0 $$ But I am also familiar with the notion of the Zariski cotangent space, which is given by $\mathfrak{m}_{p} / \mathfrak{m}_{p}^{2}$. But in this case, it seems to be telling me that the Zariski cotangent space is precisely the same as the conormal space, which seems wrong. Have I completely misunderstood what these things are, or is there some flaw in my reasoning above? Is this simply expressing the fact that for a closed point, the conormal space is isomorphic to the cotangent space (in some kind of perfect pairing)?

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    $\begingroup$ It actually seems pretty intuitive to me that the tangent space of the ambient scheme at a point should be the same as the normal bundle of the point in the scheme --- intuitively the normal bundle measures how you can wiggle the point (subscheme), which is the same as the tangent space at that point.. another somewhat relevant example is that the tangent space of the hilbert scheme at a point (corresponding to a closed subscheme of a projective scheme) is isomorphic to the global sections of the normal bundle of the closed subscheme (Geometry of Schemes VI.2.3) $\endgroup$ – loch Dec 21 '17 at 1:14
  • $\begingroup$ @user37864 fair enough, that does make sense I suppose. But is my reasoning in this question generally correct? $\endgroup$ – Luke Dec 21 '17 at 5:56
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    $\begingroup$ I think it's right $\endgroup$ – loch Dec 21 '17 at 15:35

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