Tangent bundle on a smooth scheme My question is, how might I construct the tangent bundle on a smooth scheme?
It is clear how to define the tangent space at a point: the Zariski tangent space.
It is also clear what we should do in the setting of manifolds: we assemble the tangent spaces at all the stalks and put a topology on it. This, along with the evident projection map, give a vector bundle.
But the situation for schemes is perhaps more difficult.
The other question I have is about why this construction is not more well known or used in algebraic geometry.
 A: In algebraic geometry, Often people work with the cotangent sheaf, which is easily defined locally over $Spec\,A$ as the module $\frac{I}{I^2}$ where $I$ is the kernel of multiplication map $A\otimes A\to A$(or more directly as the module generated by $df,f\in A$ and with the usual relations) and then gluing these modules to define a sheaf over $X$. If you like to have a scheme instead of a sheaf it is very standard thing to get a bundle:you look at $Spec Sym\Omega$(if our scheme is smooth this is locally an affine space whose coordinates are given by the local parameteres on that open subset exactly like the usual cotangent bundle).
but often the sheaf viewpoint is more useful than the associated scheme: In differential geometry, it is harder to define tensors and connections,but these are very natural when you work with cotangent sheaf, it is also easier to talk about deformation theory in the language of cotangent sheaf, and even In differential geometry people use the language of sheaves to talk about cohomology.
In summary, cotangent Sheaf is a standard thing in algebraic geometry and defined in the usual way, It is a standard construction to get an actual bundle associated with a sheaf, But for most of the things you need In algebraic geometry, the cotangent sheaf is more useful. Of course, you can start with the dual of cotangent sheaf and get the tangent sheaf and then tangent bundle but often the cotangent sheaf appears more naturally.
A: Question: "My question is, how might I construct the tangent bundle on a smooth scheme? It is clear how to define the tangent space at a point: the Zariski tangent space."
Answer: If $k$ is a field and $k \rightarrow A$ is a $k$-algebra, let $I\subseteq A\otimes_k A$ be the kernel of the multiplication map $m:A\otimes_k A \rightarrow A$
defined by $m(a\otimes b):=ab$. It follows $\Omega:=\Omega^1_{A/k}:=I/I^2$ is the module of Kahler differentials of $A$. If $\mathfrak{m} \subseteq A$ is a maximal ideal with residue field $\kappa(\mathfrak{m})\cong k$ it follows
$$\Omega^1 \otimes_A \kappa(\mathfrak{m}) \cong \mathfrak{m}/\mathfrak{m}^2$$
is the cotangent space at the ideal $\mathfrak{m}$. Hence the cotangent module has at any $k$-rational point $\mathfrak{m}$ the cotangent space as fiber.
For this reason we call $\Omega$ the "cotangent module" of $A$. When $\Omega$ is a finite rank locally free $A$-module it follows its associated vector bundle $X:=\mathbb{V}((\Omega^1)^*):=Spec(Sym_A^*((\Omega^1)^*)$ is the cotangent bundle on $S:=Spec(A)$. There is a canonical map $\pi: X \rightarrow S$ which is locally trivial in the following sense: There is an open affine cover $U_i$ of $S$ with $\pi^{-1}(U_i) \cong \mathbb{A}^r \times U_i$ where $r:=rk(\Omega^1)$. These definitions make sense for any scheme $X/S$ using the diagonal $\Delta: X \rightarrow X\times_S X$.
The dual $(\Omega^1)^*\cong Der_k(A)$ is the module of derivations. Its associated vector bundle
$$\Theta_{S/k}:=\mathbb{V}(Der_k(A)^*)$$
is the tangent bundle of $S$. When $Der_k(A)$ is locally free there is for any $k$-rational $\mathfrak{m}$ point an isomorphism
$$Der_k(A)\otimes_A \kappa(\mathfrak{m}) \cong Hom_{\kappa(\mathfrak{m})}(\mathfrak{m}/\mathfrak{m}^2, \kappa(\mathfrak{m})),$$
hence the fiber of the module of derivations is the Zariski tangent space.
