Dimension of the tangent space in algebraic geometry In my algebraic geometry course, we are studying tangent space and there is something I don't really understand. We define the tangent space as follows.

Let $a \in X$ be an variety. By chosing an affine neighborhood of $a$ we assume that $X \subset \mathbb A^n$ and that $a = 0$ is the origin. Then
$$T_aX = V(f_1:f \in I(X))$$
is the tangent space of $X$ at $a$, where $f_1$ denotes the linear term of $f\in I(X)$.

Thereafter, my teacher said that in the above definition of the tangent space, we had embedded $X$ in $\mathbb A^n$ but there are many other ways to embbed $X$ in some $\mathbb A^k$ and it is not at all obvious that the tangent space has the same dimension in these different embeddings. And then we showed the following statement that gives a definition of the tangent space that is independant of the embedding we chosed.

The tangent space is isomorphic to be the dual space of $I_a/I_a^2$, where $I_a$ is the unique maximal ideal in $\mathcal O_{X, a}$ (the stalk of the regular function at $a$).

Here are my questions: Why does we need this statement to ensure that the dimension of the tangent space do not depend on the embedding ? The tangent space is an affine variety, so it exists on its own and does not depends on any embedding right ? Moreover, as it is a vector space, the dimensons (the &quot;topological&quot; one and the &quot;vector space&quot; one) coincide and the dimension as a vector space is strictly related to the space itself, not on any embedding.. Why isn't the invariance of the dimension obvious ?
Well in your definition the tangent space is constructed from an embedding of $X$, so there is no guarantee that the tangent space that is constructed from another embedding is isomorphic to this one.
Question: &quot;Why does we need this statement to ensure that the dimension of the tangent space do not depend on the embedding?&quot;
Answer: If $X:=V(I) \subseteq \mathbb{A}^n_k$ where $I \subseteq k[x_1,..,x_n]$ is your variety and if $X :=V(J) \subseteq \mathbb{A}^m_k$ with $J\subseteq k[y_1,..,y_m]$ it follows you have two embeddings
$$i: X \rightarrow \mathbb{A}^n_k\text{  and  }j: X \rightarrow \mathbb{A}^m_k$$
You get two definitions of the tangent space $T_x(X)$ at $x$: One using $I$ and one using $J$. Denote these two spaces by $T^I_x(X)$ and $T^J_x(X)$.
Example: Let $f(x,y):=x^2+y^2-1$ and let $I:=(f) \subseteq k[x,y]$. Let $p:=(1/\sqrt{2},1/\sqrt{2})\in S:=V(I)$ and make the change of coordinates
$$x:=u+1/\sqrt{2}, v:=y+1/\sqrt{2}$$
to get
$$f(u,v)=\sqrt{2}(u+v)+u^2+v^2$$
with $f^*:=\sqrt{2}(u+v)$. Let $I^*:=(f^*)=(u+v)\subseteq k[u,v]$. Define $C_p(S):=Spec(k[u,v]/I^*)$. With this construction you have made a coordinate
change and in the new coordinates $u,v$ it follows the original point $p$ corresponds to the origin $(0,0)$ and $C_p(S)$ is the &quot;tangent line&quot; of $S$ at $p$ viewed as an algebraic variety/scheme. There is an isomorphism
$$C_p(S) \cong Spec(k[t])\cong \mathbb{A}^1_k$$
since $k[u,v]/(u+v) \cong k[t]$ is an isomorphism of rings: map $t$ to the class of $u$ or $v$.
You may check that there is an isomorphism $\mathfrak{m}/\mathfrak{m}^2 \cong k\{t\}$ when $\mathfrak{m}:=(x-\frac{1}{\sqrt{2}}, y-\frac{1}{\sqrt{2}})$. Hence $Sym_k^*((\mathfrak{m}/\mathfrak{m}^2)^*) \cong k[x]$ with $x:=t^*$ the dual basis. Hence
$$Spec(Sym_k^*(\mathfrak{m}/\mathfrak{m}^2)^*)) \cong \mathbb{A}^1_k \cong C_p(S).$$
Note: You have defined the tangent space $T^I_x(X)$ as an algebraic &quot;variety&quot;.
Given any vector space $V$ of dimension $n$ over $k$, we define the &quot;affine space of $V$&quot; (scheme theoretically) as
$$\mathbb{A}^n(V):=Spec(Sym_k^*(V^*)).$$
If $V:=k\{e_1,..,e_n\}$ is a basis for $V$ and $x_i$ the dual basis it follows
$Sym_k^*(V^*) \cong k[x_1,..,x_n]$ hence $Spec(Sym_k^*(V^*)) \cong \mathbb{A}^n_k$ is affine $n$-space over $k$.
When your lecturer says
&quot;The tangent space is isomorphic to be the dual space of Ia/I2a, where Ia is the unique maximal ideal in OX,a (the stalk of the regular function at a).&quot;
he means there is an isomorphism
$$T_a^I(X) \cong Spec(Sym_k^*(I_a/I_a^2)).$$
There cannot be an &quot;isomorphism&quot; $T_a^I(X) \cong (I_a/I_a^2)^*$ since $T_a(X)$ is an&quot;algebraic variety&quot; and $(I_a/I_a^2)^*$ is a $k$-vector space.
Since the local ring $\mathcal{O}_{X,x}$ is independent of embedding you get using this construction isomorphisms (let $I_a:=\mathfrak{m}_x$)
$$T^I_x(X) \cong Spec(Sym_k^*((\mathfrak{m}_x/\mathfrak{m}_x^2)^*)) \cong \mathbb{A}^d_k \cong T^J_x(X)$$
since $\mathfrak{m}_x/\mathfrak{m}_x^2$ has dimension $d$ over $k$.
It follows the two &quot;tangent spaces&quot; are isomorphic.
Question: &quot;Why isn't the invariance of the dimension obvious?&quot;
Answer: The definition of $T^I_x(X)$ involves the ideal $I$. It is not immediate that there is an isomorphism $T^I_x(X) \cong T^J_x(X)$.
Again: Mumford's book &quot;The red book of varieties and schemes&quot; devotes much space to explaining definitions in algebraic geometry. There is an intuitive explanation of the definition of the tangent space and tangent cone.