Abstract algebraic definition of dual tangent spaces I know that if $(M,\mathcal{A})$ is a smooth manifold, the dual tangent space at $p\in M$ can be defined as $$ T^*_pM=I_p/I_p^2, $$ where $I_p$ is the ideal of the ring $C^\infty(M)$ consisting of smooth functions that vanish at $p$ and $I_p^2$ is the second power of this ideal.
I understand that this definition is useful, because, unlike other definitions, this one can be generalized to situations when you don't have the smooth structure $\mathcal{A}$, however this definition is so unintuitive I have a hard time grasping it.


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*According to wikipedia, the product of ideals $A$ and $B$ is defined as $$ AB=\{a_1b_1+...a_nb_n|\ a_i\in A,b_i\in B,n\in\mathbb{N}\}, $$ I assume the point of this definition is that by demanding that the functions vanish, we make sure that the functions' Taylor expansion has no zeroth order terms, and since the elements of the product ideal are second order expressions, I assume the quotient is needed to get rid of second order terms.


Seems logical, since by the usual definition we can infer that the contangent space is generated by differentials of functions, which are, of course, the first order part. Am I correct in this?


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*If I am correct on the first point, what about higher than second order terms? Shouldn't we need to quotient those out too?

*Thinking about it as I write this post, elements of $I_p^2$ are second order algebraic expressions from $C^\infty(M)$, but they are not polynomials, a Taylor expansion, however is polynomial. I think I don't understand it now.

*This line of thought seems to depend on the functions having Taylor expansions. However I know this definition is generalizable to algebraic varieties as well as locally ringed spaces in general. Without having a structure that allows Taylor expansions, how do we know that this grasps the concept of "functions having the same first order behaviour at $p$" for those cases as well?

*How can we see that this quotient space is finite $n$ dimensional? If the proof is not particularily pleasant, I don't expect anyone to actually post it here, but a reference to a manifold theory textbook that utilizes this definition heavily enough to also contain related proofs would be nice.
I would appreciate any response, even if it doesn't address these points directly, if it can help me see intuitively why this particular quotient space has the same meaning as the usual definition of dual tangent space.
EDIT: Since there have been misunderstandings, I wish to clarify: In my last bullet point I am not asking how to prove that the tangent/cotangent space is $n$ dimensional for an $n$ dimensional manifold. I am asking how to see that the space $I_p/I_p^2$ is $n$ dimensional, or alternatively, how to see that it is isomorphic to the cotangent space (defined by any alternative means).
 A: Not a full answer, but too long for a comment.
Terms of order greater than $2$ will be quotiented out automatically. Recall that an ideal is closed under multiplication by ring elements. So, if $x^2 \in I^2_p$, then $x^{2+k}=x^2x^k\in I^2_p$ also. Intuitively, if $x^2=0$ in the quotient, then $x^{2+k}=x^2x^k=0x^k=0$.
This does not in fact rely on full Taylor expansion; it seems that just a few terms are sufficient, like in Peano form: $f(x)=f(p)+f'(p)(x-p)+o(|x-p|^2)$. The non-trivial part is to show that $o(|x-p|^2)$ equals $I^2_p$.
The fact that for $n$-dimensional manifold the tangent space is $n$-dimensional follows from the chart that provides us with coordinates $x_1, \dots x_n$: the vectors $\frac{\partial}{\partial x_k}$ will form a basis.
A: I have learned the answer to my question quite a while ago, posting it here now.

We may write any $C^\infty$ function on $\mathbb R^n$ as $$ f(x)=f(x_0)+\partial_\mu f(x_0)(x^\mu-x_0^\mu)+(x^\mu-x^\mu_0)(x^\nu-x^\nu_0)R_{\mu\nu}(x), $$ where $R_{\mu\nu}(x)$ is a smooth remainder.
Now if $p\in M$ is any point, $U\subseteq M$ is a chart domain with chart map $\varphi$, and $\hat U\subseteq \mathbb R^n=\varphi(U)$, then $C^\infty(U)$ and $C^\infty(\hat U)$ are isomorphic (and our results are local, so we can use $C^\infty(U)$ instead of $C^\infty(M)$), so we can write any function $f\in C^\infty(U)$ as $$ f=f(p)+\partial_\mu f(p)(x^\mu-x^\mu(p))+(x^\mu-x^\mu(p))(x^\nu-x^\nu(p))R_{\mu\nu}, $$ where $\partial_\mu$ are the coordinate vector fields associated with the chart, and $x^\mu$ are now the coordinate functions.
We can project from $C^\infty(M)$ onto $I_p$ by $f\mapsto f-f(p)$, and we can project from $I_p$ to $I_p/I_p^2$ by $f\mapsto [f]$, so we have a projection from $C^\infty(M)$ to $I_p/I_p^2$ as $$ f\mapsto\mathrm df_p:=[f-f(p)]. $$
In the above form of the function, $f(p)$ is a constant that gets chopped of when we move from $C^\infty(M)$ to $I_p$, $\partial_\mu f(p)(x^\mu-x^\mu_0)$ is an element of $I_p$ that gets mapped to $\partial_\mu f(p)[x^\mu-x^\mu_0]=\partial_\mu f(p)\mathrm dx^\mu_p$, and the term involving the remainder $R_{\mu\nu}$ is in $I_p^2$, and thus it dies when taking the quotient.
Therefore, we have $$ \mathrm df_p=[f-f(p)]=\partial_\mu f(p)\mathrm dx^\mu_p, $$ which shows that the elements $$ \mathrm dx^\mu_p=[x^\mu-x^\mu(p)] $$ generate $I_p/I_p^2$, thus this space is finite dimensional. It is also easy to show that the elements $\mathrm dx^\mu_p$ are linearly independent, so $I_p/I_p^2$ is also finite dimensional.

For intuitive interpretation, we mention two points:


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*The cotangent space (at a point) essentially consists of first-order derivatives of functions (at that point). The construction outlined in this answer clearly constructs the space of first derivatives in a purely algebraic manner.

*If $v\in T_pM$ is a tangent vector, then $v$ vanishes on constants, and $v$ vanishes on elements of $I_p^2$. We could try to define the dual space of $T_pM$ to be simply $C^\infty(M)$ by $f:v\mapsto v(f)$, but this space is too big, because two different functions may act the same way on all vectors. If we define $\mathrm df_p(v)=v(f)$, then this is independent of the representative $f$, it only depends on $[f-f(p)]$, thus by going from $C^\infty(M)$ to $I_p/I_p^2$, we cut out those parts of functions that tangent vectors are insensitive to.

