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Let $k$ be an algebraically closed field and let $\varphi :k^n\to k^r$ be a map such that there exist $f_1,\cdots ,f_r\in k[x_1,\cdots ,x_n]$ such that for every $a\in k^n$, $\varphi (a)=(f_1(a),\cdots ,f_r(a))$.

For $a\in k^n$ such that $f_i(a)=0$ for every $i$, define $\text{J}\varphi \left ( a \right )\in k^{n\times r}$ as $\left [\displaystyle \frac{\partial f_j}{\partial x_i}\left (a \right )\right ]_{ij}$. Let $\mathfrak{m}_a\subset k[x_1,\cdots ,x_n]/\langle f_1, \dots, f_r \rangle $ be the maximal ideal $\langle x_1-a_1,\cdots ,x_n-a_n \rangle$.

Prove that $\frac{\mathfrak{m}_a}{\mathfrak{m}_a^2}\cong \text{Coker}\left (\text{J}\varphi \left ( a \right ) \right )$ as $k$-vector spaces.

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    Have you tried reading the proof of I 5.1 in Hartshorne? – 2017-02-27
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    I also recommend reading the proof of Theorem 3.14 in Hulek. – 2017-02-27
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    In both books they use definitions we did not see in class. Also, I could not find the word $\text{Coker}$ in any of them. I would like to have a proof of the statement with the vocabulary I used in my post. Which function $\mathfrak{m}_a\to \text{Coker}\left (\text{J}\varphi \left ( a \right ) \right )$ should I define? – 2017-02-27
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    Coker means cokernel. If $f : V \to W$ is a linear map between vector spaces, then $Coker(f) / W / Ker(f)$. – 2017-02-27
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    Surely you want $m_a \subset k[x_1, \dots , x_n]/(f_1, \dots , f_r)$? Not $m_a \subset k[x_1, \dots , x_n]$. – 2017-02-28
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    Or perhaps $m_a $ is in the localisation of the ring I just wrote down? – 2017-02-28
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    Try defining the map the other way. Define a map $k^n \to m_a / m_a^2$ sending $(c_1, \dots c_n) $ to the equivalence class of $c_1 (x_1 - a_1)+ \dots + c_n (x_n - a_n)$ in $m_a / m_a^2$. Show that this map is surjective, and show that its kernel is equal to $Im(J(f))$. – 2017-02-28
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    Yes, I needed $f_i(a)=0$, sorry about that. The map is surjective since the equivalence class of $x_i-a_i$ is on the image for every $i$. But I do not see why its kernel equals the image of $\text{J}\varphi (a)$. – 2017-02-28
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    Your definition of $m_a$ still isn't right. You need to quotient $k[x_1, \dots, x_n ]$ by $(f_1, \dots , f_r)$. And are you sure you're not looking at the local ring? – 2017-02-28
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    Maybe I wrote down the theorem on my notebook in an incorrect way. What I am trying to understand is the following: let $k$ be an algebraically closed field, let $V\subset k^n$ a variety, $p\in V$, $R=\frac{k[x_1,\cdots ,x_n}{I(V)}$ where $I(V)=$ and define $\varphi (a)$ as $\varphi (a)=(f_1(a),\cdots ,f_r(a))$. Then the following statements are equivalent: 1) The rank of $\text{J}\varphi (p)$ equals $n-1$. 2) $\text{dim}_k\frac{\mathfrak{m}_p}{\mathfrak{m}_p^2}=1$ 3) $\mathfrak{m}_pR_{\mathfrak{m}_p}$ is a principal ideal 4) $R_{\mathfrak{m}_p}$ is a PID – 2017-03-01
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    5) $R_{\mathfrak{m}_p}$ is integrally closed and $\text{dim }V=1$ Maybe I commit a mistake when copying the statement in class... – 2017-03-01
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    Great. Let me ask this question for the third time: In the expression $m_p/m_p^2$, does $m_p$ refer to the maximal ideal in $R_{m_p}$ or does it refer to the maximal ideal in $R$? – 2017-03-01
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    My professor uses the notation $\mathfrak{m}R_{\mathfrak{m}}$ to refer the maximal ideal in $R_{\mathfrak{m}}$, so $\mathfrak{m}_p$ should refer to the maximal ideal in $R$. – 2017-03-01

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What is $m_a$ exactly? It's the ideal containing all regular functions on the variety $(f_1 = \dots = f_r = 0) \subset A^n$ that vanish at $a$. In other words, it contains all regular functions such that the constant term of their Taylor expansion around $a$ vanish.

And what is $m_a^2$? It's the ideal containing all the regular functions whose Taylor series around $a$ have vanishing constant AND linear terms.

So $m_a/m_a^2$ is the vector space of "linear truncations of regular functions on $(f_1 = \dots = f_r = 0) \in \subset A^n$ vanishing at $a$". In other words $m_a / m_a^2$ is spanned by $x_1 - a_1, \dots , x_n - a_n$ as a $k$-vector space.

Except...

$x_1 - a_1, \dots , x_n - a_n$ are not linearly independent in $m_a/m_a^2$! This is because we are quotienting by the ideal $(f_1, \dots , f_r)$. So really, $m_a / m_a^2$ is the vector space generated by $x_1 - a_1, \dots , x_n - a_n$, quotiented by the subspace generated by the linear terms of the Taylor expansions of $f_1, \dots , f_r$.

The Taylor expansion of $f_j(x)$ is $$ f_j(x) = \partial_1 f_j(a) (x_1 - a_1) + \dots + \partial_n f_j(a) (x_n - a_n)+ {\rm \ higher \ order \ }. $$ (Remember that $f_j(a) = 0$.)

So $m_a / m_a^2$ is the vector space generated by $x_1 - a_1, \dots , x_n - a_n$, quotiented by the subspace generated by linear combinations of $$\partial_1 f_1(a) (x_1 - a_1) + \dots + \partial_n f_1(a) (x_n - a_n), \\ \vdots \\ \partial_1 f_r(a) (x_1 - a_1) + \dots + \partial_n f_r(a) (x_n - a_n).$$

To finish off, consider the linear map $k^n \to m_a / m_a^2$, mapping $$ (c_1, \dots c_n) \mapsto c_1 (x_1 - a_1) + \dots + c_n (x_n - a_n).$$ It is surjective, and its kernel is spanned by the vectors $$(\partial_1 f_1(a) , \dots , \partial_n f_1(a) ), \ \dots \ , (\partial_1 f_r(a) , \dots , \partial_n f_r(a) ).$$ This kernel is clearly ${\rm Im \ }J (\varphi (a))$. So ${\rm Coker \ J(\varphi(a))} = m_a / m_a^2$.

Finally, as you pointed out in your comment, $m_a / m_a^2 \cong m_aA_a / (m_a A_a)^2 $. So we've shown ${\rm Coker \ J (\varphi(a))} = m_a A_a / (m_a A_a)^2$. Both of these vector spaces are representations of the cotangent space at $a$. The expression ${\rm Coker \ J (\varphi(a))}$ is closer to your intuitive notion of the cotangent space, but the expression $m_aA_a / (m_a A_a)^2$ has the advantage that it is intrinsic to the local ring at the point of the variety.

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    Hi Kenny, isn't the maximal ideal the maximal ideal of the localization of the coordinate ring at the point in question? Explicitly, if we call $M=(x_1-a_1,\dots,x_n-a_n)$ and the coordinate ring is $A$: $$ m_a=MA_M=\left\{\frac{g}{h}:g\in A, h\in A-m\right\} $$ – 2017-02-28
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    That's what I asked solomeo in a comment under the question. – 2017-02-28
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    The coordinate ring has a maximal ideal. The local ring also has a maximal ideal. It's not clear which we're talking about. – 2017-02-28
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    So I answered on the assumption that we're not localising. – 2017-02-28
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    Obviously this whole story is more geometrically useful if we do localise. Essentially the same argument works. – 2017-02-28