Let $Y \subset \mathbb{A}^n$ be an affine variety and $\frak{a}_p$:=$(x_1-a_1,…,x_n-a_n)$ be the ideal of the point $p=(a_1,..,a_n)$ of $Y$.
For any maximal ideal $\frak{m}$ of the local ring $\mathcal{O}_p$ of $p\in Y$, we have
$ \frak{m}/\frak{m}^2 \cong \frak{a}_p/(b + \frak{a}_p^2),$
where $\frak{b}$ is the ideal $I(Y)$ of $Y$.
I cannot understand this isomorphism.
In the following, I explain where it is used.
THEOREM.
Let $Y \subset \mathbb{A}^n$ be an affine variety. Let $P \in Y$ be apoint. Then $Y$ is nonsigular at $P$ if and only if the local ring $\mathcal{O}_{p,Y}$ is a regular local ring.
PROOF.
Let $P$ be the point $(a_1,…,a_n)$ in $\mathbb{A}^n$, and let $\frak{a}_p=(x_1-a_1,…,x_n-a_n$ be the corresponding maximal ideal in $A:=k[x_1,…,x_n]$.
We define a linear map $\theta: A \to k^n$ by
$\theta(f):=(\frac{\partial f}{\partial x_1},…,\frac{\partial f}{\partial x_n})$ for any $f \in A$. Because $\theta(x_i-a_i),i=1,…,n$ form a basis of $k^n$ and that $\theta(\frak{a}^2)=0$, $\theta$ induces an isomorphisms $ \theta': \frak{a}/\frak{a}^2$ $\to k^n$ .
Now let $\frak{b}$ be the ideal of $Y$ in $k[x_1,…,x_n]$ and let $f_1,…,f_t$ be a set of generators of $\frak{b}$. Then the rank of the Jacobian matrix $(\partial f_i/ \partial x_j(p))$ is just the dimension of $\theta(\frak{b})$. Using the isomorphism $\theta'$,
dim $\theta(\frak{b})$ = dim $\frak{a}_p/(b + \frak{a}_p^2).$
On the other hand, the local ring $\mathcal{O}_p$ of $P$ on $Y$ is obtained from $A$ by divideing by $\frak{b}$ and localizeing at the maximal ideal $\frak{a}_p$ Thus if $\frak{m}$ is the maximal ideal of $\mathcal{O}_p$, we have
$ \frak{m}/\frak{m}^2 \cong \frak{a}_p/(b + \frak{a}_p^2).$
PROOF CONTINUES.
Best Answer
$\newcommand{\ideal}[1]{\mathfrak{#1}}$ The isomorphism is very simple: Let $A=k[x_1,\ldots,x_n]$ be the coordinate ring of $\mathbb{A}^n$ and $\ideal{b} \subseteq A$ be the ideal $I(Y)$ of the variety $Y$. Furthermore let $p \in \mathbb{A}^n$ be a closed point which lies in $Y$ too. Let $\ideal{m} \subseteq A$ be the ideal $I(p)$ of this point in $\mathbb{A}^n$. Then $A/\ideal{b}$ is the coordinate ring of $Y$ and $A_\ideal{m} = \mathcal{O}_{\mathbb{A}^n,p}$ and $(A/\ideal{b})_\ideal{m} = \mathcal{O}_{Y,p}$ are the local coordinate rings.
Now the maximal ideal of $(A/\ideal{b})_\ideal{m} = C$ is
$$(\ideal{m} + \ideal{b}) C = \ideal{m} C.$$
Its square in $C$ is $(\ideal{m}^2 + \ideal{b}) C$. Note that for every $\ideal{a} \subseteq A$, we have $\ideal{a} C = (\ideal{a} + \ideal{b}) C$.
Now
$$ \ideal{m}C/\ideal{m^2} C =\ideal{m}C/(\ideal{m}^2+\ideal{b})C = (\ideal{m}/(\ideal{m}^2+\ideal{b})) C$$
For the last step one uses that we have $(I/J) R/K = (I (R/K)) / (J (R/K))$ for $K \subseteq I,J$ and $(I/J) R_K = I R_K/ J R_K$ for a prime ideal $K \supseteq I,J$ and the chain $A \to A/\ideal{b} \to (A/\ideal{b})_\ideal{m} = C$.
Note: What you call $\ideal{a}_p$ I call $\ideal{m}$ or $\ideal{m} (A/\ideal{b})$ when viewed as a point in $Y$. What you call $\ideal{m}$ is $\ideal{m} C$ for me.
Note added for clarification: The $A$-module $\ideal{m}/(\ideal{m}^2 + \ideal{b})$ is an $A/\ideal{b}$-module because of $\ideal{b} \ideal{m} \subseteq \ideal{m}^2 + \ideal{b}$ and by an analogous argument also an $A/\ideal{m} = k$-module, all compatible with $A \to A/\ideal{b} \to A/\ideal{m}$. As $\ideal{m}/(\ideal{m}^2 + \ideal{b})$ is a $k$-module via $A \to A/\ideal{m}$ it does not change under localization with $\ideal{m}$. So we have
$$ \begin{multline}\ideal{m}/(\ideal{m}^2+\ideal{b}) = \ideal{m}_\ideal{m}/(\ideal{m}^2 + \ideal{b})_\ideal{m} = \\ = (\ideal{m}_\ideal{m}/\ideal{b}_\ideal{m})/( (\ideal{m}^2 + \ideal{b})_\ideal{m}/\ideal{b}_\ideal{m}) = \ideal{m} C/ (\ideal{m}^2 + \ideal{b}) C = \ideal{m} C/\ideal{m}^2 C. \end{multline}$$
Especially the second and third quality follow from
$$I/J = (I/K)/(J/K) = (I R/K)/(J R/K)$$
for $K \subseteq J \subseteq I \subseteq R$ ideals of $R$. It is an equality of $R/K$-modules.
In total, this seems to be what is stated in Hartshorne, p. 32. (remember my change in notation).