How can we form $I/(I + \mathfrak{m}^2)$? I am trying to do an exercise in FOAG 2013.

"Suppose $(A, \mathfrak{m}, k)$ is a regular local ring of dimension $n$, and $I ⊂ A$ is an ideal of $A$ cutting out a regular local ring of dimension $d$. Let $r = n − d$. Show that $Spec A/I$ is a regular embedding in $Spec A$.
Hint: show that there are elements $f_1 ,\dots , f_r$ of $I$ spanning the k-vector space $I/(I + \mathfrak{m}^2 )$. Show that the quotient of $A$ by both $(f_1 , \dots , f_r )$ and $I$ yields dimension $d$ regular local rings. Show that a surjection of integral domains of the same dimension must be an isomorphism."

How can it be right in asking to consider the quotient $I/(I+ \mathfrak{m}^2)$? Why is $I + \mathfrak{m}^2$ a submodule of $I$?
 A: First we note that $\mathfrak{m}^2 \subseteq \mathfrak{m}^2 + I \subseteq \mathfrak{m}$ so the third isomorphism theorem says
$$(\mathfrak{m}/\mathfrak{m}^2)/(( \mathfrak{m^2} + I)/\mathfrak{m}^2) \cong \mathfrak{m}/(\mathfrak{m}^2 + I).$$
The right hand side is of course $(\mathfrak{m}/I)/(\mathfrak{m}/I)^2$ since $(\mathfrak{m}/I)^2 = (\mathfrak{m}^2 + I)/I$ and so  
$$\begin{eqnarray*}  (\mathfrak{m}/I)/(\mathfrak{m}/I)^2 &=& (\mathfrak{m}/I)/( (\mathfrak{m}^2 + I)/I) \\
&=& \mathfrak{m}/(\mathfrak{m}^2 + I)\end{eqnarray*}$$
by the third isomorphism theorem. So we get
$$\begin{eqnarray*} \dim_k(( \mathfrak{m^2} + I)/\mathfrak{m}^2) &=& \dim_k (\mathfrak{m}/\mathfrak{m}^2) - \dim_k (\mathfrak{m}/(\mathfrak{m}^2 + I)) \\
 &=& n -d \\
&=& r.\end{eqnarray*}$$
Now choose elements $x_1,\ldots,x_r \in I$ so that their images in $(\mathfrak{m}^2 + I)/\mathfrak{m}^2$ form a basis for this $k$-vector space. Since 
$$\dim_k (( \mathfrak{m}^2 + (x_1,\ldots,x_r))/\mathfrak{m}^2) = \dim_k( \mathfrak{m}^2 + I)/\mathfrak{m}^2$$
we get that $\dim A/(x_1,\ldots,x_r) = \dim A/I$ since $A/(x_1,\ldots,x_r)$ is a regular local ring. It follows $(x_1,\ldots,x_r)$ since we have a surjective map of rings of the same dimension which must be an isomorphism. Finally, we may complete $x_1,\ldots,x_r$ to a minimal generating set of $\mathfrak{m}$ showing $I$ is generated by the regular sequence $(x_1,\ldots,x_r)$.
