Krull's Principal Ideal Theorem for tangent spaces In "Foundations of Algebraic Geometry" by Ravi Vakil (page $333$, problem $12.1.B$) there is the following problem 

Suppose $A$ is a ring, and $m$ a maximal ideal. If $f ∈ m$, show that the Zariski tangent space of $A/f$ is cut out in the Zariski tangent space of $A$ by $f $ mod $(m ^2 )$.

At this point my question is the following : What is the corresponding precise mathematical statement in terms of isomorphism of vector spaces? Does it mean the following isomorphism of vector spaces:
$(m/(f)/(m/(f))^2)^* \cong (m/m^2 -f+m^2)^*$ ?
(I mean what is the precise mathematical statement of the part which says "cut out in the Zariski tangent space"?)
Any help from anyone is welcome.
 A: $f$ gives an element of the cotangent space $m/m^2$. Since the tangent space is the dual of the cotangent space, we can evaluate elements of the tangent space on $f$. "The tangent space of $A/f$ is cut out by $f$" means that the tangent space of $A/f$ at $m$ inside the tangent space of $A$ at $m$ is exactly the stuff that gives zero when evaluated on $f$. 
Another way to state this is that we have the following maps:


*

*$f\in m$ defines a map $ev_f:(m/m^2)^*\to A/m$ given by $x\in (m/m^2)^* \mapsto x(\overline{f})\in k$ where $\overline{f}$ represents the class of $f$ in $m/m^2$

*$i:\operatorname{Spec} A/f \to \operatorname{Spec} A$, the standard closed embedding corresponding to the ideal $(f)\subset A$

*$di_m: T_m \operatorname{Spec}A/f\to T_m \operatorname{Spec}A$, the map of tangent spaces at the point $m$ corresponding to the closed embedding $i$
The sentence "the tangent space of $A/f$ is cut out by $f$" means that $im(di_m)=\ker(ev_f)$.
A: KReiser's answer is very thorough, but I would like to add a supplementary answer which I suspect is probably closer to what Vakil had in mind.

The Zariski cotangent space of $A$ at $\frak m$ is the $A/\frak m$-vector space $\frak m/\frak m^2$.
The Zariski cotangent space of $A/(f)$ at $\mathfrak m/(f)$ is the $A/\frak m$-vector space $\frac{\mathfrak m/(f)}{(\mathfrak m/(f))^2}$.
But note that
$$\frac{\mathfrak m/(f)}{(\mathfrak m/(f))^2} = \frac{\mathfrak m/(f)}{(\mathfrak m^2 +(f))/(f)} \cong \frac{\mathfrak m}{\mathfrak m^2 +(f)}\cong \frac{\mathfrak m/\mathfrak m^2}{(f \pmod {\mathfrak m^2})}.$$
So, the Zariski cotangent space of $A/(f)$ at $\mathfrak m/(f)$ is given by modding out the Zariski cotangent space of $A$ at $\frak m$ by $f \pmod {\frak m^2}$; in other words, it is cut out by $f \pmod {\frak m^2}$.
Now since the map on cotangent spaces $$\frac{\frak m}{\frak m^2} \to \frac{\mathfrak m/\mathfrak m^2}{(f \pmod {\mathfrak m^2})}$$
is surjective, then the dual map (which is the map on tangent spaces)
$$\left(\frac{\mathfrak m/\mathfrak m^2}{(f \pmod {\mathfrak m^2})}\right)^\vee \to \left(\frac{\frak m}{\frak m^2}\right)^\vee$$
is injective.
So, the Zariski tangent space of $A/(f)$ at $\mathfrak m/(f)$ is "cut out" in the Zariski tangent space of $A$ at $\mathfrak m$ by $f \pmod {\frak  m^2}$.
