Krull's Principal Ideal Theorem for tangent spaces

$f$ 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 $im(di_m)=\ker(ev_f)$.