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.

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

• is the map $di_m: T_m \operatorname{Spec}A/f\to T_m \operatorname{Spec}A$ is same as the obvious map between $(m/(f)/(m/(f))^2)^* \to (m/m^2 )^*$? Commented Sep 3, 2019 at 6:41
• Yes, that's correct. Commented Sep 3, 2019 at 6:55
• @KReiser I've been working on this problem for a few hours now and I don't know how to make sense of your answer. At this point, Vakil has not addressed the content in your three bullets. So, I'm wondering how to make sense of this exercise with the material we have available; which is just the definition of cotangent space $\frak m / \frak m^2$, and the definition of tangent space given as $(\frak m / \frak m^2)^\vee$. Commented Dec 4, 2021 at 1:04
• One thing I have worked out is that $\frac{\mathfrak{m}/(f)}{(\mathfrak{m}/(f))^2)} \cong \frac{\mathfrak m}{\mathfrak{m}^2 +(f)}$. If $f \in \frak m^2$, then $\frac{\mathfrak{m}/(f)}{(\mathfrak{m}/(f))^2)} \cong \frak m/ \frak m^2$ and if $f \notin \frak m^2$, then the dimension of $\frac{\mathfrak{m}/(f)}{(\mathfrak{m}/(f))^2)}$ is one less than the dimension of $\frak m/\frak m^2$. Commented Dec 4, 2021 at 1:06
• @user46372819 I disagree with your assertion that Vakil "has not addressed" the stuff I'm talking about. All of the content of the bullet points follows from material previously discussed in the text in combination with standard facts from algebra about dual maps of vector spaces which might reasonably be covered in a previous course. It seems you came to the correct conclusion anyways, so what else is there to discuss? Commented Dec 4, 2021 at 1:16

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 $$f \pmod {\frak m^2}$$.