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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.

user26857
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HARRY
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  • the following link may be of interest: https://math.stackexchange.com/questions/4325846/how-do-i-formally-show-that-the-zariski-tangent-space-of-the-intersection-of-two/4327312#4327312 – hm2020 Dec 08 '21 at 11:52

2 Answers2

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

KReiser
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  • is the map $di_m: T_m \operatorname{Spec}A/f\to T_m \operatorname{Spec}A$ is same as the obvious map between $(m/(f)/(m/(f))^2)^* \to (m/m^2 )^*$? – HARRY Sep 03 '19 at 06:41
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    Yes, that's correct. – KReiser Sep 03 '19 at 06:55
  • @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 $(\frak m / \frak m^2)^\vee$. – user5826 Dec 04 '21 at 01:04
  • 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 $\frak m/\frak m^2$. – user5826 Dec 04 '21 at 01:06
  • @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? – KReiser Dec 04 '21 at 01:16
  • @KReiser At this point, I guess the only bullet we've seen is the second, but the other bullets have not been addressed yet. Can you point me to a section or page number? – user5826 Dec 04 '21 at 04:02
  • @user46372819 The first bullet point is a translation of the second sentence of the original problem - Vakil could perhaps be more explicit about it, and indeed figuring out what this means is an important part of the original question here. The map described in the third bullet is first mentioned on page 101 in aside 3.1.2. – KReiser Dec 04 '21 at 04:28
  • @user46372819 There's a surjection $m/m^2\to m/(m^2+(f))$ by $x\mapsto x+(f)$. If $f$ is nonzero modulo $m^2$, you can find a basis for the cotangent space including $f$. Write down the matrix for the map: it's given by zeroing out the $f$-coordinate. Now take the dual: this turns in to an injection of tangent spaces, where the map is given by the transpose matrix. So the tangent space to $A/f$ is given by those things which have $f$-dual coordinate zero. That's the stuff which evaluates to zero on $f$, and this is exactly what it means to be "cut out by $f$". – KReiser Dec 05 '21 at 21:24
  • @KReiser Thank you. I was able to work through your answer after trying some examples. – user5826 Dec 05 '21 at 21:38
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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 $f \pmod {\frak m^2}$.

user5826
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