I am reading Basic Algebraic Geometry 1 by Shafarevich (3rd edition) and I couldn't understand the following portion on pg 222:

Namely, in Section 5.2 we defined the module of differentials $\Omega_{A}$ for any commutative ring $A$ (as an algebra over a subring $A_0$, that we take to be the field $k$ in what follows). We saw that $\Omega_{A}$ is generated as an $A$-module by d$t$ for $t \in A$. Thus it is generated as an Abelian group by $u\text{d}t$ for $u$, $t \in A$. The relations between these generators are all obtained from the relations (3.57) by multiplying by $u \in A$. Thus they are of the form \begin{equation} u\text{d}(f+g) = u\text{d}f + u\text{d}g \;\; \text{and}\;\; u\text{d}fg = uf\text{d}g + ug\text{d}f \; \text{for } u, f, g \in A. \end{equation} We must add to these relations expressing that $\Omega_{A}$ is an $A$-module, namely \begin{equation} (u+v)\text{d}f = u\text{d}f + v\text{d}f. \end{equation} It follows that any function $F(u,f) \in k$ of a $u,f \in A$ vanishing on the subgroup generated by these relations defines a linear function on $\Omega_{A}$.

What is troubling me is the last line. First of all, $F$ is a function from $F:A\times A \to k$, so how does it define a (linear) function on $\Omega_{A}$. I think $F(u,f)$ is to be interpreted as $F(u\text{d}f)$. Is that correct?

Secondly, even if I interpret it this way, how do I show that $F$ defines a linear function on $\Omega_{A}$. If I am not mistaken, I'm required to show that

$F(u_1\text{d}f_1+u_2\text{d}f_2) = F(u_1\text{d}f_1) +F(u_2\text{d}f_2) \; \text{and} F(c\ u\text{d}f) = cF(u\text{d}f), c \in k$

What I understand from $F$ "vanishing on the subgroup generated by these relations" is

$F(u\text{d}(f+g) - u\text{d}f - u\text{d}g) = 0$, etc...

but I don't see this leading to linearity.

  • $\begingroup$ I think the relation $df = 0$ if $f\in k$ should be added. $\endgroup$ – Captain Lama Apr 13 '16 at 21:14
  • $\begingroup$ @CaptainLama Yes. I just quoted what's in the book. But even then, how do we show linearity? $\endgroup$ – Seven Apr 13 '16 at 21:16
  • $\begingroup$ In my opinion, either he implicitly assumes that $F: A\times A\to k$ is already $k$-linear in the first variable, or you don't have $k$-linearity, but just $\mathbb{Z}$-linearity. $\endgroup$ – Captain Lama Apr 13 '16 at 21:25
  • $\begingroup$ That could be a possibility. I was having doubts if I have understood $F$ "vanishing on the subgroup generated by these relations" correctly. Isn't that just what I said in the end? If that's the case, I'll just assume he means $F$ linear and move on. $\endgroup$ – Seven Apr 13 '16 at 21:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.