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I'm reading Fulton's algebraic curves book on page 105 and I didn't understand this proof:

1.Why if $R=k[X_1,\ldots,X_n]$, then $\Omega_k(R)$ is generated (as R-módule) by the differentials $dx_1,\ldots dx_n$?

2.Why if $\Omega_k(K)$ is a vector space of finite dimension over $K$, generated by $dx_1,\ldots,dx_n$ gives us the result?

If someone could help me with some of these questions I would be grateful.

Thanks.

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  1. Every element of $R$ can be written as $G(x_1,\ldots,x_n)$ for $G\in k[X_1,\ldots,X_n]$. But since $d$ is a derivation of $R$ into $\Omega_k(R)$, we have $$d(G(a_1,\ldots,x_n))=\sum_{i=1}^n G_{X_i}(x_1,\ldots,x_n)dx_i.$$ Now every element of $\Omega_k(R)$ is of the form $\sum_i r_i d(s_i)$ for some $r_i\in R$ and some $s_i\in R$ by definition of the module $\Omega_k(R)$. However since we just showed $d(s_j)$ is in the span of the $dx_i$ for all $j$, we have that $\sum_i r_id(s_i)$ is also in the span of the $dx_i$. Thus every element of $\Omega_k(R)$ is in the span of the $dx_i$, or equivalently, $\Omega_k(R)$ is generated as an $R$ module by $dx_1,\ldots,dx_n$.

  2. $K$ is the field of fractions of $R=k[x_1,\ldots,x_n]$ by definition. Since every differential of a fraction can be written in terms of differentials of elements of $R$, $\Omega_k(K)$ is generated as a $K$-module by $dx_1,\ldots,dx_n$, since the differentials of elements of $R$ are. Notice now that a module over a field is called a vector space, and since this is a finitely generated module over $K$, it is a finitely generated vector space over $K$, which implies that it is finite dimensional over $K$.

Edit:

So the proof is actually just the one sentence following the word Proof. In other words, the other paragraphs are just comments that are important when he develops the rest of the material. Let me see if I can illuminate this proof.

  1. First we define $\varphi':F\to M$ by saying $\varphi'(\sum x_i[y_i])=\sum x_i D(y_i)$.
  2. Since $D$ is a derivation, for any element $n$ of $N$, $\varphi'(n)=0$. Since the elements of $n$ are just the elements that must map to 0 under any derivation. They express the properties of a derivation.
  3. Since $\varphi'(N)=0$, and $\Omega_k(R) = F/N$, there is a unique map $\varphi : \Omega_k(R) \to M$ defined by $\varphi(\sum x_i dy_i)=\varphi'(\sum x_i[y_i])=\sum x_iD(y_i)$. Where $[y_i]$ can be any representative of $dy_i$, but in this case we just choose $[y_i]$ for simplicity. This is well defined since any representative of $dy_i$ is of the form $[y_i]+n$ for some $n\in N$, and $\varphi'([y_i]+n)=\varphi'([y_i])+0=\varphi'([y_i])$, so the value of $\varphi$ is independent of any choice of representatives.
  4. In particular, we now have $D(x)=\varphi(dx)$ for all $x\in R$.
  5. Now that we have shown existence, uniqueness follows since if we have $\varphi(dx)=D(x)$ for all $x\in R$, then $\varphi(\sum x_idy_i)=\sum x_i D(y_i)$, so $\varphi$ is determined completely by this property.
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  • $\begingroup$ Thank you very much for your answer. In the part 2, what I didn't understand is why the fact $\Omega_k(K)$ is a vector space of finite dimension over $K$, generated by $dx_1,\ldots,dx_n$ implies $D(x)=\varphi (dx)$? Thank you again!!! $\endgroup$ – user42912 May 21 '15 at 18:39
  • $\begingroup$ @user42912 sorry, misread that! Let me fix that! $\endgroup$ – jgon May 21 '15 at 18:40
  • $\begingroup$ @user42912 Edited to hopefully better address your question. $\endgroup$ – jgon May 21 '15 at 18:55
  • $\begingroup$ Thank you very much. You are very kind!!! this will help me a lot! $\endgroup$ – user42912 May 21 '15 at 19:21
  • $\begingroup$ @user42912 no problem, glad to be helpful! $\endgroup$ – jgon May 21 '15 at 19:23

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