# Algebraically independent polynomials iff linearly independent differentials

This is an exercise question in Appendix A of Introduction of Algebraic Geometry, Justin R Smith. I am looking for an intuition for the solution.

if $$k \rightarrow K$$ is an extension of fields of char $$0$$ and $$f_1, \ldots, f_m \in K$$ show that $$\{f_1, \ldots, f_m \}$$ are algebraically independent iff $$\{df_1, \ldots, df_m \} \in \Omega_{K/k}$$ are linearly independent.

• My guess is that this is because if $R = k[t_1,\ldots,t_m]$, then $\Omega_{R/k} = dt_1R \oplus dt_2R \oplus \dots \oplus dt_mR$ as an $R$-module. Mar 22, 2019 at 8:48
• As a reference, a more general version of this is proved in Eisenbud as Theorem 16.14.
– jgon
Mar 25, 2019 at 22:09

This could be an outline of a possible proof.

First,let's suppose that $$f_1 \dots f_m$$ are algebraically indipendent. Let us consider a basis of trascendence $$B=\{f_1 \dots f_m, g_1 \dots g_r\}$$, so that we have the inclusions: $$k \subseteq k(f_1 \dots f_m, g_1 \dots g_r) \subseteq K ,$$ with the last one being finite separable. Let's call the intermediate field $$F$$.

We have $$\Omega_{F /k}= F df_1 \bigoplus \dots \bigoplus F dg_r$$.Because of $$K \supseteq F$$ finite and separable, we have $$\Omega_{F/K}=0$$.

Using the standard exact sequences for Kahler differentials, one obtains $$\Omega_{K /k}=K df_1 \bigoplus \dots \bigoplus K dg_r .$$ One immediately obtain the linear indipendence of the differentials.

Let's do the other arrow. Let's call $$L$$ the field over $$k$$ generated by $$f_1 \dots f_m$$ and let us suppose that $$f_1 \dots f_m$$ are not algebraically indipendent.

Let's consider $$B=\{f_{i_1} \dots f_{i_s}\}$$ a maximal indipendent subset. Our hypothesis implies that $$|B| . One can see that $$B$$ is a trascendence basis for $$L$$. With exactly the same proof above, one has that $$\Omega_{L/k}$$ is an $$L$$ vector spae of dimension less than $$m$$ so that $$df_i$$ are not linearly indipendent. Having $$L \subseteq K$$ one obtains a contradiction.

Let me add just one more comment. It is pretty important that the characteristic is $$0$$. Let's take $$k=\mathbb{F}_p$$ and $$K=\mathbb{F}_p(t)$$ for some prime $$p$$. If we take $$f_1=t^p$$, one has that $$f_1$$ is not algebraic over $$k$$, while $$df_1=0$$.

When $$k$$ is algebraically closed:

If the polynomials are algebraic dependent, then there exists some polynomial $$F$$ such that $$F(f_1,\dots, f_m)=0$$. Take the derivative with respect to each variable in the domain of $$f_i$$. It follows from the chain rule.

If the polynomials are algebraically independent, then there exists no polynomial $$F$$ such that $$F(f_1,\dots, f_m)=0$$. Then the image of the map defined by $$f=(f_1,\dots, f_m)$$ is dense in $$\mathbb{A}^m(k)$$, and by Bertini's theorem the differential is surjective on an open subset.