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.

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    $\begingroup$ 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. $\endgroup$ Mar 22, 2019 at 8:48
  • $\begingroup$ As a reference, a more general version of this is proved in Eisenbud as Theorem 16.14. $\endgroup$
    – jgon
    Mar 25, 2019 at 22:09

2 Answers 2


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 $|B| <m$. 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.


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