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Algebraically independent polynomials iff linearly independent differentials

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1

$begingroup$

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.

algebraic-geometry commutative-algebra jacobian computational-algebra

share|cite|improve this question

asked Mar 22 at 2:48

ZoeyZoey

30829

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  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.
  $endgroup$
  – red_trumpet
  Mar 22 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 at 22:09
  

  
  
  
  

add a comment | 

1

$begingroup$

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.

algebraic-geometry commutative-algebra jacobian computational-algebra

share|cite|improve this question

asked Mar 22 at 2:48

ZoeyZoey

30829

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  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.
  $endgroup$
  – red_trumpet
  Mar 22 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 at 22:09
  

  
  
  
  

add a comment | 

$begingroup$

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.

algebraic-geometry commutative-algebra jacobian computational-algebra

share|cite|improve this question

asked Mar 22 at 2:48

ZoeyZoey

30829

$endgroup$

share|cite|improve this question

asked Mar 22 at 2:48

ZoeyZoey

30829

share|cite|improve this question

asked Mar 22 at 2:48

ZoeyZoey

30829

asked Mar 22 at 2:48

ZoeyZoey

30829

asked Mar 22 at 2:48

ZoeyZoey

30829

2 Answers
2

active

oldest

votes

2

$begingroup$

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 $k=mathbbF_p$ and $K=mathbbF_p(t)$ for some prime $p$. If we take $f_1=t^p$, one has that $f_1$ is not algebraic over $k$, while $df_1=0$.

share|cite|improve this answer

answered Mar 23 at 20:58

Tommaso ScognamiglioTommaso Scognamiglio

581412

$endgroup$

add a comment | 

0

$begingroup$

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 $mathbbA^m(k)$, and by Bertini's theorem the differential is surjective on an open subset.

share|cite|improve this answer

answered Mar 23 at 21:48

MR_QMR_Q

746

$endgroup$

add a comment | 

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2

$begingroup$

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 $k=mathbbF_p$ and $K=mathbbF_p(t)$ for some prime $p$. If we take $f_1=t^p$, one has that $f_1$ is not algebraic over $k$, while $df_1=0$.

share|cite|improve this answer

answered Mar 23 at 20:58

Tommaso ScognamiglioTommaso Scognamiglio

581412

$endgroup$

add a comment | 

2

$begingroup$

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 $k=mathbbF_p$ and $K=mathbbF_p(t)$ for some prime $p$. If we take $f_1=t^p$, one has that $f_1$ is not algebraic over $k$, while $df_1=0$.

share|cite|improve this answer

answered Mar 23 at 20:58

Tommaso ScognamiglioTommaso Scognamiglio

581412

$endgroup$

add a comment | 

$begingroup$

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 $k=mathbbF_p$ and $K=mathbbF_p(t)$ for some prime $p$. If we take $f_1=t^p$, one has that $f_1$ is not algebraic over $k$, while $df_1=0$.

share|cite|improve this answer

answered Mar 23 at 20:58

Tommaso ScognamiglioTommaso Scognamiglio

581412

$endgroup$

share|cite|improve this answer

answered Mar 23 at 20:58

Tommaso ScognamiglioTommaso Scognamiglio

581412

answered Mar 23 at 20:58

Tommaso ScognamiglioTommaso Scognamiglio

581412

answered Mar 23 at 20:58

Tommaso ScognamiglioTommaso Scognamiglio

581412

0

$begingroup$

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 $mathbbA^m(k)$, and by Bertini's theorem the differential is surjective on an open subset.

share|cite|improve this answer

answered Mar 23 at 21:48

MR_QMR_Q

746

$endgroup$

add a comment | 

0

$begingroup$

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 $mathbbA^m(k)$, and by Bertini's theorem the differential is surjective on an open subset.

share|cite|improve this answer

answered Mar 23 at 21:48

MR_QMR_Q

746

$endgroup$

add a comment | 

$begingroup$

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 $mathbbA^m(k)$, and by Bertini's theorem the differential is surjective on an open subset.

share|cite|improve this answer

answered Mar 23 at 21:48

MR_QMR_Q

746

$endgroup$

share|cite|improve this answer

answered Mar 23 at 21:48

MR_QMR_Q

746

answered Mar 23 at 21:48

MR_QMR_Q

746

answered Mar 23 at 21:48

MR_QMR_Q

746