$k[epsilon]to k$ is not a flat ring homomorphism

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Let $k$ be a field and $k[epsilon]:=k[x]/(x^2)$ the ring of dual numbers. Suppose that we have a map
$$
k[epsilon]to k,
$$
$$
a+bepsilon mapsto a.
$$
How can we see that this map not flat, i.e., does not make $k$ into a flat $k[epsilon]$-module? It seems that $k$ is a torsion-free module. I would particularly appreciate a direct algebraic proof.

abstract-algebra

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asked Jul 24 at 15:20

Jimmy R

1,185618

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Have you tried checking some exact sequences?
  – Dzoooks
  Jul 24 at 15:25
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Dzooks I have tried $0to (t^2) to k[t] to k[epsilon]to 0$, but this did not get me anywhere
  – Jimmy R
  Jul 24 at 15:26
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  What about $0to k to k[epsilon] to k to 0$?
  – xarles
  Jul 24 at 15:30
  
  

  
  
  
  

add a comment | 

up vote
0
down vote

favorite

Let $k$ be a field and $k[epsilon]:=k[x]/(x^2)$ the ring of dual numbers. Suppose that we have a map
$$
k[epsilon]to k,
$$
$$
a+bepsilon mapsto a.
$$
How can we see that this map not flat, i.e., does not make $k$ into a flat $k[epsilon]$-module? It seems that $k$ is a torsion-free module. I would particularly appreciate a direct algebraic proof.

abstract-algebra

share|cite|improve this question

asked Jul 24 at 15:20

Jimmy R

1,185618

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Have you tried checking some exact sequences?
  – Dzoooks
  Jul 24 at 15:25
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Dzooks I have tried $0to (t^2) to k[t] to k[epsilon]to 0$, but this did not get me anywhere
  – Jimmy R
  Jul 24 at 15:26
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  What about $0to k to k[epsilon] to k to 0$?
  – xarles
  Jul 24 at 15:30
  
  

  
  
  
  

add a comment | 

up vote
0
down vote

favorite

up vote
0
down vote

favorite

Let $k$ be a field and $k[epsilon]:=k[x]/(x^2)$ the ring of dual numbers. Suppose that we have a map
$$
k[epsilon]to k,
$$
$$
a+bepsilon mapsto a.
$$
How can we see that this map not flat, i.e., does not make $k$ into a flat $k[epsilon]$-module? It seems that $k$ is a torsion-free module. I would particularly appreciate a direct algebraic proof.

abstract-algebra

share|cite|improve this question

asked Jul 24 at 15:20

Jimmy R

1,185618

Let $k$ be a field and $k[epsilon]:=k[x]/(x^2)$ the ring of dual numbers. Suppose that we have a map
$$
k[epsilon]to k,
$$
$$
a+bepsilon mapsto a.
$$
How can we see that this map not flat, i.e., does not make $k$ into a flat $k[epsilon]$-module? It seems that $k$ is a torsion-free module. I would particularly appreciate a direct algebraic proof.

abstract-algebra

share|cite|improve this question

asked Jul 24 at 15:20

Jimmy R

1,185618

share|cite|improve this question

share|cite|improve this question

asked Jul 24 at 15:20

Jimmy R

1,185618

asked Jul 24 at 15:20

Jimmy R

1,185618

asked Jul 24 at 15:20

Jimmy R

1,185618

1,185618

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Have you tried checking some exact sequences?
  – Dzoooks
  Jul 24 at 15:25
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Dzooks I have tried $0to (t^2) to k[t] to k[epsilon]to 0$, but this did not get me anywhere
  – Jimmy R
  Jul 24 at 15:26
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  What about $0to k to k[epsilon] to k to 0$?
  – xarles
  Jul 24 at 15:30
  
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Have you tried checking some exact sequences?
  – Dzoooks
  Jul 24 at 15:25
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Dzooks I have tried $0to (t^2) to k[t] to k[epsilon]to 0$, but this did not get me anywhere
  – Jimmy R
  Jul 24 at 15:26
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  What about $0to k to k[epsilon] to k to 0$?
  – xarles
  Jul 24 at 15:30
  
  

  
  
  
  

Have you tried checking some exact sequences?
– Dzoooks
Jul 24 at 15:25

Have you tried checking some exact sequences?
– Dzoooks
Jul 24 at 15:25

@Dzooks I have tried $0to (t^2) to k[t] to k[epsilon]to 0$, but this did not get me anywhere
– Jimmy R
Jul 24 at 15:26

@Dzooks I have tried $0to (t^2) to k[t] to k[epsilon]to 0$, but this did not get me anywhere
– Jimmy R
Jul 24 at 15:26

1

1

What about $0to k to k[epsilon] to k to 0$?
– xarles
Jul 24 at 15:30

What about $0to k to k[epsilon] to k to 0$?
– xarles
Jul 24 at 15:30

add a comment | 

2 Answers
2

active

oldest

votes

up vote
1
down vote



accepted

I'm not sure what direct algebraic means exactly, but I feel this is direct:

Hint: $k[epsilon]$ is local:

What does that say about the f.g. flat modules?

What submodules of $k[epsilon]$ can be summands?

From this you should be able to see why.

share|cite|improve this answer

answered Jul 24 at 15:28

rschwieb

99.7k1192226

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks! By 'direct algebraic' I meant something not taking Spec or Tor.
  – Jimmy R
  Jul 24 at 15:31
  

  
  
  
  

add a comment | 

up vote
2
down vote

To spell out @xarles's suggestion in the comment section above: the $k[epsilon]$-linear map $(epsilon) to k[epsilon]$, $$aepsilon mapsto aepsilon,$$ with $ain k[epsilon]$, is injective. However, if one tensors with $k = k[epsilon]/(epsilon)$, the resulting map is the zero map, with a non-zero domain: that is, $(epsilon) otimes_k[epsilon] k simeq k$, while its image in $k[epsilon]otimes_k[epsilon] k$ is $0$.

In other words, one can move $epsilon$ across the tensor product in the image (where $epsilonotimes 1 = 1 otimes epsilon = 0$), but not in the domain.

share|cite|improve this answer

answered Jul 24 at 16:00

peter a g

2,9701614

add a comment | 

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
1
down vote



accepted

I'm not sure what direct algebraic means exactly, but I feel this is direct:

Hint: $k[epsilon]$ is local:

What does that say about the f.g. flat modules?

What submodules of $k[epsilon]$ can be summands?

From this you should be able to see why.

share|cite|improve this answer

answered Jul 24 at 15:28

rschwieb

99.7k1192226

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks! By 'direct algebraic' I meant something not taking Spec or Tor.
  – Jimmy R
  Jul 24 at 15:31
  

  
  
  
  

add a comment | 

up vote
1
down vote



accepted

I'm not sure what direct algebraic means exactly, but I feel this is direct:

Hint: $k[epsilon]$ is local:

What does that say about the f.g. flat modules?

What submodules of $k[epsilon]$ can be summands?

From this you should be able to see why.

share|cite|improve this answer

answered Jul 24 at 15:28

rschwieb

99.7k1192226

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks! By 'direct algebraic' I meant something not taking Spec or Tor.
  – Jimmy R
  Jul 24 at 15:31
  

  
  
  
  

add a comment | 

up vote
1
down vote



accepted

up vote
1
down vote



accepted

I'm not sure what direct algebraic means exactly, but I feel this is direct:

Hint: $k[epsilon]$ is local:

What does that say about the f.g. flat modules?

What submodules of $k[epsilon]$ can be summands?

From this you should be able to see why.

share|cite|improve this answer

answered Jul 24 at 15:28

rschwieb

99.7k1192226

I'm not sure what direct algebraic means exactly, but I feel this is direct:

Hint: $k[epsilon]$ is local:

What does that say about the f.g. flat modules?

What submodules of $k[epsilon]$ can be summands?

From this you should be able to see why.

share|cite|improve this answer

answered Jul 24 at 15:28

rschwieb

99.7k1192226

share|cite|improve this answer

share|cite|improve this answer

answered Jul 24 at 15:28

rschwieb

99.7k1192226

answered Jul 24 at 15:28

rschwieb

99.7k1192226

answered Jul 24 at 15:28

rschwieb

99.7k1192226

99.7k1192226

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks! By 'direct algebraic' I meant something not taking Spec or Tor.
  – Jimmy R
  Jul 24 at 15:31
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks! By 'direct algebraic' I meant something not taking Spec or Tor.
  – Jimmy R
  Jul 24 at 15:31
  

  
  
  
  

Thanks! By 'direct algebraic' I meant something not taking Spec or Tor.
– Jimmy R
Jul 24 at 15:31

Thanks! By 'direct algebraic' I meant something not taking Spec or Tor.
– Jimmy R
Jul 24 at 15:31

add a comment | 

up vote
2
down vote

To spell out @xarles's suggestion in the comment section above: the $k[epsilon]$-linear map $(epsilon) to k[epsilon]$, $$aepsilon mapsto aepsilon,$$ with $ain k[epsilon]$, is injective. However, if one tensors with $k = k[epsilon]/(epsilon)$, the resulting map is the zero map, with a non-zero domain: that is, $(epsilon) otimes_k[epsilon] k simeq k$, while its image in $k[epsilon]otimes_k[epsilon] k$ is $0$.

In other words, one can move $epsilon$ across the tensor product in the image (where $epsilonotimes 1 = 1 otimes epsilon = 0$), but not in the domain.

share|cite|improve this answer

answered Jul 24 at 16:00

peter a g

2,9701614

add a comment | 

up vote
2
down vote

To spell out @xarles's suggestion in the comment section above: the $k[epsilon]$-linear map $(epsilon) to k[epsilon]$, $$aepsilon mapsto aepsilon,$$ with $ain k[epsilon]$, is injective. However, if one tensors with $k = k[epsilon]/(epsilon)$, the resulting map is the zero map, with a non-zero domain: that is, $(epsilon) otimes_k[epsilon] k simeq k$, while its image in $k[epsilon]otimes_k[epsilon] k$ is $0$.

In other words, one can move $epsilon$ across the tensor product in the image (where $epsilonotimes 1 = 1 otimes epsilon = 0$), but not in the domain.

share|cite|improve this answer

answered Jul 24 at 16:00

peter a g

2,9701614

add a comment | 

up vote
2
down vote

up vote
2
down vote

To spell out @xarles's suggestion in the comment section above: the $k[epsilon]$-linear map $(epsilon) to k[epsilon]$, $$aepsilon mapsto aepsilon,$$ with $ain k[epsilon]$, is injective. However, if one tensors with $k = k[epsilon]/(epsilon)$, the resulting map is the zero map, with a non-zero domain: that is, $(epsilon) otimes_k[epsilon] k simeq k$, while its image in $k[epsilon]otimes_k[epsilon] k$ is $0$.

In other words, one can move $epsilon$ across the tensor product in the image (where $epsilonotimes 1 = 1 otimes epsilon = 0$), but not in the domain.

share|cite|improve this answer

answered Jul 24 at 16:00

peter a g

2,9701614

To spell out @xarles's suggestion in the comment section above: the $k[epsilon]$-linear map $(epsilon) to k[epsilon]$, $$aepsilon mapsto aepsilon,$$ with $ain k[epsilon]$, is injective. However, if one tensors with $k = k[epsilon]/(epsilon)$, the resulting map is the zero map, with a non-zero domain: that is, $(epsilon) otimes_k[epsilon] k simeq k$, while its image in $k[epsilon]otimes_k[epsilon] k$ is $0$.

In other words, one can move $epsilon$ across the tensor product in the image (where $epsilonotimes 1 = 1 otimes epsilon = 0$), but not in the domain.

share|cite|improve this answer

answered Jul 24 at 16:00

peter a g

2,9701614

share|cite|improve this answer

share|cite|improve this answer

answered Jul 24 at 16:00

peter a g

2,9701614

answered Jul 24 at 16:00

peter a g

2,9701614

answered Jul 24 at 16:00

peter a g

2,9701614

2,9701614

add a comment | 

add a comment | 

 

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