Show F is isomorphic to F/{0}

  • #1
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Homework Statement


Let F be a field. Show that F is isomorphic to F/{0}

Homework Equations




The Attempt at a Solution


By the first ring isomorphic theorem, kernel of the homomorphism is an ideal which is either {0} or I. Hence F isomorphic to F/{0}

I think I misunderstood the problem can anyone check my work where I did wrong
 

Answers and Replies

  • #2
STEMucator
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For two fields to be isomorphic, there has to be a bijective homomorphism between them:

$$\phi : F \rightarrow (F - \{0\})$$

You need to define the map you used to show ##\text{ker}(\phi)## is an ideal of ##F##. You then need to show ##\ker(\phi)## really is an ideal.

Then I would be convinced.

You could also go the long way by proving the map you've defined is a bijective homomorphism between the fields, which would imply the fields are isomorphic to each other.
 
  • #3
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For two fields to be isomorphic, there has to be a bijective homomorphism between them:

$$\phi : F \rightarrow (F - \{0\})$$

You need to define the map you used to show ##\text{ker}(\phi)## is an ideal of ##F##. You then need to show ##\ker(\phi)## really is an ideal.

Then I would be convinced.

You could also go the long way by proving the map you've defined is a bijective homomorphism between the fields, which would imply the fields are isomorphic to each other.

Suppose $\phi : F\rightarrow F$ is an identity homomorphism, then $\ker(\phi)=\{0\}$ is an ideal of $F$ and hence $F/\{0\}\cong F$ by the first ring isomorphism.
 
  • #4
STEMucator
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an identity homomorphism

Are you saying ##\phi(f) = f, \forall f \in F##? If so there is a problem because ##\phi(0) = 0## and ##0 \notin (F - \{ 0 \})##.
 
  • #5
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Are you saying ##\phi(f) = f, \forall f \in F##? If so there is a problem because ##\phi(0) = 0## and ##0 \notin (F - \{ 0 \})##.

The OP means ##F/\{0\}##, which means the ring ##F## modulo ##\{0\}##. He does not mean the set theoretic difference ##F\setminus \{0\}## (in that case, the result in the OP is obviously false).
 
  • #6
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Suppose $\phi : F\rightarrow F$ is an identity homomorphism, then $\ker(\phi)=\{0\}$ is an ideal of $F$ and hence $F/\{0\}\cong F$ by the first ring isomorphism.

Correct.
 
  • #7
STEMucator
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The OP means ##F/\{0\}##, which means the ring ##F## modulo ##\{0\}##. He does not mean the set theoretic difference ##F\setminus \{0\}## (in that case, the result in the OP is obviously false).

Oh I thought the OP was intending to mean the set theoretical difference.

I never knew \setminus was the way to do that, so I guess I learned something too.

The OP is correct if they were intending modulo, but I definitely would have been a bit more explicit when asking.
 
  • #8
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Oh I thought the OP was intending to mean the set theoretical difference.

I never knew \setminus was the way to do that, so I guess I learned something too.

The OP is correct if they were intending modulo, but I definitely would have been a bit more explicit when asking.

I agree, I was very confused too.
 

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