Is F Isomorphic to Its Own Quotient by {0}?

[HaLAA]

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

 

[STEMucator]

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.

 

[HaLAA]

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.

 

[STEMucator]

an identity homomorphism

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

 

[micromass]

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).

 

[micromass]

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.

 

[STEMucator]

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.

 

[micromass]

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.