# Show F is isomorphic to F/{0}

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1. May 27, 2015

### HaLAA

1. The problem statement, all variables and given/known data
Let F be a field. Show that F is isomorphic to F/{0}
2. Relevant equations

3. 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

2. May 27, 2015

### Zondrina

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. May 28, 2015

### HaLAA

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. May 29, 2015

### Zondrina

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. May 29, 2015

### micromass

Staff Emeritus
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. May 29, 2015

### micromass

Staff Emeritus
Correct.

7. May 30, 2015

### Zondrina

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. May 30, 2015

### micromass

Staff Emeritus
I agree, I was very confused too.