Question

If $F$is a field,$R$ a nonzero ring, and $f:F\to R$a surjective homomorphism, prove that$f$ is an isomorphism.

Short answer

It can be proved$f$ is an isomorphism.

Step-by-step solution

Results used

In reference to Exercise 10 in section 6.1,

if$I$ is an ideal in a field$F$ , then $I=\left(0_F\right)$or $I=F$.

Show kernel is an ideal 

$f:F\to R$is a surjective homomorphism.

Then the kernel is an ideal of $F$.

By the above-mentioned result,

kernel is either$\left(0\right)$ or$F$ .

If the kernel is 0

If the kernel is $\left(0\right)$, then $f$is injective.

Therefore, $f$is an isomorphism.

If the kernel is F

If the kernel is$F$ ,then$\theta$ carries every element to $\left(0_R\right)$

But $R$is a non-zero ring; therefore, this is not the case.

Then kernel must be$\left(0\right)$ .

Conclusion

$f$is an isomorphism.