Homomorphism from field to ring

[Fantini]

Good afternoon! I wasn't able to get the necessary grade in abstract algebra and now I'm redoing many exercises and I would like some correction. All help is appreciated! (Smile)

Here is the question:

Show that every homomorphism of a field to a ring is one-to-one or null.

Let $\phi: F \to R$ be a homomorphism from a field $F$ to a ring $R$. Assume that $\phi \neq 0$. We have that $\phi(0_F) = 0_R$. Suppose that $\phi (a) = \phi(b)$. Since $F$ is a field, we will have $\phi(a-b) = 0_R$ if and only if $a-b=0$ and therefore $a=b$.

Cheers! (Yes)

 

[Deveno]

you're "just" missing it. what you want to PROVE is:

$\phi(x) = 0_R \iff x = 0_F$ if $\phi$ is not the 0-map.

you can't use this in the proof itself (the "if" part is easy).

now if $\phi(a) = \phi(b)$ then $\phi(a) - \phi(b) = 0_R$, and since $\phi$ is a ring-hmomorphism:

$\phi(a-b) = 0_R$ up to here, you're good. now we need to prove what i stated above.

well the set $\{x \in F: \phi(x) = 0_R\}$ is the kernel of $\phi$ and kernels are ideals (in rings).

but a field F only has two ideals, F and {0_F}. if the kernel is F, $\phi$ is the 0-map. since we are assuming $\phi$ is not the 0-map, the kernel is {0_F}. NOW we know that:

$a-b = 0_F$ so that $a = b$.

 

[Fantini]

I was close! Thanks Deveno! I will pay more attention next time. The kernel is a neat way to conclude the argument (don't know if it's the only one, but I definitely liked the idea).