Ring Homomorphism Question

Question

Let $f:F_1 → F_2$ be a ring homomorphism between fields $F_1, F_2$.
(a) Show that if $f(1)=0$ then $f=0$.
(b) Show that if $f(1)\ne0$ then $f$ is injective.

Hi, I'm not too sure how to do this question. Any help would be greatly appreciated :)

Answer 1

All ring homomorphisms send the identity of the first ring to the identity of the second ring. Since $f(1)=0$, we know that 0 is the both the multiplicative identity and the additive identity of the field $F_2$. That implies that $F_2$ is the trivial field of one element so $f$ maps all elements of $F_1$ to $0$.\

The second part is answered here.

Answer 2

1. Suppose that $f(1) = 0,$ and let $x\in F_1.$ Then $$f(x) = f(1\cdot x) = f(1)\cdot f(x) = 0\cdot f(x) = 0,$$ so that $f$ is identically $0.$
2. Suppose that $f(1)\neq 0,$ and let $x\in F_1^\times.$ We wish to show that $f(x)\in F_2^\times,$ so that the only element of $F_1$ mapped to $0$ by $f$ is $0,$ which will imply that $f$ is injective. Observe that $$ f(1) = f(x\cdot x^{-1}) = f(x)\cdot f(x)^{-1}. $$ It follows that $f(x)\neq 0,$ as $f(1)\neq 0.$

A much quicker proof of both these results is indicated by azif00's comment: the kernel of any ring homomorphism (unital or nonunital) is an ideal of the domain. The only ideals of a field are $(0)$ and the entire field itself. We have $1\in\ker f$ if and only if $f(1) = 0,$ so that $f(1) = 0$ if and only if $\ker f = F_1,$ which implies that $\ker f = (0)$ (and hence $f$ is injective) if $f(1)\neq 0.$