Reference: Ring Homomorphism $\phi:f\to S$ is injective
Referring this I have a doubt which I needed to clear. Below is my answer and query.
We know $\phi:f\to S$ be ring homomorphism, where $f$ is a field and $R$ is a ring ($S$ itself could be a ring). For all $x \in f$. The field $f$ has a unity of $1$. Suppose that $1 \in \ker \phi$. Then $\phi(1)=0$. For any $x\in f$. We use fact that $1$ is identity for multiplication in $f$ & the fact that $\phi$ is assumed to be a homomorphism to get
$$\phi(x) = \phi(x1)= \phi(x)\phi(1)=\phi(x)0=0$$
Therefore $\phi$ maps into $0$. If $1\not\in\ker \phi$. We know that if $\phi(x)=0$ for $x\neq 0$, then
$$\phi(1) = \phi(xx^{-1})= \phi(x)\phi(x^{-1})= 0\phi(x^{-1})=0$$
contradicting our assumption that $1\not\in\ker\phi$. So $x$ must be $0$. Therefore $\ker\phi=0$
I have read this in stackexchange. Referring to my above link and also to my answer. How will I prove if I there is slight change in the question that $S$ is a commutative ring with $1$ and $\phi(1) = 1$
Then how to prove that $\phi$ is injective.
Best Answer
The best way to understand why ring homomorphisms out of fields are injective is to consider the kernel as an ideal. Fields only have two ideals, the zero ideal and the whole field. Thus every ring homomorphism from a field is either zero or injective. In your case you are assuming it preserves multiplicative identity so it can't be zero (assuming $1 \neq 0$ in codomain).