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.

  • $\begingroup$ Are you asking how to prove that every ring homomorphism from a commutative ring with identity is injective? Because that is not true. For example consider evaluation at $1$ from $\mathbb{k}[x]\to \mathbb{k}$. $\endgroup$ – Seth Mar 28 '14 at 15:37
  • $\begingroup$ Yes, I am asking to prove. With the criteria given that Ø: f -> s be a ring homomorphism where f is a field, S is a commutative ring with 1 and Ø(1) = 1. $\endgroup$ – user1413 Mar 28 '14 at 15:38
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    $\begingroup$ Please use $\LaTeX$ in the future. Also I don't understand -- what are you asking exactly? The proof you write down is already correct. $\endgroup$ – Najib Idrissi Mar 28 '14 at 15:42
  • $\begingroup$ @nik my question is f -> s be a ring homomorphism where f is a field, S is a commutative ring with 1 and Ø(1) = 1. My proof is not exactly correct. It is half correct I have to prove via Ø(1) = 1. $\endgroup$ – user1413 Mar 28 '14 at 16:05

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).

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  • $\begingroup$ Thank You seth. But if I wanted to write this as 'Ø(x) = Ø(x1)= Ø(x)Ø(1)=Ø(x)0=0' then how will I write which fulfills my requirement $\endgroup$ – user1413 Mar 28 '14 at 15:48

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