How can we show that a ring map $f: R\rightarrow S$ is one-to-one iff $\ker(f)=\{0\}$? I have seen this for a while axiomatically so I am unsure of my rigor.
Ring map one-to one kernel-proof
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2This does not really have to do with rings. If $f$ is just a homomorphism of (additive) groups one already has $f~\mathrm{injective}\iff \ker(f)=\{0\}$; the part $\implies$ is immediate from the definition of injective (and homomorphism) and $f(x)=f(y)\implies f(x-y)=0$ will give the opposite implication. – 2012-07-06
3 Answers
$(\Rightarrow)$ If $f$ is injective, then $\text{ker}(f) = \{0\}$ since $f(0) = 0$ for all homomorphisms and injectivity.
$(\Leftarrow)$ Suppose $f$ is not injective. Then there exists $x \neq y$ in $R$ such that $f(x) = f(y)$. Since $x \neq y$, $x - y \neq 0$. Then $f(x - y) = f(x) - f(y) = 0$. Hence $x - y \in \text{ker}(f)$. So $\text{ker}(f) \neq \{0\}$.
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0I mean, why don't we define the kernel of a field homomorphism $f$ as $ker(f):=\{r\in R: f(r)=\text{Identity of multiplication}\}$? This is exactly the way to define the kernel of group homomorphisms. – 2018-09-21
Claim: $\ker f = \{0\}$ $\iff$ $f$ is injective
Proof:
$\Longleftarrow$: Assume $\ker f \neq \{0\}$. Then there exist $x \neq y$ such that $f(x) = 0 = f(y)$ hence $f$ is not injective.
$\implies$: Assume $\ker f = \{0\}$. Let $f(x) = f(y)$. Then $f(y) - f(x) = f(y-x) = 0$. And by assumption, $y-x = 0$. But $y-x = 0$ $\iff$ $x=y$.
Some potential help:
$f(v)=f(w)\iff f(v)-f(w)=0\iff f(v-w)=0\iff v-w\in \mathrm{Ker}\;f.$
Is it possible for $v,w$ to be distinct if the kernel is trivial? And if it's nontrivial?