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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.

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you need that $f$ is a homomorphism. –  Daniel Montealegre Jul 6 '12 at 7:45
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This 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. –  Marc van Leeuwen Jul 6 '12 at 9:43
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3 Answers

up vote 2 down vote accepted

$(\Rightarrow)$ 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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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?

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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 $y-x = 0$ $\iff$ $x=y$.

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