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Source

Theorem 6.10

Let $f : R → S$ be any homomorphism of rings and let $K = \ker f$. Then $K$ is an ideal in $R$.

Proof. We know $0 ∈ K$, so $K \neq ∅$. Let $a, b ∈ K$. Then $f(a) = f(b) = 0$, so $f(a − b) = f(a) − f(b) = 0$. For any $r ∈ R,$ $f(ra) = f(r)f(a) = f(r) · 0 = 0$. Similarly, $f(ar) = f(a)f(r) = 0$. Thus $a − b, ra$ and $ar$ are also in $K$, hence $K$ is an ideal.

I'm confusing why do we have $0 ∈ K$ at the beginning. Homomorphism doesn't give that. Kernel only contains elements such that $f(r)=0$, but not $0$ itself.

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  • $\begingroup$ If $f(0)\neq 0$, there will be contradiction. $\endgroup$ – W. mu Jun 15 '18 at 3:25
  • $\begingroup$ But why $f(0)$ must be $0$?. Do I miss some definition? $\endgroup$ – user8314628 Jun 15 '18 at 3:29
  • $\begingroup$ Since $f$ is a ring homomorphism, it is a group homomorphism between $R$ and $S$ considered as groups under addition. Such homomorphisms send the first group's identity to the second group's identity. $\endgroup$ – Jonathan Dunay Jun 15 '18 at 3:33
  • $\begingroup$ @JonathanDunay: We have $f(1_R) = f(1_R^2) = (f(1_R))^2 \Longrightarrow (f(1_R) - 1_S)f(1_R) = 0$; if $S$ has no zero divisors, we can conclude $f(1_R) = 1_S$ unless $f(1_R) = 0$, which if $f$ is non-trivial we may rule out since it implies $f(r) = f(r 1_R) = f(r) f(1_R) = 0$; if $S$ does have divisors of zero, it may not be true that $f(1_R) = 1_S$; but $f(1_R)$ is idempotent, and will be the unit on $f(R)$. $\endgroup$ – Robert Lewis Jun 15 '18 at 3:48
  • $\begingroup$ @RobertLewis When you consider the ring as a group under addition, the group identity is the additive identity of the ring $0$ not the multiplicative identity $1$ $\endgroup$ – Jonathan Dunay Jun 15 '18 at 3:50
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Homomorphism $f:R \to S$ does in fact give

$f(0_R) = 0_S, \tag 1$

since

$f(a + b) = f(a) + f(b), \; a, b \in R; \tag 2$

thus

$f(0_R) + f(0_R) = f(0_R + 0_R) = f(0_R), \tag 3$

whence

$f(0_R) = f(0_R) - f(0_R) = 0_S; \tag 4$

(4) holds no matter what $\ker R$ may be; but if we have $0_R \ne a \in \ker R$, then we may also write

$f(0_R) = f(a - a) = f(a) - f(a) = 0_S, \tag 5$

which I guess gives another way to see that (1) must bind.

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$f(0_R) = f(0_R + 0_R) = f(0_R) + f(0_R)$. Therefore, $f(0_R) + 0_S = f(0_R) + f(0_R)$. Thus $0_S = f(0_R)$.

Ring homomorphisms will always preserve additive identities (and multiplicative identities if it is a ring with unit)

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    $\begingroup$ See my comment to user8314628's answer in response to JonathanDunay's comment: $f(1_R) \ne 1_S$ necessarily. Cheers! $\endgroup$ – Robert Lewis Jun 15 '18 at 4:06
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    $\begingroup$ @RobertLewis ah thank you. pesky non-unitary rings $\endgroup$ – Good Morning Captain Jun 15 '18 at 4:23
  • $\begingroup$ Indeed they are pesky! Vermin! Cheers! $\endgroup$ – Robert Lewis Jun 15 '18 at 4:25
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For all $r \in R$, $f(r) = f(0 + r) = f(0) + f(r)$ which implies that $f(0) = 0$.

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