Why $0$ must be in the kernel of a ring?

Question

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.

Answer 1

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.

Answer 2

$f(0_R) = f(0_R + 0_R) = f(0_R) + f(0_R)$. 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)

Answer 3

For all $r \in R$, $f(r) = f(0 + r) = f(0) + f(r)$ which implies that $f(0) = 0$.