We verify the three laws for Ker(f) to be an ideal:

0 Ker(f).

If u, u' Ker(f), then u + u' Ker(f).

If r R and u Ker(f), then ru Ker(f).

Clearly, f(0) = 0, so 0 Ker(f).

Suppose u, v Ker(f). Then

f(u + v) = f(u) + f(v) = 0 + 0 = 0,

so u + v Ker(f).

Suppose r R and u Ker(f). Then

f(ru) = f(r) f(u) = f(r) 0 = 0,

so ru Ker(f).