page1883 Rings Matthias Lorentzen...mattegrisenforlag.com
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Question
Let us show that `x + ker (phi) sube [x]`. Given a homomrphism `phi: A rightarrow B`, we define an equivalence relation ~ on A by: `x ~ y Leftrightarrow phi(x) = phi(y)`. Let `y in x + ker(phi)`. Then y = x + z for some `z in ker(phi)`. `phi(y) = phi(x + z) = phi(x) + phi(z) =` ?
