Suppose $A$ and $K$ are rings with $f: A \to K$ a homomorphism.
Prove that for any $x \in a + \ker(f)$ we have $f(x)=f(a)$.
Im not sure how to start this, any help is appreciated!
2026-04-01 14:22:42.1775053362
Kernel of a homomorphism: $f(a)=f(x)$?
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Say $x = a + b$ for $b \in \text{Ker}(f)$. Then $f(x) = f(a+b) = f(a) + f(b) = f(a) + 0 = f(a)$.