When $\mathbb Z$ is a direct summand of $K_0(R)$

Question

Suppose that there is a ring homomorphism $R \to F$ where $F$ is a field.

I'm trying to verify that $\mathbb Z$ is a direct summand of $K_0(R)$.

We have an induced ring homomorphism $(K_0(R), \oplus, \otimes_R) \to K_0(F) \cong \mathbb Z$ such that $[R] \mapsto 1$. Also there is an injection $\mathbb Z \to K_0(R)$. But I am missing how this implies what I want. Any help is appreciated.

Answer 1

I guess the explanation is what follows :

Call $p$ the group homomorphism $K_0(R) \to \mathbb{Z}$. Then, calling $K=Ker(p)$, you have an exact sequence of abelian groups $$0 \to K \to K_0(R) \to \mathbb{Z} \to 0$$ which admits a section $\mathbb{Z} \to K_0(R)$, hence $K_0(R)= \mathbb{Z} \oplus K$.