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In Milnor's Introduction to Algebraic K-Theory, he writes on page 7:

Suppose that $\Lambda$ can be mapped homomorphically into a field or skew-field $F$... Then we obtain a homomorphism $j_*: K_0 \Lambda \rightarrow K_0 F \cong \mathbb{Z}$. In the commutative case, this homomorphism is clearly determined by the kernel of $j$ [where $j: \Lambda \rightarrow F$ is a homomorphism], which is a prime ideal in $\Lambda$.

I'm having trouble seeing how and why $j_*$ is determined by the kernel of $j$. I agree that ker $j$ is a prime ideal in $\Lambda$, and I know that $j_*$ is a group homomorphism which must send generators to generators, and since $\mathbb{Z}$ is cyclic, I can believe that $j_*$ will be completely determined by what values it sends to 1, but where's the connection between ker $j$ and $j_*^{-1}(1)$?

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Let $F_0$ be the field of fractions of $\Lambda/\ker j$. Then $j$ factors uniquely as a composite $\Lambda\to F_0\to F$, where the first map is canonical map. Since $F_0$ is a field, the induced map $K_0F_0\to K_0F$ can canonically be identified with the identity map $\mathbb{Z}\to\mathbb{Z}$. So $j_*$ can be identified with the map $K_0\Lambda\to K_0 F_0$, which depends only on $\ker j$.

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  • $\begingroup$ Thanks for the insight! To clarify, does the composite $\Lambda \rightarrow F_0 \rightarrow F$ do $x \mapsto \frac{x}{1} + \text{ker}(j) \mapsto j(x)$? $\endgroup$ – Sam Birns Jan 12 '16 at 4:30
  • $\begingroup$ Yes, that's right. $\endgroup$ – Eric Wofsey Jan 12 '16 at 4:33

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