Let $\mathcal{O}_L$ be the ring of integers in a number field $L$. Let $K$ be an algebraically closed field of characteristic zero, and let $f:\mathcal{O}_L \to K$ be a ring homomorphism. Why must $f$ be injective?
I know the following: if $\alpha \in \mathcal{O}_L$ has minimal monic irreducible polynomial $h_\alpha(x) \in \mathbb{Z}[x]$, then $f(\alpha)$ is also a root of $h_\alpha(x)$. That is, $f$ preserves Galois orbits of $\operatorname{Gal}(L/\mathbb{Q})$ acting on $\mathcal{O}_L$. That is, $f(\alpha) = \sigma(\alpha)$ for some $\sigma \in \operatorname{Gal}(L/\mathbb{Q})$. However, if $\alpha, \beta \in \mathcal{O}_L$ with $f(\alpha) = f(\beta)$, I don't know how to conclude that $f(\alpha) = \sigma(\alpha)$ and $f(\beta) = \sigma(\beta)$ for the same Galois automorphism $\sigma$. If I could do that, I could apply $\sigma^{-1}$ to both and conclude that $\alpha = \beta$, but I don't see how this works.
Best Answer
Every nonzero ideal of $\mathcal{O}_L$ contains an integer (if $a\in\mathcal{O}_L$ is nonzero, then its minimal polynomial has nonzero constant term and that constant term is in the ideal generated by $a$). So, if the kernel of $f$ were nontrivial, it would contain a nonzero integer, but this is impossible since every nonzero integer remains nonzero in $K$ since $K$ has characteristic $0$.