The Theorem: Suppose $K$ is a field, $\alpha\in K$ and $ev_{\alpha}: K[x]\to K$ is defined by $g(x)\mapsto g(\alpha)$. Then $\ker(ev_{\alpha})$ is a nonzero prime ideal iff $K[\alpha]=ev_{\alpha}(K[x])$ is a field.
My Question: The $\implies$ direction is straightforward, since every nonzero prime ideal in a field is maximal, and hence $K[x]/\ker(ev_{alpha})\cong K[\alpha]$ is a field. For the other direction, I don't understand why $K[x]$ is a field would imply that $\ker(ev_{\alpha})$ is a NONZERO ideal-any help would be greatly appreciated.
If $\mathrm{ker}(\mathrm{ev}_\alpha)$ is the zero ideal, then $K[\alpha]\cong K[x]/(0)\cong K[x]$. But $K[x]$ is not a field, since, for example, $x$ has no multiplicative inverse.