Clearly k[x]⊆Z(R)k[x] \subseteq Z(R)k[x]⊆Z(R). Conversely, if f=∑aixi∈Z(R)f = \sum a_{i}x^{i} \in Z(R)f=∑aixi∈Z(R), then fa=affa = affa=af for all a∈Ka \in Ka∈K shows that each a−i∈Z(K)=ka_{-}i \in Z(K) = ka−i∈Z(K)=k, and hence f∈k[x]f \in k[x]f∈k[x].
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