= Primitive-element kernel rigidity lemma
{title2=$\ker\alpha=\ker\beta$}
If primitive elements have equal restrictions along $\alpha:P\to m$ and $\beta:P\to n$, then these surjections have equal kernels. To rule out $\alpha(a)=\alpha(b)$ with unequal $\beta$ images, identify those two images by $q:n\to n-1$. The set of pairs equal under $\alpha$ and $q\beta$ maps surjectively both to $P$ by projection and to the kernel pair of $q$ by applying $\beta$. Injective restrictions force the kernel-pair matching condition on the primitive element, contradicting descent along $q$.
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