Let $\ell,p$ be two distinct primes. Let $K$ be a finite extension of $\mathbb Q_{\ell}$ with residue field $k$ of size $\# k\equiv 1\pmod p$. Let $\mathcal O$ be the ring of integers of a finite extension of $\mathbb Q_p$ with residue field $\mathbb F$.
Let $\overline \rho:G_K\to GL_2(\mathbb F)$ be an unramified representation of the absolute Galois group of $K$. Suppose $\overline\rho(Frob_K)$ has distinct eigenvalues in $\mathbb F$. Let $\chi:G_K\to \mathcal O^\times$ be an unramified character with $\overline \chi=\det\overline\rho\in \mathbb F^\times$. According to p.20 of https://www.ma.imperial.ac.uk/~tsg/Index_files/ArizonaWinterSchool2013.pdf, the universal lifting ring $R^\Box_{\overline\rho,\chi}$ of $\overline\rho$ with determinant $\chi$ (in the category of complete Noetherian local $\mathcal O$-algebra with residue field $\mathbb F$) can be taken to be $\mathcal O[[x,y,B,u]]/((1+u)^{p^m}-1)$, where $m=v_p(\#k-1)$.
However, why isn't $\mathcal O[[x,y,B,u]]/((1+u)^{\#k-1}-1)$ the universal lifting ring? While I think I understand the proof on p.31 of the note above, it seems to me that the argument there works for this ring too.
If $u$ is mapped to zero in $\mathbb{F}$, then $(1+u)^{p^m}-1$ and $(1+u)^{|k|-1}-1$ are associated in the ring $\mathcal{O}[[u]]$ – and therefore the two rings are the same.
Proof: first, $(1+u)^{|k|-1}-1$ is a multiple of $(1+u)^{p^m}-1$.
Write $|k|-1=p^mr$: the quotient is $q=\sum_{i=0}^{r-1}{(1+u)^{p^mi}}$. In particular, $q$ maps to $r$ in $\mathbb{F}$, so $q \in \mathcal{O}[[u]]^{\times}$, QED.