Group-like elements in quotients of group rings $\DeclareMathOperator\Gr{Gr}$Let $R$ be a local ring, let $A$ be a finite abelian group, and let $I$ be a Hopf ideal of the ring $R[A]$. The quotient $R[A]\twoheadrightarrow R[A]/I$ induces a map on group-like elements $f\colon \Gr(R[A])\to \Gr(R[A]/I)$.
Is the map $f$ surjective?
The answer is yes if the order of $A$ is invertible in $R$, so let's assume it's not.
(By $R[A]$ I just mean the usual ring with generators $T_a$ for $a\in A$ and relations $T_0=1$ and $T_aT_b=T_{a+b}$).
 A: I found a counter-example, so the answer is no.
Let $B=\mathbb{F}_2[a,b,c]/J$ where $J=(a,b,c)^3+(ab+ac-bc)$ and put $s=ab$ and $t=ac$. Let $A=(\mathbb{Z}/2\mathbb{Z})^2$ and define $R = B[x_{10},x_{01},x_{11}]/J'$ where
$$
   J'=(x_{10}^2-s, x_{01}^2-t, x_{11}^2-(s+t), x_{10}x_{01}-ax_{11}, x_{10}x_{11}-bx_{01}, x_{01}x_{11}-cx_{10})\,.
$$
Then $R[A]\cong R[T_{10},T_{01}]/(T_{10}^2-1, T_{01}^2-1)$ with the usual Hopf-algebra structure. Define
$$
   I=(x_{10}(T_{10}-1), x_{01}(T_{01}-1), x_{11}(T_{10}T_{01}-1))\,.
$$
I claim that element $g=1+t(T_{10}-1)$ is not equal to 1 but group-like modulo $I$. To see that $g$ is group-like, we must show that $\varepsilon(g)=1$ and $\Delta(g)=g\otimes g$, where $\varepsilon$ denotes the counit and $\Delta$ denotes the coaction. It is clear that $\varepsilon(g)=1$ so we will show that $\Delta(g)=g\otimes g$. Since
$$
    T_{\lambda}-1=(T_{\lambda-\lambda'}-1)(T_{\lambda'}-1)+(T_{\lambda-\lambda'}-1)+(T_{\lambda'}-1)\,,
$$
we have $x_{\lambda'}(T_\lambda-1)=x_{\lambda'}(T_{\lambda-\lambda'}-1)$ modulo $I$ for all $\lambda,\lambda'\in A$.
This implies that, modulo $I\otimes R[A]+R[A]\otimes I$, we have
$$
    \begin{split}
        \Delta(g)-g\otimes g 
        & = \Delta(1+t(T_{10}-1))-(1+t(T_{10}-1))\otimes (1+t(T_{10}-1)) \\
        & = t(T_{10}-1)\otimes (T_{10}-1) \\
        & = (s+t)(T_{10}-1)\otimes (T_{10}-1) \\
        & = (s+t)(T_{01}-1)\otimes (T_{10}-1) \\
        & = t(T_{01}-1)\otimes (T_{10}-1)+(T_{01}-1)\otimes s(T_{10}-1) \\ 
        & = 0\,.
    \end{split}    
$$
Hence $g$ is a group-like element which is not the image of a group-like element in $R[A]$.
