Is Mazur's deformation ring R integral?

Question

Consider the absolutely irreducible Galois representation $\overline{\rho} \colon G_{\Bbb Q} \to {\mathrm{GL}}_2({\Bbb F}_p)$. We apply the Mazur's deformation theory on the lift ${\rho} \colon G_{\Bbb Q} \to {\mathrm{GL}}_2({\Bbb Z}_p)$ of $\overline{\rho}$ such that

1. ${\mathrm{det}}\rho$ = cyclotomic character

2. Ramifying primes in ${\rho}$ are exactly the same ones that ramify in $\overline{\rho}$

3. $\rho$ is finite at $p$. I.e., the restriction of $\rho$ to ${\mathrm{G}}_{{\Bbb Q}_p}$ comes from the finite flat group scheme.

Mazur showed that there is a deformation ring $R$ finitely generated over ${\Bbb Z}_p$, and any such lift $\rho$ can be obtained from the universal Galois representation $\rho^{\mathrm{univ}} \colon G_{\Bbb Q} \to {\mathrm{GL}}_2(R)$ followed by the map ${\mathrm{GL}}_2(R) \to {\mathrm{GL}}_2({\Bbb Z}_p)$ induced by the unique ring homomorphism $\psi_{\rho} \colon R \twoheadrightarrow {\Bbb Z}_p$ corresponding to $\rho$.

Question: Assume that Mazur's deformation ring $R$ is reduced, then is $R$ necessarily integral?

Answer 1

First of all, your deformation problem is empty if $\det\bar{\rho}\neq\bar{\chi}_{cyc}$ so I am guessing that you are assuming this and likewise for the relevant restriction on $\bar{\rho}$ to $G_{\mathbb Q_{p}}$. If by $R$ integral you mean that $R$ is an integral domain, then the answer is surely no, though it might be computationally non-trivial to write down an explicit counter-example (universal deformation rings of modular forms of low level and weight tend to be domains). To see that it is very unlikely that $R$ is a domain, think about two congruent eigenforms whose residual representation satisfies all the hypotheses of your question and imagine the various possibilities.