TITLE: Fermat's Last Theorem in the cyclotomic integers.
QUESTION [26 upvotes]: Kummer proved that there are no non-trivial solutions to the Fermat equation FLT(n): $x^n + y^n = z^n$ with $n > 2$ natural and $x,y,z$ elements of a regular cyclotomic ring of integers $K$.
I am looking for non-trivial solutions to the Fermat equation FLT(p) in the cyclotomic integer ring $\mathbb{Z}[\zeta_{p}]$ for irregular primes p or any information about how the solutions must be (as a step toward constructing them).
George Lowther pointed out in an earlier discussion that by Kolyvagin's criterion any solution in $\mathbb{Z}[\zeta_{37}]$ must be in the second case.

REPLY [2 votes]: Very late response but since it is still unresolved, I will answer your question. By Tauno's answer, any solution must belong to the second case, which in $\mathbb{Z}[\zeta]$ looks like $x^p+y^p=z^p$ with $1-\zeta \mid z$ and $x,y$ both coprime with $1-\zeta$. The second case also has no solutions in this ring by another criteria of Kolyvagin (from a different paper: " Fermat Equations over Cyclotomic Fields"). 
 The general criteria is a bit involved to write up here but the prime $p=37$ satisfies a simpler criteria (which applies to both the first and second case): 1) If the index of irregularity $=1$ with $p \mid B_i$ and 2) there is a prime $l \equiv 1 (\text{mod} \ p)$ for which $x^p+y^p=z^p$ has only trivial solutions modulo $l$ and $(1-\zeta(l))^{(l-3)}$, $(1-\zeta(l))^{(i)}$ are not $p$-th powers modulo $l$, then the Fermat equation has no solutions in $\mathbb{Q}(\zeta_p)$. Here $\zeta(l)$ is a primitive $p$-th root of unity modulo $l$ (which exists since $p \mid (l-1)$) and $a^{(j)}$ is the $j$-th component of $a \in \mathbb{F}_l^{\times}/(\mathbb{F}_{l}^{\times})^p$ relative to the Eigenspace decomposition of this group for the operators $\varepsilon_j = -\sum_{a=1}^{p-1}a^j\sigma_{a}^{-1}$. 
Since the index of irregularity for $37$ is $1$ with $37 \mid B_{32}$ and $x^{37}+y^{37}=z^{37}$ has only trivial solutions modulo $l=149=1+4\times 37$, then by taking $\zeta(149)=16$, we compute that $(1-16)^{(34)}$ and $(1-16)^{(32)}$ are not $37$-th powers modulo $149$. Therefore $p=37$ satisfies the criteria.