Let $R$ be a finite ring such that the group of units of $R$, $U(R)$, has $34$ elements. I would like to find the characteristic of $R$.
Let $k:= \mathrm{Char}(R)$. If $\varphi$ denotes the Euler totient, then $\varphi(k)$ divides $34$, hence $k\in \{2,3,4,6\}$. I think it is possible to rule out some of these values, but I'm not sure how. And if we could rule them all out, even better, then there is no such ring. Any idea is appreciated.
There is no finite ring with $34$ units (I assume all rings are commutative, with $1$). First, recall the following fact about local rings: $\newcommand{\char}{\operatorname{char}}$ $\newcommand{\m}{\mathfrak{m}}$ $\newcommand{\Z}{\mathbb{Z}}$
Proposition: Let $(R,\m)$ be a local ring. Then $\char R = p^n$ for some prime $p$, $n \in \mathbb{N}$.
Proof (for completeness): If $\char(R/\m) = 0$, then $\char R = 0$. If $\char(R/\m) = p$ for some prime $p$, then under the unique ring map $i : \Z \to R$, we have $i(p) \in \m$, so $i^{-1}(\m)$ is a prime ideal containing $p$, thus $i^{-1}(\m) = (p)$, and there is an induced map $\overline{i} : \Z_{(p)} \to R$. As $\Z_{(p)}$ is a DVR, every ideal of $\Z_{(p)}$ is of the form $p^n\Z_{(p)}$, so for some $n$, $\Z_{(p)}/p^n\Z_{(p)} \hookrightarrow R$, hence $\char R = \char(\Z_{(p)}/p^n\Z_{(p)}) = p^n$ (since characteristic may be computed in any subring).
Returning to the problem, suppose $R$ is a finite ring with $34$ units, and that $|R|$ is minimal among all examples. Then $R$ is Artinian, hence $R \cong \prod_{i=1}^n R_i$ is a finite product of Artinian local rings. Then $34 = |R^\times| = \prod_{i=1}^n |R_i^\times|$. Ignoring factors with $|R_i^\times| = 1$, we may assume $|R_i^\times| \ge 2$ for each $i$, which implies $n \le 2$. If $n = 2$, then WLOG $|R_1^\times| = 17$, but no finite ring has $17$ units (if a finite ring has an odd number of units, that number must be a product of terms of the form $2^k - 1$, see e.g. the linked paper below). Thus $n = 1$, so $R$ itself is Artinian local, with maximal ideal $\m$. Then $|R| = |\m| + |R^\times| = |\m| + 34 \implies \dfrac{34}{|\m|} = |R/\m| - 1 \in \Z$, so $|\m| = 1, 2, 17, 34$.
If $|\m| = 1$, then $\m = 0$, $R$ is a field, and $|R| = 1 + 34 = 35$, but no field has $35$ elements.
If $|\m| = 2$, then $|R/\m| = 17 + 1 = 18$, but no field has $18$ elements.
If $|\m| = 17$, then $|R| = 51$, so the additive group of $R$ is a group of order $51$. The only such group is the cyclic group of order $51$, so $(R, +) \cong \Z/51\Z$, hence $\char R = 51$, which contradicts the proposition.
If $|\m| = 34$, then $|R| = 68$, $|R/\m| = 2$, so $2 \mid \char R$. But the additive group of $R$ is an abelian group of order $68$, which has an element of order $17$, so $17 \mid \char R$ also, which again contradicts the proposition.
For much more in this regard, see this paper.
