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In this recent post the original question led people to look for rigid, noncommutative rings. (Rigid means that the only endomorphisms are zero and the identity). Several (somewhat complicated) examples were constructed, and Dejan Govc asked if the ring $R={\mathbb Z}\langle X,Y\rangle/(X^3-3,Y^3-5)$ is rigid.

Now we may look at the larger ring $R'={\mathbb Q}\langle X,Y\rangle /(X^3-3,Y^3-5)$. Let $U$ denote the multiplicative group of invertible elements in $R'$. Clearly ${\mathbb Q}^{*} \subseteq U$ and $X\in U$, $Y \in U$. Denote by $U'$ the multiplicative subgroup of $U$ generated by ${\mathbb Q}^{*},X$ and $Y$. Denote by $U''$ the set of elements of finite order in $U$.

I conjecture that $U''\subseteq U'$. It is easily seen that this implies that $R$ is indeed rigid.Are there general methods to compute $U$ and $U''$ ?

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Ewan Delanoy
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  • $R'$ contains a number field $\mathbb{Q}[X] / (X^3 - 3)$, which has signature $(1,1)$ and no additional roots of unity, and thus (its subring of integers) has unit group $\mathbb{Z} \oplus \mathbb{Z}/2$. I would be surprised if a fundamental unit of this number field is in $U'$. –  Apr 27 '12 at 18:14
  • @Hurkyl : you're right, I updated my post accordingly. – Ewan Delanoy Apr 27 '12 at 18:20
  • I'm not sure the tag "polynomials" is appropriate here, as noncommutative polynomials aren't what's usually meant by the word "polynomials". –  Apr 27 '12 at 18:39

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