10

If the number of units of a finite ring is odd, then does the ring has cardinality as a power of $2$?

I think yes. For fields, it is trivial. For non-fields, it is a hard question for me. I saw a paper here that sates that an odd number is the cardinality of the group of units of a ring if it is of the form $\prod_i (2^{n_i}-1)$. But, that proof is quite lengthy, and still the ring need not be a power of $2$. Any short proof? Thanks beforehand.

vidyarthi
  • 7,315

1 Answers1

18

Consider the canonical ring morphism $\varphi \colon \mathbb{Z} \to R$. Since $\mathbb{Z}^{\times} = \{-1, 1\}$, the induced group morphism $\mathbb{Z}^{\times} \to R^{\times}$ must be trivial by Lagrange, so $\varphi(1) = \varphi(-1) = 1$. In particular, $\varphi$ factors through an injective morphism $\mathbb{F}_{2} \to R$, so $R$ is an $\mathbb{F}_{2}$-vector space, and thus must have cardinality a power of $2$.

Alex Wertheim
  • 20,788
  • why cant the induced map have two elements in the image? – vidyarthi Jul 16 '20 at 09:15
  • 2
    @vidyarthi: by hypothesis, the order of $R^{\times}$ is odd, so it cannot contain a subgroup of order $2$ (namely, the injective image of $\mathbb{Z}^{\times}$). – Alex Wertheim Jul 16 '20 at 09:18
  • The part " $\varphi$ factor through an injective morphism from $\mathbb{F}_2\to R$" is not sufficiently clear to me, though – vidyarthi Jul 16 '20 at 09:23
  • @vidyarthi: if $R$ is the zero ring, then the original claim is evident. If $R$ is nonzero, then the kernel $\varphi$ contains $2$ and is not all of $\mathbb{Z}$, and so must be $2\mathbb{Z}$. – Alex Wertheim Jul 16 '20 at 09:36
  • no, I got that the kernel is $\mathbb{Z}_2$. But how is it injective, it is certainly not injective right? – vidyarthi Jul 16 '20 at 09:41
  • 1
    @vidyarthi: For any ring homomorphism $\alpha \colon S \to T$, the canonical induced map $S/\mathrm{ker}(\alpha) \to T$ is always injective. (But as a cheap aside: a (unital) ring morphism from a field to any ring is always injective.) – Alex Wertheim Jul 16 '20 at 09:50
  • 1
    @AlexWertheim Well, from a field to any nonzero ring! :^) – Stahl Jul 16 '20 at 09:51
  • @AlexWertheim sorry, but is not $\varphi(-1)=\varphi(1)$ here? – vidyarthi Jul 16 '20 at 09:54
  • 1
    Stahl: haha, yes, careless of me! @vidyarthi: yes, $\varphi(-1) = \varphi(1)$. But $1$ and $-1$ are identified in the quotient $\mathbb{F}{2} = \mathbb{Z}/2\mathbb{Z}$, so the induced map $\mathbb{F}{2} \to R$ is injective. – Alex Wertheim Jul 16 '20 at 09:58
  • Could you elaborate on what is meant by "$\varphi$ factors through an injective morphism" and how that allows us to conclude $R$ is an $\mathbb{F}_2$-vector space? – Karambwan Nov 12 '22 at 15:41
  • @AlexWertheim: sorry, I forgot to tag you in my previous comment! – Karambwan Nov 13 '22 at 02:29