9
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ Try proving the contrapositive. $\endgroup$ – Qiaochu Yuan Jul 16 at 8:28
  • $\begingroup$ @QiaochuYuan so, if a ring is not a power of $2$, then it still can have an odd number of units right? why should always have an even number of units? Should we assume commutativity of rings here? $\endgroup$ – vidyarthi Jul 16 at 8:46
  • 1
    $\begingroup$ No, commutativity is unnecessary. It is in fact true that if a finite ring has cardinality that's not a power of $2$ then it has an even number of units, and as a hint, the proof is short. $\endgroup$ – Qiaochu Yuan Jul 16 at 8:51
  • $\begingroup$ @Bernard well, here I assume the ring is finite. Edited the poat. $\endgroup$ – vidyarthi Jul 16 at 9:48
  • 1
    $\begingroup$ The early parts of this answer also answer your question. Not sure whether call this a duplicate or not. $\endgroup$ – Jyrki Lahtonen Jul 16 at 18:20
16
$\begingroup$

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 $2$.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ why cant the induced map have two elements in the image? $\endgroup$ – vidyarthi Jul 16 at 9:15
  • 1
    $\begingroup$ @vidyarthi: by hypothesis, the order of $R^{\times}$ is odd, so it cannot contain a subgroup of order $2$ (namely, the injective image of $\mathbb{Z}^{\times}$). $\endgroup$ – Alex Wertheim Jul 16 at 9:18
  • 1
    $\begingroup$ @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.) $\endgroup$ – Alex Wertheim Jul 16 at 9:50
  • 1
    $\begingroup$ @AlexWertheim Well, from a field to any nonzero ring! :^) $\endgroup$ – Stahl Jul 16 at 9:51
  • 1
    $\begingroup$ 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. $\endgroup$ – Alex Wertheim Jul 16 at 9:58

Not the answer you're looking for? Browse other questions tagged or ask your own question.