Finite rings

**TheAbstractionist** · August 7th 2009, 04:49 AM

Given a ring $R$ with multiplicative identity, let $U(R)$ denote the set of its units.

What is the smallest positive integer $n$ such that there does not exist any finite ring such that $U(R)$ contains exactly $n$ elements?

**Swlabr** · August 7th 2009, 06:58 AM

Originally Posted by TheAbstractionist

Given a ring $R$ with multiplicative identity, let $U(R)$ denote the set of its units.

What is the smallest positive integer $n$ such that there does not exist any finite ring such that $U(R)$ contains exactly $n$ elements?

Emm...I think you natural number, not integer. As there exists no ring with identity such that the set of units has 0 elements, or -1 elements, etc...

**TheAbstractionist** · August 7th 2009, 07:33 AM

I did say positive integer. Isn’t a positive integer a natural number?

**Swlabr** · August 7th 2009, 07:39 AM

Originally Posted by TheAbstractionist

I did say positive integer. Isn’t a positive integer a natural number?

Touché - I just read what I wanted to read...my bad.

**NonCommAlg** · August 7th 2009, 12:14 PM

Originally Posted by TheAbstractionist

Given a ring $R$ with multiplicative identity, let $U(R)$ denote the set of its units.

What is the smallest positive integer $n$ such that there does not exist any finite ring such that $U(R)$ contains exactly $n$ elements?

this is a quite popular question is ring theory i guess. there's a solution which uses the structure theory of rings:

the answer is $n=5.$ you don't need to assume $R$ is finite. it's clear that for $n \leq 4$ there exist rings with $|U(R)|=n.$ now let $R$ be a ring and $U(R)=G.$ suppose $|G|=5.$ then $2=0$ in $R,$ because
otherwise $\{1,-1 \}$ would be a subgroup of $G,$ which is impossible because $|G|$ is odd. now look at the group algebra $S=\mathbb{F}_2[G].$ by Maschke's theorem $S$ is semisimple and thus by Wedderburn-Artin

theorem $S \cong M_{k_1}(D_1) \times M_{k_2}(D_2) \times \cdots M_{k_r}(D_r),$ where each $D_i$ is a division ring containing $\mathbb{F}_2.$ (here $M_t(D)$ means the ring of $t \times t$ matrices with entries from $D.$ ) since $S$ is finite, each $D_i$ is

finite. but we know that a finite division ring is a field. so each $D_i$ is a finite field extension of $\mathbb{F}_2,$ i.e. $|D_i|=2^{n_i}=N_i,$ for some $n_i \geq 1.$

also if $k_i > 1,$ for some $i,$ then $|U(M_{k_i}(D_i)|=(N_i^{k_i}-1)(N_i^{k_i}-N_i) \cdots (N_i^{k_i}-N_i^{k_i - 1}),$ which is odd only for $k_i=1.$ thus $S \cong D_1 \times D_2 \times \cdots D_r.$ but then $G=U(S) \cong D_1^{\times} \times D_2^{\times} \times \cdots \times D_r^{\times}$ and

hence: $5=|G|=(2^{n_1}-1)(2^{n_2}-1) \cdots (2^{n_r}-1),$ which is clearly impossible. Q.E.D.

Remark: the above solution uses some powerful theorems in ring theory and so you should expect to get a lot more than just answering your question out of it. so ... see if you can do that!

**TheAbstractionist** · August 8th 2009, 07:58 AM

Briliant!

Over in the AoPS/MathLinks forum at the moment, some people are trying to determine if there exists a finite ring with unity $R$ such that $|U(R)|=38.$ The latest development appears to be that if such a ring exists, it must have characteristic 4. It looks exciting – I wish I knew more ring theory so I could join in and not be left out of the fun.

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