9
$\begingroup$

Consider a finite ring $(R, +, \times)$ comprising a finite additive abelian group $(R, +)$, a finite multiplicative monoid $(R, \times)$, and a distributivity rule relating the two. Let the rank of the ring be the minimum size of a set that generates the ring (that is, a set $S$ of ring elements such that the closure of $S$ under both addition and multiplication is $R$). The rank of this ring is bounded above by the smaller of the rank of the additive group and the rank of the multiplicative monoid, since a generating set for either the group or the monoid is a generating set for the ring.

Is there a finite ring whose rank is strictly less than both the rank of its group and the rank of its monoid? In other words, is there a ring in which the combination of both addition and multiplication allows us to generate more elements than using either operation individually?

$\endgroup$
6
+50
$\begingroup$

Yes, there is. An example is $R=\mathbb{Z}_2\times\mathbb{Z}_2\times\mathbb{Z}_2$.

Indeed, the rank of $R$ as a ring is $2$: it is generated by $(1,1,0)$ and $(0,1,1)$. On the other hand, its rank as an abelian group is $3$.

It remains to be shown that the rank of the multiplicative monoid of $R$ is at least $3$. Note that a generating set of this monoid has to contain $(1,1,1)$. This element is not sufficient for generating $R$; nor is any set of the form $\{(1,1,1),(a,b,c)\}$, since this set only generates itself as a monoid. Thus the rank of the multiplicative monoid of $R$ is at least $3$.

In fact, I think the rank of $R$ as a multiplicative monoid is $4$, with generating set $\{(1,1,1), (1,1,0), (1,0,1), (0,1,1)\}$. In any case, the above argument shows that $R$ is such that its rank as a ring is strictly smaller than both the rank of its abelian group and that of its monoid.


Edit: As per Eric Wofsey's comment, if by the rank of a monoid $R$ one means the smallest cardinal of a subset $S$ such that $R$ is the smallest submonoid containing $S$, then the rank of the monoid in my example above is $3$, not $4$ (since $(1,1,1)$ will be automatically contained in any submonoid).

$\endgroup$
  • 1
    $\begingroup$ You don't need to include $(1,1,1)$ in the generators as a multiplicative monoid, since the unit is part of the structure of a monoid. $\endgroup$ – Eric Wofsey Feb 10 '16 at 23:23
3
$\begingroup$

Let me generalize Pierre-Guy Plamondon's example. Let $p$ be a prime and let $R=\mathbb{F}_p^n$. Then the additive rank of $R$ is $n$. If $x\in R$ has exactly one coordinate which is $0$ and $x=yz$, then at least one of $y$ and $z$ must have that same coordinate as its only coordinate which is $0$. Thus the multiplicative rank is at least $n$, since any generating set must contain an element which has exactly one coordinate which is zero, for each possible coordinate. (For $p=2$ the rank is exactly $n$; for $p>2$ it will be $2n$, since the group of units has rank $n$, and any set of multiplicative generators must also contain a set of generators for the units.) But I claim the rank of $R$ as a ring is only $\lceil\log_p n\rceil$. For each $i<\lceil\log_p n\rceil$ let $x_i\in R$ be the element whose $k$th coordinate is the $i$th base $p$ digit of $k$. Then the set $\{x_i:0\leq i<\lceil\log_p n\rceil\}$ separates coordinates, and it follows that it generates $R$ as a ring (in general, if $K$ is a field, then (unital) $K$-subalgebras of $K^n$ are determined by the coordinates they separate). Conversely, it is easy to see that no subset of $R$ of cardinality less than $\lceil\log_p n\rceil$ can separate coordinates, and thus no smaller set can generate the ring.

In particular, taking $p\geq n$, this gives rings which are generated by a single element as a ring but require arbitrarily large numbers of elements to generate either additively or multiplicatively.

$\endgroup$
  • $\begingroup$ Can you clarify what "separating coordinates" means? $\endgroup$ – argentpepper Feb 11 '16 at 15:48
  • $\begingroup$ A subset $S\subseteq R^n$ separates coordinates if for each $i,j\in\{1,\dots,n\}$ with $i\neq j$, there exists $s=(s_1,\dots,s_n)\in S$ such that $s_i\neq s_j$. $\endgroup$ – Eric Wofsey Feb 11 '16 at 19:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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