# Is there a finite ring whose rank is smaller than the rank of its group and its monoid?

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?

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).

• 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. – Eric Wofsey Feb 10 '16 at 23:23

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.

• Can you clarify what "separating coordinates" means? – argentpepper Feb 11 '16 at 15:48
• 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 $s_i\neq s_j$. – Eric Wofsey Feb 11 '16 at 19:56