# How many rings exist when ring is subspace of finite dimentional vector space?

Suppose we have ring $R[M]$ over monoid $M$ in real number $R$.The number of generators for the monoid, is finite. Now suppose that every ring element $r$ has decomposition in finite linear basis of n-elements. So ring is subspace of n-dimensional vector space.

How many rings, depending on n has this property? Could You give me some references ( books, preprints, articles) when such theorem is stated or proved?

Motivation: Here monoid ring and some structure within it - how is it called? in a comment to the question, Scott Carnahan wrote:

The existence of the decomposition of elements of R[M] implies the ring is a subspace of a 4-dimensional real vector space. There are many such rings, but only finitely many monoid rings of dimension less than 5.

Clarification of the problem:

I have finitely presented noncommutative monoid with unity and two generators $g_1,g_2$: M = F/Rel where $Rel = \{g_1^2 = e , g_2^2 = g_2 \}$, $e$ is unit element and $F$ is free monoid over two generators. Because of relations $Rel$ every elemet in monoid has form for example $g = g_1g_2g_1g_2...g_1g_2$ ( alternating finite sequence with subscripts 1212... or 2121...). Different monoid elements contains different number of multiplications. It is very simple although infinite multiplicative structure.

Then I consider monoid ring over reals $R[M]$. Every element in $R[M]$ has form:

(1) $t = r_1g_1 + r_2g_2 r_3g_1g_2+r_4g_2g_1+ r_5g_1g_2g_1 + ...+r_p g_ig_kg_i...g_s+ ...$ and so on. $r_i \in R$ and $g_i \in M$.

Note that in general monomial element $g_ig_k...g_s$ every subscript has value in $\{1,2\}$ and no two following each other subscripts are the same ( they alternate like in sequence like $1212..$ or $2121..$. Of course this is standard ring definition.

In structure, I would like to describe You here, I have strange additional property: there is element $g_3$ in ring $R[M]$ ( but it is not monoid element!) which allows following decomposition:

For every $r \in R[M]$ we have

$r= r_0 e + r_1 g_1 +r_2 g_2 +r_3 g_3$

Look: there are only four terms in decomposition, even if You decompose general ring element in the form of (1). However after such decomposition I may only multiply such elements and not add them. In some way it looks like vector space. From the other side such decomposition may be treated as another monoid. So in fact decomposition as above, I trying to treat as some kind of "parametrization" of ring elements. Indeed it is element of some vector space, but there are further requirements on coefficients $r_i$ ( which I do not need to describe here, they in fact are part of monoid definition).

As far as I know this is not standard ring property - maybe I am wrong. If I think about for example polynomial ring (that in simple case is real ring over multiplicative monoid generated by one generator $x$) such decomposition is not possible.

• Is your question is equivalent to asking: how many monoids are there which have $n$ elements? – Mariano Suárez-Álvarez Feb 8 '10 at 15:40
• @Mariano: probably not! But in fact I do not know... – kakaz Feb 8 '10 at 18:44

## 1 Answer

Your $M$ is infinite, so the ring monoid $R[M]$ is infinite dimensional. In particular, there does not exist an element $g_3$ with the property you state, for otherwise $R[M]$ would be simultaneously 4 dimensional.

• So there are two possibilities: I made mistake in calculus on 2x2 matrices ( performed within Sage software), or I miss some important elements of my construction to describe here. Does it has any influence that decomposition do not agree with ring structure? That is $r+r'$ ( regarded as decomposed) is not ring element whilst r*r' is? I do not want to describe the whole big picture of my construction because it is rather reach structure over 2x2 matrix monoid and may be not interesting for anyone... – kakaz Feb 8 '10 at 19:21
• @kakaz, it is impossible to tell which of the two options holds, becaus eyou have not given us any information. In any case, an infinite dimensional vector space is not a 4-dimensional vector space. – Mariano Suárez-Álvarez Feb 8 '10 at 19:46
• Yes Yu are right, and I am sorry for that. But details are too complicated to write it here. I have monoid M of 2x2 matrices with $det=1$ and this gives some nonlinear constraints on $r_i$ coefficients. So even if decomposition exists and it reflects some ring structure ( multiplicative one) it is not any type of "ring of 4 generators". Probably it would be better to say that I found function from $R[M]$ to another monoid (where elements are parametrised as in post). But it is interesting that it is so small.... Maybe I will try to describe it in detail in the future. Thanks a lot! – kakaz Feb 9 '10 at 7:24