# Example of a subalgebra of infinite-generated algebra which cannot be extended to a maximal subalgebra

First of all, sorry for my English, especially mathematical one. The problem is that I know how to call such things in Ukrainian but unfortunately did not manage to find proper translations to English. So, I will give some brief definitions of terms in order to not confuse you with the names.

So, an algebra (maybe, partial) is a set of elements together with a set of operations defined on these elements $(A, \Omega)$.

Let $B \subset A$. Then a closure $[B]_f$ of $A$ by an $n$-ary operation $f \in \Omega$ is a set defined by two rules:

1. $B \subseteq [B]_f$
2. $\forall (a_1, a_2, \ldots, a_n) \subset [B]_f$ if $f(a_1, a_2, \ldots, a_n)$ is defined then $f(a_1, a_2, \ldots, a_n) \in [B]_f$

A closure $[B]$ of $A$ is $B^0 \cup B^1 \cup B^2 \ldots$ where $B^0 = B$, $B^{i+1} = \bigcup_{f \in \Omega}[B^i]_f$.

The definition of subalgebra and therefore algebra extension is obvious, I think.

$B$ is a system of generators (I believe, it is close to the basis) for algebra $(A,\Omega)$ if $[B] = A$.

And, finally, subalgebra $(B,\Omega)$ is called maximal subalgebra of $(A,\Omega)$ if there is no subalgebra $(C,\Omega)$ such that $B \subset C \subset A, B \neq A, C \neq A, C \neq B$.

It can be proved that for a algebra with existing finite system of generators each its subalgebra can be extended to some maximal subalgebra. But for algebras with infinite systems of generators, this statement is not always true. And I need a counter-example. Therefore, I need an example of infinite-based algebra and a subalgebra which is impossible to extend to some maximum subalgebra.

-
I thought about infinite-based vector spaces and replacements of all vectors with i-th coordinate equal to some number (maybe, to all numbers <=i) for each i, but I didn't manage to come up with something. – ikostia Feb 26 '12 at 13:07
Unfortunately, every proper subspace $W$ of a vector space $V$ is contained in a maximal (I assume by this you also mean it doesn't equal the whole thing) subspace; just take a vector $v \in V$ not in $W$, let $B$ be a basis of $W$, extend $B \cup \{ v \}$ to a basis of $V$, and consider the span of this basis minus $v$. – Qiaochu Yuan Feb 26 '12 at 13:10
Your "creators" are usually called "generators". – g.castro Feb 26 '12 at 13:25
Anyway, I would recommend you try thinking about groups instead of vector spaces. In fact there is a relatively straightforward example of an abelian group which has no maximal proper subgroups whatsoever. – Qiaochu Yuan Feb 26 '12 at 13:26
@g.castro: thanks – ikostia Feb 26 '12 at 13:31

The answer is an abelian group of rational numbers $Q$. It has no maximal subgroups at all. Proof can be found here. Thanks, Qiaochu Yuan, for the direction.

-

EDITED: Hint: Consider the operation $f(n) = n-1$ on the natural numbers (with $f(0)$ arbitrary).

The original hint (Consider an algebra without any operations, or with only a trivial operation) was completely wrong.

-
Hm, identity operation cannot give you any new elements. This means, that if you have a system B, then the closure of this system by identity operation would give you B again. This in turn means that adding new elements to B means nothing but adding new elements, so if you just take any subset B of algebra base A, you can add all but one of the A\B elements to B and you will get a maximal subalgebra. I believe, that very alike thought have place with constant operation. Where am I wrong? – ikostia Feb 26 '12 at 13:51
Thank you. I sketched a new construction. – g.castro Feb 26 '12 at 20:25
This one seems to be really cool and easy to understand! Thanks! – ikostia Feb 27 '12 at 9:33
I'm willing to edit your answer in purpose to add a full proof. Please, respond if it is okay. Or, if you would like, you can do it by yourself. – ikostia Feb 27 '12 at 9:40
Sure, go ahead. – g.castro Feb 27 '12 at 10:07