Given a ring $A$ we know that its center $C=\{x : yx= xy \;, \forall y \in A\}$ is a well defined subset of $A$. Now I want define a set that, intuitively, is '' The set of all elements that commute between them'', but I have some trouble to write a good definition. I have written: $ B=\{x : xy=yx \;, \forall y \in B\}$, but this does not seems to me a good definition because the set that I want define is present also in the definition at left side.

I suppose that, if $B$ can be defined, then $C \subset B \subset A$, and I am interested to know in what cases those inclusions become equalities. But, Is the set $B$ really definable?

Added after the comments.

For a finite set of elements $\{ a_i\}$ with $ 1\le i \le n$ we say that the $a_i$ commute between them if $ [a_i,a_j]=0\;,\forall i,j\,: 0\le i,j \le n$. It seems to me that this definition is good (and, maybe, it can be extended to a countable set of indices) but the problem is to extend it to a possibly not countable set $B$.

  • $\begingroup$ Since you haven't defined $B$, how are we supposed to know what elements you want $B$ to contain? What does "elements that commute between them" mean? $\endgroup$ – bof Feb 19 '15 at 9:01
  • $\begingroup$ Can you elaborate on your description of $B$? I do not quite grasp what it should contain. $\endgroup$ – Regret Feb 19 '15 at 9:02
  • $\begingroup$ My problem is exactly to give a good definition of $B$. I intuitively think to a set such that all his elements commutes between them, but not necessarily with all elements of $A$. $\endgroup$ – Emilio Novati Feb 19 '15 at 9:15
  • $\begingroup$ @Emilio: What do you mean by the word them here? $\endgroup$ – Regret Feb 19 '15 at 9:18
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    $\begingroup$ @Emilio, If I understand your question correctly, you are wanting to study the maximal commutative subrings of your ring $A$. The problem is that there could be more than one of these, so you can't define such a thing as a single set... $\endgroup$ – Nick Gill Feb 19 '15 at 9:52

If I understand you correctly, you're looking for a set $B \subset A$ such that $$\forall x,y \in B, xy = yx, \tag{$\star$}$$ and such that $B$ is in some sense "maximal". Let $$P = \{ B \subset A : \forall x,y \in B, xy = yx \}.$$

This set $P$ is partially ordered by inclusion (ie. we say $B \le B' \iff B \subset B'$). It is also not empty, because $\{1\} \in P$. Suppose that you have a chain $C$ in $P$, that is a totally ordered subset of $P$. Then $$\mathcal{B} = \bigcup_{B \in C} B$$ is an element of $P$. Indeed, let $x, y \in \mathcal{B}$, then $x \in B$ and $y \in B'$ where $B, B' \in C$. Since $C$ is a chain, either $B \subset B'$ or $B' \subset B$; in both cases, $x$ and $y$ commute. It's also clear that $\mathcal{B}$ is an upper bound for $C$.

Thus every chain has an upper bound. So by Zorn's lemma*, $P$ has at least one maximal element. In other words, there exists a subset $B_0 \subset A$ such that:

  • $\forall x,y \in B_0, xy = yx$;
  • $\forall z \not\in B_0, \exists x \in B_0 \text{ st. } xz \neq zx$.

It's possible that $B_0$ is bigger than the center of $A$.
For example, take $A = \Bbbk\{x,y\}$, the free associative algebra on two generators. Then the center of $A$ is just $\Bbbk$, but the subset $\Bbbk\{x\} \subset A$ of polynomials in $x$ satisfies $\forall P, Q \in \Bbbk\{x\}, PQ = QP$. And as you can see, $\Bbbk\{y\}$ satisfies the same property: it's possible that there are different maximal $B_0$ satisfying the condition $(\star)$. You cannot define a single subset $B_0$ of "elements that commute with each other" in general.

(Prompted by the comments.) Such a $B_0$ is necessarily a subring of $A$.
Since $1x = x1$ for all $x$, it's clear that $1 \in B_0$ (otherwise $B_0 \subsetneq B_0 \cup \{1\}$ is not maximal). Similarly, if $x,y \in B_0$, then for all $z \in B_0$, $$\begin{align} (x+y)z & = xz + yz = zx + zy = z(x+y) \\ (xy)z & = xzy = z(xy), \end{align}$$ and so $x+y \in B_0$ and $xy \in B_0$ (otherwise $B_0$ is not maximal: $B_0 \subset B_0 \cup \{xy, x+y\}$). Thus $B_0$ is a subring.

It is also true that $A$ is the union of all the possible $B_0$.
If $a \in A$, consider $P_a = \{ B \in P \mid a \in B \}$. This set is again not empty (eg. $\{a\} \in P_a$: it is true that $a \cdot a = a \cdot a$!), and every chain still has an upper bound by the previous argument. Thus $P_a$ has at least one maximal element $B_a$, which contains $a$. It's also clear that $B_a$ is a maximal element of $P$, ie. it is "a $B_0$". So every element of $A$ is in one of the $B_0$.

* Yes, this argument requires the axiom of choice. I don't know enough about set theory to say what the precise relation between AC and this is, though.

  • $\begingroup$ Thank you for your answer that fits my question. If I well understand there are different $B_0$ and the intersection of those is $C$. About their union we can say nothing? I.e. there is a way to characterize the rings $A$ for which that union is a proper subset of $A$ or coincides with $A$? $\endgroup$ – Emilio Novati Feb 19 '15 at 10:46
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    $\begingroup$ I think every element is in a commutative subring (e.g. the one generated by itself) and so the union of maximal commutative subrings will always equal $A$. $\endgroup$ – Nick Gill Feb 19 '15 at 14:01

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