2
$\begingroup$

I'm having a hard time understanding the proof (as presented for example in A Course in Universal Algebra of the GTM series) of the theorem by Magari which states that a variety $V$ with a nontrivial member always contains a simple algebra.

The proof goes as follows. Let $X:=\{x,y\}$ and consider the free $V$-algebra $\mathbf{F}$ on $X$. Let $$S:=\{p(\bar{x}) : p(x) \text{ is a term with variable } x\} \subseteq F$$ (By $\bar{x}$ I mean the equivalence class of $x$ under the congruence generated by the equations of $V$). Let $\theta(S)$ be the smallest congruence on $\mathbf{F}$ which contains $S$.Then two cases are considered. If $\theta(S) \neq F^2$, then via Zorn's lemma we get the desired result. If $\theta(S) = F^2$, then the ingredients are

  • there is a finite $S_0 \subseteq S$ such that $(\bar{x},\bar{y}) \in \theta(S_0)$;
  • $\mathbf{S}$, the subalgebra of $\mathbf{F}$ with carrier $S$ is nontrivial;
  • $\theta(S_0) = S^2$.

How are we supposed to conclude form here?

Any help is appreciated, thanks!

$\endgroup$
3
  • $\begingroup$ This might be a stupid question, but why can't we just take $A$ to be a nontrivial algebra in the variety $V$ and use Zorn's lemma on the poset of proper congruences on $A$ to find a maximal one ${\sim}$ (by proper, I mean not equal to $A^2$). Won't $A/{\sim}$ be simple? $\endgroup$ Oct 21, 2023 at 19:52
  • $\begingroup$ @AlexKruckman But first, in order to apply ZL, we must prove that that in the set of proper congruences there is an upper bound (still proper) of each chain. $\endgroup$
    – amrsa
    Oct 21, 2023 at 20:04
  • $\begingroup$ @amrsa Thanks, I see what I was missing. $\endgroup$ Oct 21, 2023 at 21:28

1 Answer 1

3
$\begingroup$

In the first case, let $\theta_0$ be a maximal element in $[\theta(S),F^2] \setminus \{F^2\}$ (see Edit 1).
It follows that $\mathbf F/\theta_0$ is simple.

In the other case, if $P = [\Delta_S,S^2] \setminus \{S^2\}$, where $\Delta_S$ is the least congruence of $\mathbf S$, then any chain in $P$ has an upper bound in $P$ (see Edit 2).
This is because $S = \theta(S_0)$ and $S_0$ is finite.
By Zorn's Lemma, there is a maximal element in $P$, say $\theta_0$.
Then $\mathbf S/\theta_0$ is simple.


Edit 1. In the first case, I forgot to justify the existence of such $\theta_0$, which is by Zorn's Lemma too.
The reasoning is by noticing that for $\theta \in [\theta(S),F^2]$, we have $(\bar x, \bar y) \in \theta$ iff $\theta = F^2$.
So if $C$ is a chain in $[\theta(S),F^2] \setminus\{F^2\}$, then $(\bar x, \bar y) \notin \eta$, for each $\eta \in C$, whence $(\bar x, \bar y) \notin \bigcup C$, and therefore $\bigcup C \in [\theta(S),F^2] \setminus\{F^2\}$.
Now, we're in conditions to apply Zorn's Lemma.

Edit 2. If $(\alpha_i)_{i\in I}$ were a chain in $P$ with $$\bigcup_{i\in I}\alpha_i = S^2 = \theta(S_0),$$ then, as $S_0$ is finite, $$S^2 = \bigcup_{i=1}^m\alpha_i,$$ for some choice of elements $\alpha_1,\dots,\alpha_m$ in the chain. But then $S^2 = \alpha_{i_0}$, the largest element of these. By contrapositive, if $C$ is a chain all of whose elements are strictly below $S^2$, then so is its union; so the chain has an upper bound in $P$.

$\endgroup$
4
  • 1
    $\begingroup$ It's just the second case I don't understand. Also, possibly, there is "any chain" missing? $\endgroup$ Oct 21, 2023 at 19:41
  • 1
    $\begingroup$ @Mockingbird If $(\alpha_i)_{i\in I}$ is a chain of congruences such that $$\bigcup_{i\in I}\alpha_i = S^2 = \theta(S_0),$$ then, as $S_0$ is finite, $$S^2=\bigcup_{i=1}^m\alpha_i,$$ for some choice of elements $\alpha_1,\dots,\alpha_m$. It follows that $S^2=\alpha_{i_0}$ because it's a chain, and so not all elements of the chain were strictly below $S^2$. So by the contrapositive we're under the conditions to apply Zorn's Lemma. $\endgroup$
    – amrsa
    Oct 21, 2023 at 20:01
  • 1
    $\begingroup$ Thanks, I think I've got it now. (Minor remark, there is still some bad phrasing in the third sentence of your answer). $\endgroup$ Oct 21, 2023 at 20:12
  • $\begingroup$ Thanks. You had already mentioned it in your first comment but I missed the meaning of it. Meanwhile, since I was editing I also incorporated the reasoning in the previous comment, which I will delete later on. $\endgroup$
    – amrsa
    Oct 21, 2023 at 20:32

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .