# Every variety has a simple algebra -- proof

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!

• 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? Oct 21, 2023 at 19:52
• @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. Oct 21, 2023 at 20:04
• @amrsa Thanks, I see what I was missing. Oct 21, 2023 at 21:28

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 $$P$$.

• It's just the second case I don't understand. Also, possibly, there is "any chain" missing? Oct 21, 2023 at 19:41
• @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. Oct 21, 2023 at 20:01
• Thanks, I think I've got it now. (Minor remark, there is still some bad phrasing in the third sentence of your answer). Oct 21, 2023 at 20:12
• 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. Oct 21, 2023 at 20:32