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Hi i'm a maths undergrad currently writing an essay on Universal algebra for my second year project.

I'm primarily using "A Course in Universal Algebra" by Burris and Sankappanavar. I can't seem to understand the proof of lemma 11.8 from chapter 2. The lemma essentially covers the $\Leftarrow$ part of the proof of Birkhoff's theorem (which is stated right below the lemma).

I'm struggling to understand why a second infinite set of variables Y is introduced, as the set X of variables is already infinite.

If it is because we need an infinite set of variables where we can choose any cardinality so that it is always larger or equal to the algebra A, as per the requirements of 10.11. Then why do we need an infinite set X?

Thank you.

Lemma 11.8 If V is a variety and X is an infinite set of variables, then $V=M(Id_V(X) )$

Proof: Let $$V′ = M(IdV (X))$$. Clearly $V′$ is a variety by 11.3, $V′ ⊇ V$, and
$$Id_V ′(X) = Id_V (X)$$ So by 11.4,
$$\mathbf{F}_V ′(\overline{X}) = \mathbf{F}_V(\overline{X})$$ Now given any infinite set of variables Y, we have by 11.6
$$Id_{V′}(Y) = Id_{\mathbf{F}_{V′(X)}}(\overline{Y}) = Id_{\mathbf{F}_{V(X)}} (\overline{Y}) = Id_V(Y).$$ Thus again by 11.4, $$θ_{V′}(Y) = θ_{V}(Y)$$ hence $$\mathbf{F}_{V′}(\overline{Y}) = \mathbf{F}_{V}(\overline{Y})$$ Now for $\mathbf{A}∈ V'$ we have (by 10.11), for suitable infinite Y, $$\mathbf{A} ∈ H(\mathbf{F}_{V′}(\overline{Y}));$$ hence $$\mathbf{A} ∈ H(\mathbf{F}_V(\overline{Y}))$$ so $A ∈ V$ ; hence $V′ ⊆ V$, and thus $V′ = V$.

Suhaib Afzal
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    Maybe you could reproduce the relevant lemma here so that people don't have to go through a 330 page text to help you. – lcv Mar 22 '20 at 20:33

1 Answers1

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The reason is just to apply Corollary 10.11.
Notice that $X$ could perhaps not be large enough.

Corollary 10.11 tell us that

If $K$ is a class of algebras of type $\mathscr F$ and $\mathbf A \in K$, then for sufficiently large $X$, $\mathbf A \in H(\mathbf F_K(\overline{X}))$.

The proof starts with "let $|X|\geq |A|$..." and proceeds to prove that that is enough.
So in this case (in the proof of Lemma 11.8), you need $|Y|\geq |A|$, and it could be that $|X| < |A|$ (there's always such an algebra in any non-trivial variety—if $\mathbf B$ is a non-trivial algebra, then $\mathbf A=\mathbf B^X$ will be such an algebra), so that 10.11 wouldn't apply.

amrsa
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  • So would the proof work just as well if we had X as an finite set of variables? – Suhaib Afzal Mar 28 '20 at 15:58
  • Also i don't understand what the notation $\mathbf{B}^X$ means, could you please clarify? – Suhaib Afzal Mar 28 '20 at 16:07
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    The proof couldn't work if $X$ was finite, because the statement of the lemma is for $X$ infinite. Now, the lemma could be stated for a finite set $X$, provided the variety had a base with a finite set of equations, which is not true in general; still, it would be a weaker form of the lemma, even for that variety, because there would always be identities between terms using more variables than the elements in $X$. That's not possible if $X$ is infinite. – amrsa Mar 28 '20 at 16:26
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    $\mathbf A=\mathbf B^X$ is algebra $\mathbf B$ to the power of $X$, that is, the set of maps from $X$ to $B$, with the operations defined coordinate-wise. For example, if $X$ is finite, with $n$ elements, then $\mathbf B^X$ is just $\mathbf B^n$ (up to isomorphism). If $\mathbf B$ belongs to a variety and $|B|\geq 2$, then $\mathbf A$ also belongs to that variety, and $|A|=|B^X|\geq 2^{|X|}>|X|$. – amrsa Mar 28 '20 at 16:30
  • Could I get a source for your claim that "there's always such an algebra in any non-trivial variety". A source with that direct statement or a source for the properties of $\mathbf{B}^X$ you use will do. I need something to cite in the essay. Thank you. – Suhaib Afzal Mar 31 '20 at 15:39
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    @SuhaibAfzal I don't know of any such source, but it is so evident; actually I provided a proof. A trivial variety is one in which all algebras are trivial (they're singletons). A trivial variety satisfies all equations, and so the lemma 11.8 is trivially satisfied; for a non-trivial variety, pick a non-trivial algebra $\mathbf B$ as in my last comment, and make $\mathbf A=\mathbf B^X$, thus obtaining an algebra $\mathbf A$ with $|X|<|A|$. – amrsa Mar 31 '20 at 17:16