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So I'm reading Grätzer's Universal Algebra, and I'm at chapter 5, paragraph 31 about independance and bases. The theorem I have a question about is theorem 4 (for those who have the book), it states :

"Let $\mathfrak{A}$ be an algebra with more than one element. If $\mathfrak{A}$ has bases with different cardinal numbers, then all bases of $\mathfrak{A}$ are finite and the numbers $n$ for which there is a basis of $n$ elements form an arithmetic progression".

Let me just remind the definition of basis : a basis of an algebra $\mathfrak{A}$ is a generating set that's independent; where an independent subset $H\subset A$ is a subset that is a free generating set of the subalgebra generated by $H$ (which I will denote $[H]$) over the equational class generated by $\mathfrak{A}$, that is $\mathbf{HSP}(\{\mathfrak{A}\})$.

So Grätzer begins the proof with "It suffices to show that if we have a basis of cardinality $p$, and bases of cardinality $q$, $q+r$, then we have a basis of cardinality $p+r$". I understand up to that point (note that we proved earlier that if there is an infinite basis, then all bases have the same infinite cardinality) but what I don't understand is what follows : he lets $(a_1,...a_p)$, $(b_1,...,b_q)$ and $(c_1,..., c_{q+r})$ denote bases of the appropriate cardinality, then lets $\mathfrak{C} = [ c_1,...,c_q]$, which is thus an algebra isomorphic to $\mathfrak{A}$ (with an isomorphism $\phi$ such that $\phi(b_i) = c_i$ for $i\in \{1,...,q\}$), and he wants to show that $\{\phi(a_1),...,\phi(a_p),.c_{q+1},...,c_{q+r+1}\}$ is basis for $\mathfrak{A}$.

The point I don't understand is that he says "Obviously, this is a generating set", but I have no idea why that's obvious. My guess is that I'm missing...the obvious; but I don't see it. Would anyone care to explain ?

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    Just out of curiosity, is there a nice example of this phenomenon? I think I've seen an example before, but I can't remember it... – 2017-02-13
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    I think you can find some examples in cyclic groups. For instance, $\mathbb{Z}/6\mathbb{Z}$ has bases $\{1\}$ and $\{2,3\}$. – 2017-02-13
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    Remember that "independence" is defined with respect to a variety of algebras. The class of cyclic groups is not a variety (it's not closed under product), so you'll have to be more careful about the context. But in any case, no matter what variety containing $\mathbb{Z}/6\mathbb{Z}$ you pick, $\{2,3\}$ satisfies $2 + 2 + 2 = 3 + 3$, and not every pair of elements in $\mathbb{Z}/6\mathbb{Z}$ satisfies the identity $x+x+x=y+y$, so the set $\{2,3\}$ can't be a basis. – 2017-02-13
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    I've just located a copy of Grätzer, and the example given after Theorem 4 is the class of Jónsson-Tarski algebras, which I'm remembering now is the canonical example. Any free Jónsson-Tarski algebra on finitely many generators has a basis of every finite size $>0$. – 2017-02-13
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    You are correct about my wrong example, my bad ! I'll look it up later, thanks for the indication (I hadn't read what came after since I was stuck on this "obvious") – 2017-02-13

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Let $B$ be the proposed basis. It suffices to show that each $c_i$ for $1 \leq i \leq q$ is in the subalgebra $\langle B\rangle$ generated by $B$, since then $\langle B\rangle$ contains the whole basis $\{c_1,\dots,c_{q+r}\}$.

But there is some term $t$ such that $b_i = t(a_1,\dots,a_p)$, so $c_i = \phi(b_i) = \phi(t(a_1,\dots,a_p)) = t(\phi(a_1),\dots,\phi(a_p))$, and $c_i\in \langle B\rangle$.

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    Oh so it was obvious indeed...I must have been tired while reading this part ! Thanks ! – 2017-02-13