I'm trying to show the Tarski's Theorem in Universal algebra. The theorem states that V=HSP, where V,H,S,P are operators between classes of algebras, V(K) is the smallest variety containing K, H(K) is the class of algebras which are homomorphic images of some members of K, S(K) is the class of algebras which are subalgebras of some member of K and P(K) is the class of algebras which elements are direct product of a nonempty family of algebras in K. The proof start like that: Since HV=SV=IPV=V and I $\leq$ V (I(K) $\subset$ V(K) for all classes of algebras K) it follows that .... I don't understand how to prove these relations.
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Is that Tarski's theorem? How is that related to Birkhoff's theorem? – bof Apr 06 '21 at 11:21
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Yes, it's Tarski's theorem. You need it to prove one of the "arrows" of Birkhoff theorem, i.e. equational class $\implies$ variety. In particular you need it to prove that $V(K)$ and $K$ satisfy the same identities, which implies that if $K=M(\Sigma)$ for a set of identities $\Sigma$ then $V(K) \models \Sigma$ and so you have $V(K) \subset M(\Sigma) = K$ and the other containment is obvious so $K=V(K)$. – SilvioM Apr 06 '21 at 13:34
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Notice that a variety is, by definition (in the context of Tarski's theorem), closed under $H,S,P$, and so if $K=V(K)$, that is, $K$ is a variety, then $H(K), S(K), IP(K) \subseteq K$, whence $HV, SV, IPV \leq V$, while the converse is trivial.
It also follows that $H,S,IP \leq V$, and of course, $I \leq H$.
I suppose this is enough to answer your questions.
amrsa
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