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My original question was the following:

Let $T$ denote an algebraic theory and suppose $X$ is a $T$-algebra. If for every identity $\eta$ in the language of $T$ we have that $(X \models \eta) \rightarrow (T \vdash \eta),$ is it necessarily the case that $X$ is a free algebra?

The answer is "no," because the first-order theory of $\mathbb{N}$ has non-standard models. In particular, let $T$ denote the equational theory of commutative unital semirings. Then $\mathbb{N}$ is the free $T$-algebra on $\emptyset$. Furthermore (binding the variable $\eta$ to the set of identities in the language of $T$), we have:

$$\forall \eta.(\mathbb{N} \models \eta) \rightarrow (T \vdash \eta).$$

However, there exist non-standard models of the first-order theory of $\mathbb{N}$. Let $M$ denote such a model. Then since $M$ satisfies the same identities as $\mathbb{N}$, we have that $M$ is a $T$-algebra. Furthermore:

$$\forall \eta.(M \models \eta) \rightarrow (T \vdash \eta).$$

However, $M$ is not isomorphic to $\mathbb{N}$, nor to any other free $T$-algebra.

My question now becomes, what happens if we assume $X$ is also finite?

Let $T$ denote an algebraic theory and suppose $X$ is a finite $T$-algebra. If for every identity $\eta$ in the language of $T$ we have that $(X \models \eta) \rightarrow (T \vdash \eta),$ is it necessarily the case that $X$ is a free algebra?

goblin GONE
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  • Can you explain more concretely what the condition means? Let's suppose we're talking about groups. Is $x^2 = 1$ an identity? If so, what does it mean for $X$ to satisfy it, and what does it mean for $T$ to imply it? – Qiaochu Yuan Jul 06 '14 at 03:22
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    @QiaochuYuan, let $T$ denote the theory of groups. Then yes $x^2=1$ is (shorthand) for an identity; which we could denote $x^2 \equiv 1$ for emphasis, or perhaps $\forall x.x^2=1$. By the completeness theorem for equational logic, we can write $T \vdash (x^2 \equiv 1)$ as a shorthand for the statement that every $T$-algebra (i.e. every group) satisfies the identity $x^2 \equiv 1$. So the condition $(X⊨η)→(T⊢η)$ means that if $X$ models $\eta$, then every group models $\eta$. – goblin GONE Jul 06 '14 at 03:39
  • @QiaochuYuan, to be a little more explicit, an identity is a universally quantified equation. – goblin GONE Jul 06 '14 at 03:46
  • Thanks. Let's call this property "equational freeness." Then it seems like if an equationally free object $X$ embeds into another object $Y$, then $Y$ is necessarily also equationally free. Furthermore it seems like the free group $F_2$ on two elements is equationally free in groups. So any group into which $F_2$ embeds is also equationally free, right? (I'm trying to come up with a simpler counterexample to the original question than nonstandard models of $\mathbb{N}$.) – Qiaochu Yuan Jul 06 '14 at 03:55
  • @QiaochuYuan, I like your terminology. Yep regarding equational-freeness being "upward-closed" with respect to the embeddability order. What makes you think $F_2$ is equationally free? (I'm not saying its not, its just that I don't have any intuition either way.) – goblin GONE Jul 06 '14 at 04:53
  • The free group on countably many generators embeds into it (e.g. as its commutator subgroup). Maybe there is an easier argument. – Qiaochu Yuan Jul 06 '14 at 04:59
  • @QiaochuYuan You could use the fact that every countable group is a subgroup of a $2$-generator group, which is a homomorphic image of $F_2$. But I guess this is a harder way of showing it. – bof Jul 06 '14 at 07:27

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Let $T$ be the theory of Boolean algebras, and let $X$ be an $8$-element Boolean algebra (with $3$ atoms). Every identity satisfied by $X$ is satisfied by the $2$-element Boolean algebra (which is a subalgebra of $X$), and is therefore satisfied by all Boolean algebras, since every Boolean algebra is a subalgebra of a direct product of $2$-element algebras. However, $X$ is not free, since a free Boolean algebra on $n$ generators has $2^n$ atoms and $2^{2^n}$ elements.

bof
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