Characterizing Finitely Based Abelian Mal’cev Varieties

We prove the first item by induction on $k$. The base case of $k=0$ is trivial. Now assume the claim has been proven for $n$-ary terms. Since $m$ is a Mal’cev term and commutes with every term operation by Proposition 1, the variety $\mathcal{V}$ satisfies

The last equation is by the inductive hypothesis. So we have proven the first item.

The second item follows immediately. ∎

1. Let $\mathcal{W}$ be a variety of type $\rho:=F_{\mathcal{V}}(z)\cup F_{\mathcal{V}}^{id}(x,z)\cup\{m\}.$ Define $E$ to be an interpretation of $\mathcal{W}$ in $\mathcal{V}$ as follows: For $f\in\rho$, let $E(f)$ be a term over $\sigma$ that induces $f$ in $\mathbf{F}_{\mathcal{V}}(x,y,z).$ For $\mathbf{A}\in\mathcal{V}$, let $\mathbf{A}^{E}$ be the algebra with universe $A$, unary operations $u^{\mathbf{A}}(z)$ for $u\in F_{\mathcal{V}}(z),$ binary operations $r^{\mathbf{A}}(x,z)$ for $r\in F_{\mathcal{V}}^{id}(x,z),$ and ternary operation $m^{\mathbf{A}}(x,y,z)$ for the Mal’cev term $m$ of $\mathcal{V}$. Here we slightly abuse notation when writing $u^{\mathbf{A}}$ for the term function on $\mathbf{A}$ that is induced by the term $u$ over $\sigma$ which induces the element $u(z)\in F_{\mathcal{V}}(z)$. Similarly for $r^{\mathbf{A}}$. Then $E$ is an interpretation of $\mathcal{V}$ in $\mathcal{W}:=\{\mathbf{A}^{E}\ |\ \mathbf{A}\in\mathcal{V}\}$.

Conversely, define $D$ to be an interpretation of $\mathcal{V}$ in $\mathcal{W}$ as follows: Let $f\in\sigma$. In the case that $f$ is 0-ary, let $t(z)\in F_{\mathcal{V}}(z)$ be any term such that $t(z)=f$ in $F_{\mathcal{V}}(z)$. In the case that $f(x_{1},\ldots,x_{k},z)$ is $(k+1)$-ary, define $f_{0}(z):=f(z,\ldots,z)$ and $f_{i}(x):=f(z,\ldots,z,x,z,\ldots,z)$, where $x$ appears in the $i$-th position. Then define

Then $D$ is an interpretation of $\mathcal{W}$ in $\mathcal{V}$ by Lemma 4.1. Furthermore, for any algebra $\mathbf{A}^{E}\in\mathcal{W}$, the algebra $\mathbf{A}^{ED}=\mathbf{A}$ and hence in $\mathcal{V}$.

It follows that for all $\mathbf{A}\in\mathcal{V}$, we have $\mathbf{A}^{ED}=\mathbf{A}\in\mathcal{V}$ and $\mathbf{A}^{EDE}=\mathbf{A}^{E}\in\mathcal{W}$. The latter means that $\mathbf{B}^{DE}=\mathbf{B}$ for all $\mathbf{B}\in\mathcal{W}.$ Thus $\mathcal{V}$ and $\mathcal{W}$ are equivalent.

2. Suppose that $\mathcal{V}$ has infinite type and suppose by way of contradiction that $B$ is a finite basis for $\mathcal{V}$. Then there exists a basic operation $f\in\sigma$ such that $f$ does not occur in $B$. Say $f$ has arity $k$. Since $\mathcal{V}$ is abelian Mal’cev, by Proposition 1, we know that

However, $B$ cannot prove this fact since $f$ never occurs in $B$, contradicting $B$ being a basis.

3. If $\sigma$ is finite, you need finitely many unary terms and idempotent binary terms with $m$ to interpret every basic operation in $\sigma$. Let us denote said unary terms as $u_{1},\ldots,u_{\ell}$, where $\ell\in\mathbb{N}$. Now define $r_{i}(x,z)=m(u_{i}(x),u_{i}(z),z)$ for $i\leq\ell$. Let $r_{i}(x,z)$ be the rest of the needed idempotent binary terms for $\ell<i\leq n$, where $n\geq\ell$. Then we need to show that the terms $u_{1},\ldots,u_{\ell},r_{1},\ldots,r_{n}$ satisfy Definition 5.1-6. That would complete the proof.

Equation 1 is satisfied because $m$ is assumed to be a Mal’cev term. Equations 2,4, and 5 are satisfied by Proposition 1. Equation 3 is satisfied since $r_{i}\in F_{\mathcal{V}}^{id}(x,z)$ for $i\leq n$. Lastly, Equation 6 holds by construction. ∎

We prove that $\mathbf{R}_{\mathcal{V}}$ is a ring and that $\mathbf{M}_{\mathcal{V}}$ is an $\mathbf{R}_{\mathcal{V}}$-module by collecting proofs of properties of $+,-,\cdot$. [2, Lemma 5.6] already shows that $\langle F_{\mathcal{V}}(x,z),+,-,z\rangle$ is an abelian group. Let $a,b,c\in F_{\mathcal{V}}(x,z)$ and $r,s\in F_{\mathcal{V}}^{id}(x,z).$

Now we show that $\langle F_{\mathcal{V}}^{id}(x,z),+,-\rangle$ is a subgroup. Clearly, $z\in F_{\mathcal{V}}^{id}(x,z)$. We have

So $r-s\in F_{\mathcal{V}}^{id}(x,z)$ and hence $\langle F_{\mathcal{V}}^{id}(x,z),+,-,z\rangle$ is an abelian group.

Multiplication is associative since

We show now that multiplication distributes over addition if the left factor is in $F_{\mathcal{V}}^{id}(x,z)$. We have, since $r$ is idempotent and commutes with $m$ by Proposition 1, that

Now we show right distributivity. We have

Clearly $1(x,z):=x$ is a multiplicative identity. So $\mathbf{R}_{\mathcal{V}}$ is a ring and $F_{\mathcal{V}}(x,z)$ forms an $\mathbf{R}_{\mathcal{V}}$-module with submodules $F_{\mathcal{V}}^{id}(x,z)$ and $F_{\mathcal{V}}(z)$. To see 3, note that every element $a$ in $F_{\mathcal{V}}(x,z)$ can be written uniquely as a sum of an idempotent term and a unary term in the following way

4. We have by [2, Lemma 5.6] that $m(a,b,c)=a-b+c.$ Let $a,b\in F_{\mathcal{V}}(x,z)$ and $r\in F_{\mathcal{V}}^{id}(x,z)$. Then, we have, using the idempotence of $r$,

∎

Using that $m$ and $m^{\prime}$ commute by Proposition 1, we have

Hence, the two terms induce the same term operation in $\mathcal{V}$. ∎

1. Every basic operation commutes with the Mal’cev term $m$. So $\mathcal{U}$ is an abelian Mal’cev variety by Proposition 1.

2. Let $\mathcal{U}^{\prime}$ be the variety of type $\{r_{1},\ldots,r_{n},m\}$ satisfying identities 1-4 of $\mathcal{U}$. Note that this makes $\mathcal{U}^{\prime}$ an idempotent reduct of $\mathcal{U}$ and $+,-$ are still term operations of $\mathcal{U}^{\prime}$ and $z$ is still the additive identity. Also, $\mathcal{U}$ and $\mathcal{U}^{\prime}$ are abelian varieties. The unique Mal’cev term is $m(x,y,z)=x-y+z$ by Lemma 7. It commutes with every basic operation by Proposition 1.

First, we claim

Clearly, $F_{\mathcal{U}^{\prime}}(x,z)\subseteq F_{\mathcal{U}}^{id}(x,z)$. All that is left is to show $F_{\mathcal{U}^{\prime}}(x,z)\supseteq F_{\mathcal{U}}^{id}(x,z)$. Note that every $s\in F_{\mathcal{U}}^{id}(x,z)$ has the form

for some $t\in F_{\mathcal{U}}(x,z)$. We then induct on $t$ to show that $s\in F_{\mathcal{U}^{\prime}}(x,z)$.

For the first base case, suppose $t=x$. Then $s=x-z=x\in F_{\mathcal{U}^{\prime}}(x,z)$. Now, as a second base case, suppose $t=z$. Then $s=z-z=z\in F_{\mathcal{U}^{\prime}}(x,z)$.

For the induction step, we consider several cases. First, suppose that $t=u_{i}(v(x,z))$ for some $i\leq\ell$ and some $v\in F_{\mathcal{U}}(x,z)$. We have

which is in $F_{\mathcal{U}^{\prime}}(x,z)$ since $v(x,z)-v(z,z)\in F_{\mathcal{U}^{\prime}}(x,z)$ by the inductive hypothesis.

For another induction case, suppose that $t=r_{i}(v,w)$ for some $i\leq n$ and some $v,w\in F_{\mathcal{U}}(x,z)$. Then

which is in $F_{\mathcal{U}^{\prime}}(x,z)$ since $v(x,z)-v(z,z),w(x,z)-w(z,z)\in F_{\mathcal{U}^{\prime}}(x,z)$ by the inductive hypothesis.

For the last induction step, suppose that $t=m(v,w,y)$ for some $v,w,y\in F_{\mathcal{U}}(x,z)$. Using Lemma 2.4, we have

which is in $F_{\mathcal{U}^{\prime}}(x,z)$ since $v(x,z)-v(z,z),w(x,z)-w(z,z),y(x,z)-y(z,z)\in F_{\mathcal{U}^{\prime}}(x,z)$ by inductive hypothesis.

So we have shown $F_{\mathcal{U}^{\prime}}(x,z)=F_{\mathcal{U}}^{id}(x,z)$. It follows that $\mathbf{R}_{\mathcal{U}}=\mathbf{R}_{\mathcal{U}^{\prime}}.$

Now we claim

We will write elements of $T$ in boldface to better distinguish them from elements of $F_{\mathcal{U}^{\prime}}(x,z).$

Clearly,

defines a ring homomorphism. To show $\psi$ is bijective, we will make $T$ an algebra in $\mathcal{U}^{\prime}$. Then we will construct a homomorphism

such that $\tau\circ\psi$ is the identity on $T$ and $\psi\circ\tau$ is the identity on $F_{\mathcal{U}^{\prime}}(x,z)$. We first have to define operations $r_{1}^{\mathbf{T}},\ldots,r_{n}^{\mathbf{T}},m^{\mathbf{T}}$ on $T$ such that $\mathbf{T}:=\langle T,r_{1}^{\mathbf{T}},\ldots,r_{n}^{\mathbf{T}},m^{\mathbf{T}}\rangle\in\mathcal{U}^{\prime}$. For $\mathbf{s}_{1},\mathbf{s}_{2},\mathbf{s}_{3}\in T$ and $i\leq n$ define

From the definition, it is clear that $m^{\mathbf{T}}$ is a Mal’cev operation that commutes with itself and the idempotent affine operations $r_{i}^{\mathbf{T}}$ for all $i\leq n$. Hence $\mathbf{T}$ is in $\mathcal{U}^{\prime}$.

Now we show that $\mathbf{T}$ and $\mathbf{F}_{\mathcal{U}^{\prime}}(x,z)$ are isomorphic. Let $\tau\colon\mathbf{F}_{\mathcal{U}^{\prime}}(x,z)\to\mathbf{T}$ be the homomorphism of algebras in $\mathcal{U}^{\prime}$ defined by $x\mapsto\mathbf{1}$ and $z\mapsto\mathbf{0}$. We show that $\tau\circ\psi$ is the identity map on $T$ and that $\psi\circ\tau$ is the identity map on $F_{\mathcal{U}^{\prime}}(x,z).$

We show that

via induction on $\mathbf{s}$. For the base case, we have $\tau(\psi(\mathbf{0}))=\tau(z)=\mathbf{0}$ and $\tau(\psi(\mathbf{1}))=\tau(x)=\mathbf{1}$. For $i\leq n$, we have