# Is a variety of algebras still a variety if you add in another operation or drop one?

Let $L$ be a language and let $t(x_1, x_2, . . . , x_n)$ be a term over $L$. Let the language $L_1$ be obtained from $L$ by adding a new n-ary functional symbol $g$. For any algebra $A$ over $L$ we define the algebra $A^♯$ over $L_1$ with the same carrier set by defining the operation $g$ on $A^♯$ as follows: $g(a_1, a_2, . . . , a_n) = t(a_1, a_2, . . . , a_n).$ Conversely, if $A$ is an algebra over $L_1$ then the algebra $A^*$ over $L$ is obtained from $A$ by dropping the operation $g$.

Prove the following:

(a) If V is a variety of algebras over $L$ then the class $V^♯$ of algebras consisting of all algebras $A^♯$ for $A ∈ V$, is a variety of algebras over $L_1$.

(b) If $W$ is a variety of algebras over $L_1$ then the class $W^*$ of algebras consisting of all algebras $A^*$ for $A ∈ W$, is a variety of algebras over $L$.

For part (a), I said that if $V$ is a variety then $\forall A \in V$ all of the identities hold over $L$ for all operations in $A$ so in $V^\#$, which consists of all identities from $V$ and the one additional identity $g(a_1, a_2, . . . , a_n) = t(a_1, a_2, . . . , a_n)$, $\forall A \in V$ the identities will still hold over $L_1$ and teh new identity will also hold over $L_1$ for $A^\#$ so $V^\#$ will be a variety over $L_1$.
Is that correct and is it missing any details?

For part (b), Let $W$ be a variety of algebras over $L_1$, then $\forall A \in W$ all of the identities hold over $L_1$, which means all but $g(a_1, a_2, . . . , a_n) = t(a_1, a_2, . . . , a_n)$ will hold over $L$ since thats how $L_1$ was built. This means that the class $W^*$ consisting of all algebras $A^*$ for $A \in W$ is a variety over $L$. Again, is this correct and is it too vague/completely missing the point? This is the first time I've seen varieties in my life so I am struggling a bit to get a grasp on them

• You haven't made it clear if $L_1$ is $L$ with an extended signature or $L$ with an extended signature and an additional identity which essentially states $g=t$. The setup as you've portrayed it does not suggest that $L_1$ contains the additional identity, but your arguments do assume that. If $L_1$ is $L$ with an extended signature but without the additional identity, then (b) is not true. If you were asked to prove these, then presumably $L_1$ is supposed to contain the addition identity, and you just didn't clearly convey the setup. In that case, your arguments are reasonable enough. Commented May 21, 2018 at 6:55
• @DerekElkins Sorry I'm a little confused by what you're saying. The set up provided is verbatim what I was given so its safe to say that I was wrong. How would I go about proving these in that case? Commented May 21, 2018 at 7:00

The variety $V^{\#}$ as defined doesn't really have any new identity.
When you say that $g(a_1,\ldots,a_n)=t(a_1,\ldots,a_n)$, you are only defining a new operation.
This new operation $g$ is what is called a term operation on the type of $V$ (that is, $L$), and it follows that any identity that was true in $V$, is still true in $V^{\#}$, just because the operations involved in any identity of $V$ are still available.
So if $u\approx v$ is an identity of $V$, then $u$ and $v$ are terms made from the operations defined in $V$.
But each algebra of $V^{\#}$ is an algebra of $V$ expanded with a new operation $g$, and all the other operations have the same outcome. So the same identities are satisfied.
Notice also that there is no new identity in $V^{\#}$ (except trivial ones involving $g$), because any identity in which $g$ occurs, had already a correspondent in $V$, replacing $g$ with the term $t$.
Thus the class $V^{\#}$ can be axiomatized by the same identities that work for $V$.

This is false. For example, let $V$ be the variety of abelian groups.
As one of its members, we have $\langle \mathbb Z,+,-,0\rangle$.
Here, if we drop the unary operation "$-$" (the inverse) and the nulary "$0$" (neutral element), we have $\langle \mathbb Z,+\rangle$ which is a semigroup (the algebraic reduct of semigroup of the given group).
But $\langle \mathbb N,+\rangle$ is still a commutative semigroup, and it is clear that it is not the algebraic reduct of any group.
Note: The algebraic reduct of a given algebraic type is the name given to your construction in (b), that is, if $\mathbf A = \langle A, F \rangle$ is an algebra, where $F$ is a set of operations on $A$, then an algebraic reduct of $\mathbf A$ is any algebra $\mathbf A' = \langle A, F'\rangle$ with $F' \subseteq F$.