The statement
Call a class of algebras (of a fixed signature) an equational class if it is exactly the class of all algebras satisfying some set Σ of identities. Call it a variety if it is closed under H, S and P from the previous guide. Birkhoff's theorem (1935) says these two notions coincide: a class is equational if and only if it is a variety. One direction we essentially did already; the depth is in the converse.
Why the converse holds: free algebras do the work
Let V be a class closed under H, S, P. Define Σ to be *all* identities satisfied by every algebra in V. Clearly every member of V satisfies Σ. The hard part is the reverse inclusion: any algebra B satisfying Σ must already be in V. The proof builds the free algebra F_V(X) on a large enough generating set X, realises B as a homomorphic image of F_V(X), and shows F_V(X) ∈ V using exactly the three closure properties.
- Build the V-free algebra F = F_V(X) with X large enough to surject onto B. Concretely F is a subalgebra of a product of members of V (via all maps into V-algebras), so F ∈ SP(V) ⊆ V.
- Pick generators surjecting onto B; the universal property extends this to a surjection F ↠ B. Hence B ∈ H(V) ⊆ V — *provided* the surjection is legitimate, which is where Σ enters.
- B satisfies Σ — every identity true throughout V. This guarantees the relations holding in B are *at least* those forced in F, so the map F ↠ B is a well-defined homomorphism. That closes the loop: B ∈ H(SP(V)) ⊆ V.
The slogan: V = HSP(K) for the variety GENERATED by a class K.
Worked instance. Let K = { Z } as a group (one infinite cyclic group).
Claim: HSP(Z) = the variety of ALL abelian groups.
P(Z): products Z^I are abelian.
S: subgroups of abelian groups are abelian.
H: quotients of abelian groups are abelian -- e.g. Z/nZ in H(Z).
So HSP(Z) is contained in the abelian-group variety.
Conversely every abelian group A is a quotient of a free abelian group
Z^(X) = a direct sum (subalgebra of a product) of copies of Z, so
A in H(S(P(Z))).
Hence HSP(Z) = abelian groups, defined by the single identity
x + y approx y + x (on top of the group identities).
Contrast: the class of FIELDS is NOT a variety.
A product of two fields, e.g. F2 x F2, has the zero-divisor (1,0)(0,1)=(0,0),
so it is not even an integral domain -- P fails. No set of identities can
define fields.What it buys you, and what it forbids
Birkhoff hands you a *decision criterion*. To prove some class is not equational, you need only exhibit one failure of closure: a subalgebra, quotient, or product that escapes the class. Fields fail at products (a product of fields has zero divisors); torsion-free abelian groups fail at quotients (Z is torsion-free but Z/2Z is not); finite groups fail at infinite products. Each non-example is a single coordinate-level computation, not a model-theoretic argument.