To conclude this chapter, we consider some simple ideas about comparing and changing alge- bras.
The natural numbers are contained in the integers. The integers are contained in the rational numbers. The rational numbers are contained in the real numbers. We are used to expressing these ideas using sets and the subset relation, thus:
N⊆Z⊆Q⊆R.
Furthermore, we are used to the idea that the operations on these numbers are equivalent. Since R extends Q, Q extends Z, Zextends N, we expect for m,n ∈N,
n+Nm, n+Zm, n+Qm, and n+Rm
to be the same number. In fact, this is not correct: there are tricky details concerning repre- sentations of these numbers to take care of before we can make these statements precise and correct. For example, depending on the precise definitions of these sets, we have to convert between number representations, say, with transformations
n ∈N7→+n ∈Z z ∈Z7→+z
1 ∈Q p
4.7. SUBALGEBRAS 109
This process introduces conceptual and technical complications that we will sort out much later (when we write about homomorphisms in Chapter7). However, intuitively, the idea seems clear.
When calculating with these numbers we use slightly different operations and, therefore, we use different signatures and algebras. For example, we might use
(N;0;n+1,n+m,n.m) (Z;0,1;x +y,−x,x.y) (Q;0,1;x +y,−x,x.y,x−1)
(R;0,1;x +y,−x,x.y,x−1).
In this section, to express the idea that two algebras share data and operations, such as Qand R, we will introduce the idea of a
subalgebra.
In the next section, to express the idea that one Σ-algebra has more data sets and operations than another, we will introduce the concept of an
expansion
and, conversely, that an algebra has fewer data sets and operations than another, we will introduce the concept of a
reduct.
4.7.1
Examples of Subalgebras
Let us look at a simple example of a subalgebra to introduce the idea. Consider an algebra of integers made from addition and subtraction.
LetΣ be the signature: signature Integers sorts int
constants 0 : →int
operations + : int×int →int
− : int →int
endsig
LetA be the standard Σ-algebra of the integers based on the set Z={. . . ,−2,−1,0,1,2, . . .} of numbers in decimal notation:
A= (Z;0Z; +Z,
−Z).
Consider the subset
of all even integers, i.e., integers divisible by 2. If we apply the constant and operations 0Z,+Z,−Z
of A to even integers, we get even integers.
First note that 0 is even. Let 2z1 and 2z2 be any even integers, where z1,z2 ∈Z. Then,
2z1 +Z2z2 =2(z1 +Zz2)
is also even. Let 2z be any even integer where z ∈Z. Then −Z
2z =2(−Z
z) is also even.
We say that the even integers Even are
closed under the operations
of A. Now because the set Even is closed, we can make a Σ-algebra B = (Even;0Z; +Z,−Z)
that is contained in the Σ-algebra
A= (Z;0Z; +Z,
−Z).
This B we call a Σ -subalgebra of A. Notice that the subset
Odd ={. . . ,−3,−1,1,3, . . .}
of all odd numbers isnot closed under the operations of A. For example, 1 +Z3 =4.
4.7.2
General Definition of Subalgebras
Let us formulate the idea of a subalgebra in general.Definition (Subalgebra) LetA be an S-sortedΣ-algebra. An S-indexed family of subsets B =hBs ⊆As |s ∈Si.
forms a Σ-subalgebra of A if
(i) the subsets Bs contain the constants of A named in Σ; and
(ii) the subsets Bs are closed under the operations of A named in Σ. I.e., for each f ∈ Σ,
applying each operationfA of A to elements of B produces elements ofB: for
4.7. SUBALGEBRAS 111 B A b1 bn b f b2. . . Figure 4.1: A is a subalgebra of B.
Given these conditions, we can make aΣ-algebra usingB and the operations ofA, as illustrated in Figure 4.1.
If B is a Σ-subalgebra of A then we may omit reference to Σ and simply say that B is a subalgebra of A, writing
B ≤A.
If B is a subalgebra of A but B 6=A then we say that B is a proper subalgebra of A (or A is a proper extension of B), and write
B < A.
Example (Integers) We can easily generalise the example of the even integers. Let Σ and A be as in Section 4.1.1. Let n ∈N and define for n >0,
nZ={z ∈Z| z is divisible by n }. A typical element of nZ has the form
nz
for some z ∈ Z. We show that nZ is the carrier of a subalgebra using the definition in Section 4.7.2.
Claim For any n >0 , B = (nZ;0Z; +Z,−Z) is a subalgebra of A= (Z;0Z; +Z,−Z).
Proof ClearlynZ⊆Z. We must check the two closure conditions for constants and operations. (i) There is only one constant symbol 0 ∈Σ. In B it is interpreted as the standard integer
zero of A which is divisible by n and 0Z
∈nZ
so nZ is closed under the constants of A.
(ii) There are two operations +,− ∈ Σ. Addition + is interpreted in B as the standard integer addition of Aand
nz1 +Znz2 =n(z1 +Zz2)
∈nZ
So nZ is closed under addition of A. Subtraction − is interpreted in B as the standard integer subtraction of A and
−Z
nz =n(−Z
z) ∈nZ So nZ is closed under subtraction of A.
2