Subrings

6.2.1 Definition. Let (R, +, ⋅) a ring. A nonempty subset S of R is a called a subring of R if S is a ring with respect to the same operations of R. □

For S to be a subring of the ring R, S, it must be an additive subgroup of the group (R, +), i.e. (S, +) must be a group.

In the ring ℤ of integers, for example, the subset E of even integers is a ring, but the subset 0 of odd integers is not, as we saw in Examples 2 and 3.

The following is a criterion, necessary and sufficient for S to be a subring.

6.2.2 Proposition (Subring Criterion). Let (R, +, ⋅) a ring. A nonempty subset S of R is a subring of R if and only if S is closed under subtraction and under multiplication that is:

a − b ∈ S, ab ∈ S ∀a,b ∈ S (6.2.1)

Proof. If S is a subring the conditions above are certainly satisfied. Viceversa suppose that for all a and b in S, we have a − b ∈ S, ab ∈ S: we have to prove that (S, +) is an additive group i.e. that it satisfies:

a + b ∈ S
0 ∈ S
−a ∈ S

(the associativity with respect to + is automatically verified). Since condition (6.2.1) is true for every a,b ∈ S it also hold true for a = b hence a − a = 0 ∈ S. If we take the elements a and 0, it follows that 0 −a = −a ∈ S. It is also true that given a,b ∈ S also a and −b are in S and thus follows a−(−b) = a + b ∈ S.□

Example of subrings

Many rings live inside other rings. For example, (ℤ, +, ·) is a subring of (ℚ, +, ·), (2ℤ, +, ·) is a subring of (ℤ, +, ·).
We have seen different example of rings as

ℤ[x] = {all polynomials in the variable x whose coefficients are integers}
ℝ[x] = {all polynomials in the variable x whose coefficients are real numbers}

Let a ∈ ℂ. We indicate as

ℤ[a]

the intersection all subrings of ℂ, containing ℤ and a. For example, in the case of a = i, we have

ℤ[i] = {a + ib ∈ ℂ | a,b ∈ ℤ}

So inside the ring ℂ, there is the subring ℤ[i]. Geometrically this consists of the lattice of points in the complex plane whose coordinates are integers. The elements of ring ℤ[i] are known as Gaussian integers. We shall explore ℤ[i] thoroughly, but just to show the difference with ℤ, consider the multiplication of (1+2i)(1+2i); what do you get is 5. Notice that 5 in ℤ is a prime, which cannot be factored; yet we have just seens that in ℤ[i], which contains ℤ, 5 is no longer prime!.

We obserde that ℤ[i] is obtained by adjoining a root of the equation x² + 1 = 0 to ℤ, hence the name of quadratic extension of ℤ. The general quadratic extension of ℤ is formed by adjoining to ℤ a root of a quadratic equation:

x² + ax + b = 0, where a,b ∈ ℤ

Thus this extension ring is ℤ[√d], with nonsquare d, so that we know that d is an irrational number, e.g. ℤ[√2]. You can think of ℤ[√d] as polynomials with integers coefficients and powers of √d. Of course since (d)² = d is an integer, we really don't have to worry about any powers higher than the first: for example take ℤ[√2] and consider 4x³ − 5x² + x − 3 and replace each x with √2 we get,

4(√2)³ −5(√2)² + (√2)¹ − 3 = 16√2 − 10 + √2 − 3 = 17√2 −13.

So we really have the following definition

ℤ[√d] = {a + b √d ∈ ℝ | a,b ∈ ℤ}

ℤ[√2] = {a + b √2 ∈ ℝ | a,b ∈ ℤ}

And by similar reasoning

ℤ[∛2] = {a + b ∛2 + c(∛2)² ∈ ℝ | a,b,c ∈ ℤ}

Similarly since i² −1 ∈ ℤ, we get

ℤ[i] = {a + bi ∈ ℝ | a,b ∈ ℂ}

Every ring posseses two subrings, known as trivial subrings: the whole ring and the ring containing only 0.
In R = ℤ, S = nℤ (n ∈ ℕ) is a subring of ℤ. Take a,b ∈ nℤ; let a = nx, b = ny, for some integers x,y ∈ ℤ. Then

a−b = n(x − y) ∈ nℤ

ab = n (nxy) ∈ nℤ

so, nℤ passes the test of Proposition 6.1.4.

Proposition 4.18 shows that any subring of ℤ is of this form.
Let R a ring and a ∈ R. Let

S_a = {x ∈ R | xa = ax}

S_a is a subring of R.
In the ring 𝕂[x] the subset of polynomials with costant term equal to zero is a subring.
The intersection of an arbitrary family of subrings of a ring R is a subring of R.

6.2.3 Proposition. Every subring of ℤ is of the form nℤ, with n ∈ ℕ.

Proof. We previouslt showed that nℤ is a subring of ℤ. Now we show that all subrings are of this form. Let S be a subring of ℤ. Certainly 0 ∈ S. If S = {0}, the S = 0ℤ is of the required form. Suppose it is not. If n ∈ S, the also −n ∈ S; so S must contain positive integer. Let m be the smallest positive integer in S. We will prove that S = nℤ. Since S ⊇ nℤ, it is sufficien to prove to opposite inclusion. Dividing h by m we have h = mq + r with 0 ≤ r < m. Since r ∈ S it must necessairly be r = 0 to avoid the contradiction of m being the smallest positive integer in S. Hence any element of S is a multiple of m.□

We've seen that there exist integral domain which are not fields such as ℤ. However there are no finite integral domains which are not fields as the following propisition proves.

6.2.4 Proposition. A finite integral domain (an integral domain with only a finite number of elements) is a field.

Proof. Let D be a finite integral domain with unity denoted by 1 and D* = {a₁,...,a_n} the nonzero elements of the integral domain. We must show that every nonzero element of the integral domain has a multiplicative inverse. Consider an arbitrary element a ∈ D* and define

Ψ: D* ⟶ D*

a₁ ⟶ aa₁

by Ψ(a_i) = aa_i. If Ψ(a_i) = Ψ(a_j) then aa_i = aa_j, so by the cancellation property, a_i = a_j hence the n products aa_i are all distinct and the set {aa₁, ...,aa_n} coincides with the set D*. Thus we see that Ψ is a one-to-one function. Since D* is finite, the mapping Ψ is onto D*. But one of the elements of D* is the multiplicative identity 1. Therefore, Ψ(a_k) = 1 for some a_k ∈ D*; that is aa_k = 1. This shows that a has a multiplicative inverse and hence D is a field. Note that since D is commutative that aa_j = a_ja.□

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