Subrings

Given a ring (R,+,*), a subring (S,+,*) is a non-empty subset S of the set R that satisfies all the properties of a ring with respect to the same operations + and *.

In the case of subrings, the first operation (addition) must form an additive group (S,+) with the subset S.

How to determine if a subset forms a subring

To verify whether the subset S⊂R forms a subring (S,+,*) of the ring (R,+,*), it is sufficient to check that (S,+) is a group.

Note: The first operation of rings is referred to as the "additive operation" (or addition), even though it may not strictly be addition. Therefore, we often say that (S,+) is an additive group. Since (R,+,*) is a ring, the structure (R,+) is also an additive group. Hence, (S,+) is a subgroup of (R,+).

An example

Consider the ring formed by the set of real numbers R with addition (+) and multiplication (·)

$$ (R,+,·) $$

We want to check if the set of integers Z forms a subring (Z,+,·) of the ring (R,+,·) under the same operations.

$$ (Z,+,·) $$

The first condition is satisfied since Z is a subset of R

$$ Z \subset R $$

Now, we need to verify whether the algebraic structure (Z,+,·) satisfies the properties of rings.

• The set of integers Z is closed under the operations + and ·
• Addition is commutative in the set of integers Z $$ \forall \ a,b \in Z \ \ \ \ a+b=b+a $$
• Addition is associative in Z $$ \forall \ a,b,c \in Z \ \ \ \ (a+b)+c = a+(b+c) $$
• The additive identity, 0, exists in Z $$ \forall \ a \ \in Z \ \ \ \ a+0 = 0+a = a $$
• Each element in Z has an additive inverse (-a) $$ \forall \ a \ \in Z \ \ \ \ a+(-a) = (-a)+a = 0 $$
• Multiplication is associative in Z $$ \forall \ a,b,c \in Z \ \ \ \ (a \cdot b) \cdot c = a \cdot (b \cdot c) $$
• Multiplication distributes over addition in Z $$ \forall \ a,b,c \in Z \ \ \ \ a \cdot (b+c) = a \cdot b + a \cdot c $$ $$ \forall \ a,b,c \in Z \ \ \ \ (a + b ) \cdot c = a \cdot c + b \cdot c $$

The algebraic structure (Z,+,·) satisfies all the properties of a ring under both addition and multiplication.

Therefore, the algebraic structure (Z,+,·) is a subring of (R,+,·).

Example 2

The set of polynomials P[x]={xⁿ+...+x=b} of any degree n over the field of real numbers R forms a ring (R,+,·) under addition and multiplication.

$$ (P[x],+,·) $$

The subset P₀⊂P, consisting of polynomials with a zero constant term P₀[x]={x₂+...x=0}, forms a subring

$$ (P_0[x],+,·) $$

A faster verification method

Given a ring (R,+,*) and a subset S of R, to check if it forms a subring (S,+,*), ensure it satisfies these conditions: $$ \forall \ a,b \in S \ \ \ \ a-b \in S $$ $$ \forall \ a,b \in S \ \ \ \ ab \in S $$

Example

Consider the ring (R,+,·) made up of the set of real numbers R, under addition and multiplication

$$ (R,+,·) $$

We now check if the subset of integers Z satisfies these two conditions

$$ Z \subset R $$

The first condition is met

If we take any two integers, their difference a-b is an integer.

$$ \forall \ a,b \in Z \ \ \ \ a-b \in Z $$

The second condition is also met

If we take any two integers, their product is an integer.

$$ \forall \ a,b \in Z \ \ \ \ ab \in Z $$

Thus, the set Z is a subring (Z,+,·) of the ring (R,+,·).

Proof. Assuming the following statements are true: $$ \forall \ a,b \in S \ \ \ \ a-b \in S $$ $$ \forall \ a,b \in S \ \ \ \ ab \in S $$ where S is a subset of R, and R forms a ring (R,+,*) under addition and multiplication. From these assumptions, we can deduce the existence of an additive identity in S, because if we set a=b, the result is 0: $$ a-b = 0 \ \ \ \ if \ a=b$$. Since the additive identity (0) exists in S, every element in S must also have an additive inverse: $$ 0-a = -a \in S \ \ \ \ \forall \ a \in S $$ Since every element in S has an additive inverse, S is closed under addition: $$ a-(-b) = a+b \in S \ \ \ \ \forall \ a,b \in S $$ From this, we can also deduce that the associative property holds. Thus, the structure (S,+) forms an additive group, and this is enough to conclude that (S,+,*) is a subring of (R,+,*).

Trivial subrings

Example

Consider the ring (R,+,·) made up of the set of real numbers R under addition and multiplication

$$ (Z,+,·) $$

The subset S ={0}⊂R forms a trivial subring (S,+,·) of the ring (R,+,·).

$$ (S,+,·) $$

The set S={0} satisfies all the ring operations

• The set S={0} is closed under both addition and multiplication $$ 0 + 0 = 0 \in S $$ $$ 0 \cdot 0 = 0 \in S $$
• Addition is commutative in the set S={0} $$ \forall \ a,b \in S \ \ \ \ a+b=b+a $$
• Addition is associative in the set S={0} $$ \forall \ a,b,c \in Z \ \ \ \ (a+b)+c = a+(b+c) $$
• The additive identity exists in S={0} and is zero $$ \forall \ a \ \in S \ \ \ \ a+0 = 0+a = a $$
• Each element in S={0} has an additive inverse, which is also zero. If a=0, the inverse is -a=0 $$ \forall \ a \ \in S \ \ \ \ a+(-a) = (-a)+a = 0 $$
• Multiplication is associative in the set S={0} $$ \forall \ a,b,c \in S \ \ \ \ (a \cdot b) \cdot c = a \cdot (b \cdot c) $$
• Multiplication is distributive over addition in S={0} $$ \forall \ a,b,c \in S \ \ \ \ a \cdot (b+c) = a \cdot b + a \cdot c $$ $$ \forall \ a,b,c \in S \ \ \ \ (a + b) \cdot c = a \cdot c + b \cdot c $$

 

Thus, the subset S={0} forms a trivial subring (S,+,·) of the ring (R,+,·) under the same operations.

Note: The same applies to the improper subset S=R, which is identical to the set R. In this case, the sets S and R contain the same elements, so if R forms a ring under addition and multiplication, so does S=R.

And so forth.