Ideals of a Ring

An ideal of a ring (R,+,*) is a subset I∈R that forms an additive subgroup (I,+) of R, such that for every i∈I and every r∈R, either $$ i \cdot r \in I $$ or $$ r \cdot i \in I $$ holds. In the first case, where i·r ∈ I, we call I a right ideal; in the second, where r·i ∈ I, it is a left ideal.

If both conditions are satisfied, I is called a two-sided ideal (or bilateral ideal).

When the ring (R,+,*) is commutative, the distinction becomes unnecessary, and we simply refer to I as an ideal.

Note: Every left or right ideal of a ring R is also a subring of R. However, the converse does not hold: a subring of R is not necessarily a left or right ideal.

A Practical Example
Trivial Ideals
Remarks

A Practical Example

Example 1

Consider the ring of integers:

$$ (Z,+,*) $$

The subset P⊂Z consisting of all even integers is an ideal of the commutative ring (Z,+,*), since the product of any even number with any integer is again even:

$$ \forall \ p \in P \ , \ z \in Z \ \Rightarrow p \cdot z = z \cdot p \in P $$

Moreover, P forms an additive subgroup (P,+) of Z: it is closed under addition, addition is associative, 0 (the additive identity) belongs to P, and for every p∈P, the additive inverse -p also belongs to P.

Example 2

For any natural number n∈N, the subset nZ∈Z - that is, all integer multiples of n - is an ideal of the ring (Z,+,*).

For instance, when n = 2:

$$ 2Z = \{ ..., -8, -6, -4, -2, \ 0, \ 2, \ 4, \ 6, \ 8, ... \} $$

This subset is an additive group (2Z,+): it satisfies associativity, contains the additive identity (0), and every x∈2Z has an additive inverse -x∈2Z.

Additionally, the product of any element from 2Z with any integer from Z remains in 2Z:

$$ \forall \ x \in 2Z \ , \ z \in Z \ \Rightarrow x \cdot z = z \cdot x \in 2Z $$

Thus, 2Z is indeed an ideal of the ring (Z,+,*).

Example 3

The set of n-by-n square matrices whose last column is zero forms a left ideal in the ring of all n-by-n matrices. However, it is not a right ideal.

For example, consider the set of 2×2 matrices: $$ M_2 = \begin{pmatrix} a & b \\ c & d \end{pmatrix} $$ and the subset of matrices with a zero last column: $$ M'_2 = \begin{pmatrix} e & 0 \\ f & 0 \end{pmatrix} $$ The product M₂M'₂ results in a matrix that still lies in M'₂, hence M'₂ is closed under left multiplication and is a left ideal: $$ M_2 \cdot M'_2 = \begin{pmatrix} ae + bf & 0 \\ ce + df & 0 \end{pmatrix} $$ On the other hand, $$ M'_2 \cdot M_2 = \begin{pmatrix} ae & eb \\ af & bf \end{pmatrix} $$ is generally not in M'₂, since the resulting matrix does not have a zero last column. Therefore, M'₂ is not a right ideal.

Example 4

Let K[x] denote the ring of all polynomials with real coefficients (i.e., K = ℝ).

The set K₀[x] of polynomials whose constant term is zero forms an ideal of the ring (P,+,*), for the following reasons:

K₀[x] is a subset of K[x]: $$ K_0[x] \subset K[x] $$
K₀[x] forms an additive subgroup: it is closed under addition, contains the zero polynomial (which has constant term zero), and is closed under taking additive inverses.
The product of any polynomial in K[x] with any polynomial in K₀[x] is still in K₀[x].
Example: Consider $$ x^2 + 1 \in K[x] $$ and $$ x^2 + x \in K_0[x] $$ Their product is: $$ (x^2 + 1)(x^2 + x) = x^4 + x^3 + x^2 + x \in K_0[x] $$ which again has zero constant term.

Hence, K₀[x] is an ideal of the ring (P,+,*).

Trivial Ideals

Every ring (R,+,*) always has two trivial ideals: the subset containing only the additive identity {0}, and the ring itself R.

Example

Consider the commutative ring of real numbers:

$$ (R,+,*) $$

The trivial ideals in this ring are:

The full set R⊆R, which is an ideal since it is an additive group and closed under multiplication.
The singleton set {0}, which is also an ideal, as it trivially satisfies all the required properties.

Remarks

Some additional observations about ideals:

The kernel (Ker φ) of a ring homomorphism from (R,+,*) to (R',+,*) is always a two-sided ideal.
Proof: For any k∈Ker_φ and r∈R, we have $$ \phi(k) \cdot \phi(r) = 0 \cdot \phi(r) = 0 $$ $$ \phi(r) \cdot \phi(k) = \phi(r) \cdot 0 = 0 $$ Therefore, $$ 0 \in \ Ker \ \phi $$ showing that Ker_φ is closed under both left and right multiplication by elements of R.

And so on.