Consider the ideal <0 > of Z. Then <0 > is
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We begin with the definition of an ideal in a commutative ring. Ideals in non-commutative
rings can be defined but we will not study them here.
Definition 1.1.1. Let R be a commutative ring. A subset of I of R is an ideal of R if
(i) 0R ∈ I,
(ii) if a, b ∈ I then a − b ∈ I, and
(iii) if a ∈ I and r ∈ R then ar = ra ∈ I.
Observe that the set I = {0R} is an ideal of R. Also, the entire ring R is an ideal of R.
These two ideals are called the trivial ideals. An example of non-trivial ideal is the set of
even integers.
2Z = {. . . , −4, −2, 0, 2, 4, . . . }.
This is an ideal in Z because if a, b are even integers, and r is any integer, we have a − b
is even and ar is even. Now the even integers are also a subring of Z. There is a relation
between ideals and subrings, namely, all ideals are subrings but not all subrings are ideals.
We recall the test for a subring of a ring. See Lemma 2.2.4.
Subring Test Let R be a ring. A subset of S of R is a subring of R if
(1) S is not empty,
(2) if a, b ∈ S then −a ∈ S and a + b ∈ S, and
(3) if a, b ∈ S then ab ∈ S.
Some texts will have 0R ∈ S instead of S is not empty for condition (1). The two
conditions are in fact equivalent. Furthermore condition (2) in the test for a subring and
condition (ii) in the definition for an ideal are also equivalent conditions. Let us state and
prove this formally.
Lemma 1.1.2. Let S and I be as above. Then (i) S is closed under subtraction and (ii) I
is closed under addition and negation.