3.4 Ideal
Definition 3.4.1 (IDEAL). A subring \(A\) of a ring \(R\) is called a \((\) two-sided\()\) ideal of \(R\) if for every \(r\in R\)
and every \(a\in A\) both \(ar\) and \(ra\) are in \(A\).
A subring \(A\) of \(R\) is an ideal if it absorbs elements of \(R\). i.e \(A\, r\subseteq A\, , \hspace {0.2cm} r\, A \subseteq A\,\) for all \(r\in R\).
Theorem 3.4.2 (IDEAL TEST). A non-empty subset \(A\) of a ring \(R\) is an ideal of \(R\) if and only if
- 1.
- \(a - b\in A\) for all \(a\, ,\, b \in A\)
- 2.
- \(ra\) and \(ar\) are in \(A\) for all \(a\, , \, b\in A\, , \,\, r\in R\).
Proof. If \(A\) is an ideal then it is an additive subgroup, so it is closed under differences, and it absorbs multiplication by \(R\) on both sides; that is conditions (1) and (2).
Conversely, suppose (1) and (2) hold. Condition (1) is the subgroup test for the additive group: \(A\) is non-empty, so taking \(a=b\) gives \(0\in A\), then \(a=0\) gives \(-b\in A\), and \(a-(-b) = a+b\) gives closure under addition. So \(A\) is an additive subgroup. Condition (2) is exactly the absorbing property. Hence \(A\) is an ideal. □
Note. Only differences need checking, not sums and inverses separately — that is the economy of the subgroup test, and it is why the criterion is stated this way. Note also that (2) demands absorption on both sides; in a non-commutative ring a subset absorbing on one side only is a one-sided ideal, and the quotient construction fails for it.
- 1.
- For any ring \(R\), \(\, \{0\}\) and \(R\) are ideals in \(R\cdot \{0\}\) is called the trivial ideal.
- 2.
- Let \(R= \mathbb {Z}\) and \(n\mathbb {Z} = \big \{0\, ,\, \pm n\, , \, \pm 2n\, , \cdots \cdots \cdots , \pm \big \}\) is an ideal of \(\mathbb {Z}\)
- Clearly, \(0\in n\mathbb {Z}\,\) so \(\,n\mathbb {Z}\neq \emptyset \).
- Let \(a\, , \, b\in n\mathbb {Z}\), then \(a = kn\, , \, b = ln\,\) where \(k\, , \, l\in \mathbb {Z}. \, a - b = kn - ln = n(k-l) \in n\mathbb {Z}\)
- \(ra = r(kn) = (rk)n\in n\mathbb {Z}\) and \(\, ar = (kn)r = (kr)n\in n\mathbb {Z}\)
Thus \(\, n\mathbb {Z}\,\) is an ideal in \(\mathbb {Z}\).
- 3.
- \(R = \mathbb {Z}[x]\), \(\, I = \) subset of all polynomials with even constant term (has integral coefficients)
- \(x + 2 \in I\) so \(I\neq \emptyset \)
- Let \(f(x)\, , \, g(x) \in I\), then \(f(x)\, , \, g(x)\) are of the form; \[f(x) = 2k + a_1x + a_2 x^2 + \cdots \cdots + a_n x^n\, , \, \, a_i\in \mathbb {Z}\] \[g(x) = 2l + b_1 x + b_2 x^2 + \cdots \cdots + b_n x^n\, ,\,\, b_i\in \mathbb {Z}\] Then \(\, f(x) - g(x) = 2(k-l) + (a_1 - b_1)x + (a_2 - b_2)x^2 + \cdots \cdots + (a_n - b_n) x^n \in I\)
- Consider \(\, r = h(x) \in R\,\) given by \(h(x) = m + c_1 x + c_2 x^2 + \cdots \cdots + c_n x^n\, , \, c_i\in \mathbb {Z}\) \[f(x)h(x) = \big (2k + a_1 x + a_2 x^2 + \cdots \cdots + a_n x^n\big )\big (m + c_1 x + c_2 x^2 + \cdots \cdots + c_nx^n\big )\] has constants term 2km. Thus \(f(x)h(x) \in I\). Thus \(I\) is an ideal of \(\mathbb {Z}[x]\).
Exercise 3.4.4. Show that the subring \(\mathbb {Z}\) is not an ideal in \(\mathbb {Q}\).
- Clearly \(\, \mathbb {Z} \neq \emptyset \, \) since \(0\in \mathbb {Z}\)
Let \(\, r = \dfrac {a}{b}\in \mathbb {Q}\, , \hspace {0.2cm} x\, , \, y\in \mathbb {Z}\)
- 1.
- \(x - y \in \mathbb {Z}\hspace {0.5cm} \forall \, x\, , \, y \in \mathbb {Z}\)
- 2.
- \(rx = \dfrac {a}{b}\, x\not \in \mathbb {Z}\hspace {0.5cm}\forall \, x\in \mathbb {Z}\, , \, \dfrac {a}{b}\in \mathbb {Q}\)
Likewise
\(x r = x\cdot \dfrac {a}{b}\not \in \mathbb {Z}\hspace {0.5cm} \forall \, x \in \mathbb {Z}\)
\(\therefore \,\, \mathbb {Z}\) is not an ideal of \(\mathbb {Q}\).
Proof. Suppose \(I\) is ideal in \(F\) and \(I\neq \{0\}\). We show that \(I = F\). Let \(a\in I\, , \, a\neq 0\). Then \(a\) has an inverse \(a^{-1} \in F\) since \(F\) is a field.
Let 1 be the unity in \(F\), then \(1 = a\, a^{-1} \in I\) since \(I\) is an ideal. But if \(r\) is any element of \(F\), then \(r = 1\cdot r\in I\). Then \(I = F\).
□
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