Ideal

Definition m. Let \(R\) be a commutative ring and \(\emptyset \ne I \subseteq R\). Then \(I\) is an ideal of \(R\) if:

The sum \(a + b\) is in \(I\) for all \(a, b \in I\).
The product \(ar\) is in \(I\) for all \(a \in I\) and \(r \in R\).

The ideal \(I\) is proper if \(I \ne R\).

Remark c. Definition m says that if \(R\) is a commutative ring, and \(\emptyset \neq I \subseteq R\), then \(I\) is an ideal of \(R\) if:

The set \(I\) is closed under addition.
The set \(I\) is closed under multiplication by any \(r \in R\).

Proposition p. Let \(I\) be an ideal of a commutative ring \(R\). Then:

The equation \(a \cdot 0 = 0 \cdot a = 0\) holds for all \(a \in R\).
The element \(0\) is in \(I\).
If \(1 \in I\) then \(I\) is not proper.

Proof. (i) Suppose \(a \in R\). Then \(a + a \cdot 0 = a \cdot 1 + a \cdot 0 = a(1 + 0) = a \cdot 1 = a\). Adding the additive inverse of \(a\) on both sides gives \(a \cdot 0 = 0\). Since \(R\) is commutative, \(0 \cdot a = 0\) also holds. (ii) By the ring axioms, \(0 \in R\), and by the ideal definition, \(0 \cdot a = 0 \in I\) for any \(a \in I\). (iii) Since \(1\) is the multiplicative identity, \(1 \cdot r = r \in I\) for all \(r \in R\), and thus \(I = R\).

Remark dm. There is an analogy from modular arithmetic that illustrates an intuitive view of ideals. Consider the ring \(\mathbb{Z}_n\) of integers modulo \(n \in \mathbb{Z}\). The exact set of integers that we identify with \(0\) in \(\mathbb{Z}_n\) is the set \(n\mathbb{Z} = \{nm \mid m \in \mathbb{Z}\}\). This set meets the criteria for being an ideal of \(\mathbb{Z}\) and its elements behave like \(0\) in \(\mathbb{Z}\): adding two elements of \(n\mathbb{Z}\) yields another element of \(n\mathbb{Z}\) and multiplying any element of \(n\mathbb{Z}\) again yields an element of \(n\mathbb{Z}\).

Considering our definition of rings, an ideal might not be a ring itself. For example, consider the ring of integers and its ideal consisting of even numbers. This ideal is not a ring since it has no multiplicative identity.

Definition bw. Let \(I\) be an ideal of a ring \(R\). The coset of \(I\) defined by \(a \in R\) is the set \(\{b + a \mid b \in I\}\), denoted \(I + a\) or \([a]\).

Remark bi. The addition of cosets is defined by \((I + a) + (I + a’) = I + (a + a’)\) and multiplication by \((I + a)(I + a’) = I + aa’\). The set of all cosets forms a ring under these operations (Shoup 2009). This ring is denoted \(R / I\) and called the quotient ring.

Definition bs. Let \(\{f_1, \ldots, f_s\} \subset \mathbb{F}[\mathbf{x}]\) be a set of polynomials. Then \[\langle f_1, \ldots, f_s \rangle = \left\{\sum_{i=1}^s h_i f_i \mid h_1, \ldots, h_s \in \mathbb{F}[\mathbf{x}]\right\}.\] The set \(\langle f_1, \ldots, f_s \rangle\)is an ideal of \(\mathbb{F}[\mathbf{x}]\) called the ideal generated by \(\{f_1, \ldots, f_s\}\). The set \(\{f_1, \ldots, f_s\}\) is a basis of \(\langle f_1, \ldots, f_s \rangle\).

Proof. Assume \(f = \sum_{i=1}^s p_i f_i\) and \(g = \sum_{i=1}^s q_i f_i\) are polynomials, and let \(h \in \mathbb{F}[\mathbf{x}]\). Then \[f + g = \sum_{i=1}^s (p_i + q_i) f_i \quad \text{and} \quad hf = \sum_{i=1}^s (hp_i) f_i\] show that \(\langle f_1, \ldots, f_s \rangle\)meets the criteria for being an ideal of \(\mathbb{F}[\mathbf{x}]\).

Remark dm provides an intuitive view of ideals through modular arithmetic. Another analogy comes from linear algebra where the definition of subspaces can be likened to the definition of ideals of polynomial rings. Both are closed under addition. Subspaces are closed under multiplication by scalars while ideals of polynomial rings are closed under multiplication by polynomials.

Proposition dc. Let \(I \subseteq \mathbb{F}[\mathbf{x}]\) be an ideal, and let \(\{f_1, \ldots, f_s\} \subset \mathbb{F}[\mathbf{x}]\) be a set of polynomials. Then \(\langle f_1, \ldots, f_s \rangle \subseteq I\) if and only if \(\{f_1, \ldots, f_s\} \subseteq I\).

Proof. (\(\implies\)) Assume \(\langle f_1, \ldots, f_s \rangle \subseteq I\). Each \(f_i\) can be constructed as \(f_i = 0 \cdot f_1 + \cdots + 1 \cdot f_i + \cdots + 0 \cdot f_s\), and hence \(\{f_1, \ldots, f_s\} \subseteq I\). (\(\impliedby\)) Assume \(\{f_1, \ldots, f_s\} \subseteq I\) and choose any \(f \in \langle f_1, \ldots, f_s \rangle\) so that \(f = h_1 f_1 + \cdots + h_s f_s\) where each \(h_i \in \mathbb{F}[\mathbf{x}]\). We see that \(f \in I\) since \(I\) is an ideal and so \(\langle f_1, \ldots, f_s \rangle \subseteq I\).

Example dp. Consider the ideals \(\langle x, y \rangle\)and \(\langle x + y, x - y \rangle\)in the polynomial ring \(\mathbb{Q}[x, y]\). We will show that \(\langle x, y \rangle = \langle x + y, x - y \rangle\). We see that \(x + y \in \langle x, y \rangle\)and \(x - y \in \langle x, y \rangle\), so by Proposition dc, \(\langle x + y, x - y \rangle \subseteq \langle x, y \rangle\). Similarly, both \(x = \frac{1}{2}(x + y) + \frac{1}{2}(x - y)\) and \(y = \frac{1}{2}(x + y) - \frac{1}{2}(x - y)\) are in \(\langle x + y, x - y \rangle\), so by Proposition dc, \(\langle x, y \rangle \subseteq \langle x + y, x - y \rangle\) and the equality follows.

References

Shoup, Victor. 2009. A Computational Introduction to Number Theory and Algebra. Cambridge university press.