Let \(R\) be a commutative ring with identity \(1 \ne 0\).
For \(r \in R\), show that the multiplication map \[\mu_r: R \to R, \quad x \mapsto r x\] is an \(R\)-module endomorphism of \(R\).
An element \(r \in R\) is called a zero-divisor if \(\mu_r\) is not injective (i.e. \(\ker \mu_r \ne \{0\}\)). Show that if \(r\) is a non-zero zero-divisor (\(r \ne 0\)), then the kernel \(\ker \mu_r\) and the image \(\operatorname{im} \mu_r\) each consist entirely of zero-divisors.
Let \(n \ge 2\) be an integer. Show that if \(R\) has exactly \(n\) zero-divisors, then \(|R| \le n^2\).
Show that up to isomorphism, there are exactly two commutative rings with identity having precisely 2 zero-divisors.
(You may use without proof that every ring of order 4 is isomorphic to exactly one of \(\mathbb{Z}/4\mathbb{Z}\), \(\mathbb{F}_2[t]/\langle t^2 + t + 1 \rangle\), \(\mathbb{F}_2[t]/\langle t^2 - t \rangle\), or \(\mathbb{F}_2[t]/\langle t^2 \rangle\).)
Solution
Goal: Prove that \(\mu_r \in \operatorname{End}_R(R)\) in (a), that kernel and image of a non-zero zero-divisor consist of zero-divisors in (b), that \(|R| \le n^2\) in (c), and classify the rings with 2 zero-divisors in (d).
1. Part (a): \(\mu_r\) is an \(R\)-module endomorphism.
1.1. Additivity: For all \(x, y \in R\), by distributivity in \(R\):\[\mu_r(x + y) = r (x + y) = r x + r y = \mu_r(x) + \mu_r(y).\]
1.2. Scalar linearity: For all \(s \in R\) and \(x \in R\), by associativity and commutativity of multiplication in \(R\):\[\mu_r(s x) = r (s x) = (r s) x = (s r) x = s (r x) = s \mu_r(x).\]
1.3. Thus \(\mu_r\) is an \(R\)-module homomorphism from \(R\) to \(R\), so \(\mu_r \in \operatorname{End}_R(R)\).
2. Part (b): \(\ker \mu_r\) and \(\operatorname{im} \mu_r\) consist of zero-divisors.
2.1. Let \(Z(R) \subseteq R\) denote the set of all zero-divisors in \(R\).
2.2. Kernel containment \(\ker \mu_r \subseteq Z(R)\):
Let \(x \in \ker \mu_r\), so \(r x = 0\). Since \(R\) is commutative, \(x r = 0\). Since \(r \ne 0\), the non-zero element \(r\) lies in \(\ker \mu_x\), so \(\ker \mu_x \ne \{0\}\), which means \(x \in Z(R)\).
2.3. Image containment \(\operatorname{im} \mu_r \subseteq Z(R)\):
Let \(y \in \operatorname{im} \mu_r\), so \(y = r x\) for some \(x \in R\). Since \(r \in Z(R) \setminus \{0\}\), there exists a non-zero element \(z \in R \setminus \{0\}\) such that \(r z = 0\). Multiply \(y\) by \(z\): \[y z = (r x) z = (x r) z = x (r z) = x \cdot 0 = 0.\] Since \(z \ne 0\), \(z \in \ker \mu_y\), so \(\ker \mu_y \ne \{0\}\), which means \(y \in Z(R)\).
3. Part (c): \(|R| \le n^2\) if \(|Z(R)| = n \ge 2\).
3.1. Since \(n \ge 2\), \(Z(R)\) contains \(0\) and at least one non-zero zero-divisor \(r \ne 0\).
3.2. By Part (b), \(\ker \mu_r \subseteq Z(R)\) and \(\operatorname{im} \mu_r \subseteq Z(R)\).
3.3. Since \(|Z(R)| = n\), we obtain the cardinality bounds:\[|\ker \mu_r| \le n \quad \text{and} \quad |\operatorname{im} \mu_r| \le n.\]
3.4. By the First Isomorphism Theorem for modules, \(R/\ker \mu_r \cong \operatorname{im} \mu_r\).
3.5. Therefore:\[|R| = [R : \ker \mu_r] \cdot |\ker \mu_r| = |\operatorname{im} \mu_r| \cdot |\ker \mu_r| \le n \cdot n = n^2.\]
4. Part (d): Rings with precisely 2 zero-divisors.
4.1. Suppose \(|Z(R)| = 2\).
4.2. The element \(1 \in R\) is never a zero-divisor (since \(1 \cdot x = 0 \implies x = 0\)).
4.3. Thus \(1 \notin Z(R)\), which means \(R\) contains at least \(0, 1\), and the non-zero zero-divisor \(r\). Hence \(|R| \ge 3\).
4.4. By Part (c), \(|R| \le 2^2 = 4\), so \(|R| \in \{3, 4\}\).
4.5. If \(|R| = 3\), \(R \cong \mathbb{Z}/3\mathbb{Z}\) is a field, which has only 1 zero-divisor (\(Z(\mathbb{Z}/3\mathbb{Z}) = \{0\}\)), a contradiction.
4.6. Thus \(|R| = 4\).
4.7. Test the four isomorphism classes of rings of order 4:
4.7.1. \(R_1 \cong \mathbb{Z}/4\mathbb{Z}\): The elements are \(\{0, 1, 2, 3\}\). The zero-divisors are \(\{0, 2\}\) because \(2 \cdot 2 = 0\) in \(\mathbb{Z}/4\mathbb{Z}\), while \(1\) and \(3\) are units. Thus \(|Z(R_1)| = 2\).
4.7.2. \(R_2 \cong \mathbb{F}_2[t]/\langle t^2 + t + 1 \rangle\): Since \(t^2 + t + 1\) is irreducible over \(\mathbb{F}_2\), \(R_2 \cong \mathbb{F}_4\) is a field, so \(|Z(R_2)| = |\{0\}| = 1 \ne 2\).
4.7.3. \(R_3 \cong \mathbb{F}_2[t]/\langle t^2 - t \rangle \cong \mathbb{F}_2 \times \mathbb{F}_2\): The zero-divisors are \(\{(0, 0), (1, 0), (0, 1)\}\), so \(|Z(R_3)| = 3 \ne 2\).
4.7.4. \(R_4 \cong \mathbb{F}_2[t]/\langle t^2 \rangle\): The elements are \(\{0, 1, t, 1+t\}\). Since \(t^2 = 0\), \(t\) is a zero-divisor. The element \(1+t\) satisfies \((1+t)^2 = 1 + 2t + t^2 = 1\), so \(1+t\) is a unit (hence not a zero-divisor). Thus \(Z(R_4) = \{0, t\}\), so \(|Z(R_4)| = 2\).
4.8. Thus the only rings with exactly 2 zero-divisors are \(\mathbb{Z}/4\mathbb{Z}\) and \(\mathbb{F}_2[t]/\langle t^2 \rangle\).
5. Conclusion:
All parts are established, and the two rings with precisely two zero-divisors are $\mathbb{Z}/4\mathbb{Z}$ and $\mathbb{F}_2[t]/\langle t^2 \rangle$.