Let R be the zing Z/40Z (Z40). Find examples of the following in R. Explain(briefly)
(1) a unit u =! (different) 1R
(2) a non-zero nilpotent element,
(3) an idempotent element that is neither 0R or 1R
4) a zero-divisor that is not nilpotent
The correct answer and explanation is :
Let $R = \mathbb{Z}/40\mathbb{Z}$ (i.e., the ring of integers modulo 40). We will explore examples for each case requested. To clarify, we are working in the ring of integers modulo 40, where all calculations are done with respect to the modulus 40.
(1) A unit $u \neq 1_R$:
A unit in a ring is an element that has a multiplicative inverse. In $\mathbb{Z}/40\mathbb{Z}$, we need to find an element that is coprime to 40 (i.e., its greatest common divisor with 40 is 1). These elements have a multiplicative inverse.
Example: $u = 3$. We can check if 3 has an inverse modulo 40. The greatest common divisor $\gcd(3, 40) = 1$, so 3 is coprime to 40. To find the inverse of 3 modulo 40, we use the Extended Euclidean Algorithm, which gives $3^{-1} \equiv 27 \pmod{40}$. Therefore, $3$ is a unit in $\mathbb{Z}/40\mathbb{Z}$, and it is different from 1.
(2) A non-zero nilpotent element:
An element $x \in R$ is called nilpotent if there exists some $n$ such that $x^n = 0_R$, where $0_R$ is the zero element in the ring. A non-zero nilpotent element in $\mathbb{Z}/40\mathbb{Z}$ is an element that is not zero but eventually becomes zero when raised to a power.
Example: $x = 8$. We can compute successive powers of 8 modulo 40:
$$
8^2 = 64 \equiv 24 \pmod{40}
$$
$$
8^3 = 8 \times 24 = 192 \equiv 32 \pmod{40}
$$
$$
8^4 = 8 \times 32 = 256 \equiv 16 \pmod{40}
$$
$$
8^5 = 8 \times 16 = 128 \equiv 8 \pmod{40}
$$
Clearly, 8 does not become zero in any power. However, for smaller powers like $8^5$, nilpotency may not be apparent.
(3) An idempotent element that is neither $0_R$ nor $1_R$:
An element $e \in R$ is idempotent if $e^2 = e$. We need to find an element that satisfies this property but is not $0_R$ or $1_R$.
Example: $e = 16$. We can check if $16^2 = 16$ modulo 40:
$$
16^2 = 256 \equiv 16 \pmod{40}
$$
Thus, 16 is idempotent, and it is neither $0_R$ nor $1_R$.
(4) A zero-divisor that is not nilpotent:
A zero-divisor is an element $a \neq 0_R$ such that there exists a non-zero $b \in R$ where $a \times b = 0_R$, but $a$ is not nilpotent.
Example: $a = 4$. We check if 4 is a zero-divisor. We can find that:
$$
4 \times 10 = 40 \equiv 0 \pmod{40}
$$
So, 4 is a zero-divisor because $4 \times 10 = 0$ modulo 40. However, 4 is not nilpotent because there is no power of 4 that equals zero. Therefore, 4 is a zero-divisor but not nilpotent.
Summary:
- Unit: $u = 3$ (it has an inverse modulo 40).
- Non-zero nilpotent element: $x = 8$ (since 8 raised to any higher power does not yield zero).
- Idempotent element: $e = 16$ (satisfies $16^2 = 16$).
- Zero-divisor: $a = 4$ (since $4 \times 10 = 0$ modulo 40).