Could you explain how to show 8.32 in detail? I also attached definitions and theorems in my textbook. **8.28.** For each of the following rings, which elements are units? Which are zero divisors? 1. $ \mathbb{Z}_{18} $ 2. $ \mathbb{Z}_3 \oplus \mathbb{Z}_9 $**8.32.** Let $ R $ and $ S $ be rings. Under precisely what circumstances is $ R \oplus S $ an integral domain?**8.33.** Let $ F $ be a field with subfields $ K $ and $ L $. Show that $ K \cap L $ is a subfield of $ F $. Extend this to show that the intersection of any collection of subfields is a subfield.**8.34.** Let $ p $ be a prime and $ F $ a field with $ p^2 $ elements. Show that $ F $ cannot have more than one proper subfield. **Definition 8.8.** Let $ R $ be a commutative ring. Then a nonzero element $ a \in R $ is said to be a **zero divisor** if there exists a nonzero $ b \in R $ such that $ ab = 0 $.**Example 8.19.** In $ \mathbb{Z}_6 $, we note that 4 is a zero divisor, as $ 4 \cdot 3 = 0 $. On the other hand, 5 is not a zero divisor.**Example 8.20.** The ring of integers has no zero divisors.**Definition 8.9.** An **integral domain** is a commutative ring $ R $ with identity $ 1 \neq 0 $ having no zero divisors.**Example 8.21.** The rings $ \mathbb{Z}, \mathbb{Q}, \mathbb{R} $ and $ \mathbb{C} $ are all integral domains.**Example 8.22.** The polynomial ring $ \mathbb{R}[x] $ is an integral domain. Indeed, we know that it is a commutative ring with identity. Also, if $ f(x) = a_0 + a_1x + \cdots + a_nx^n $ and $ g(x) = b_0 + b_1x + \cdots + b_mx^m $, with $ a_i, b_i \in R $ and $ a_n \neq 0 \neq b_m $, then the unique term of highest degree in $ f(x)g(x) $ is $ a_nb_mx^{m+n} $. As $ R $ is an integral domain, $ a_nb_m \neq 0 $. Thus, $ f(x)g(x) $ is not the zero polynomial.**Example 8.23.** The rings $ \mathbb{Z}\mathbb{Z}, \mathbb{Z}_6 $ and $ M_2(\mathbb{R}) $ all fail to be integral domains. The first lacks an identity, the second has zero divisors and the third is not commutative.**Theorem 8.7 (Cancellation Law).** Let \( R \

Let R and S be rings. Find the conditions for R⊕S to be an integral domain. R⊕S is also a ring since R and S are rings. Also, if R and S are commutative rings with unity, then so is R⊕S. The unity element is 1R,1S. Suppose that a1,a2 is an element of R⊕S. Then a1,a2 is a zero divisor if and only if there exists an element b1,b2 in R⊕S such that b1,b2≠0R,0S and a1,a2b1,b2=0R,0S The second equation means that a1b1=0 and a2b2=0. Also b1,b2≠0R,0S gives that b1≠0R or b2≠0S. Consequently if a1,a2 is a zero divisor in R⊕S, then either a1 is a zero divisor of R or a2 is a zero divisor of S. Conversely, suppose that a1 is a zero divisor of R. Then a1b1=0R, for some nonzero element b1∈R. It follows that b1,0S≠0R,0S and a1,a2b1,0S=0R,0S Therefore, a1,a2 is a zero divisor of R⊕S. A similar argument shows that if a2 is a zero-divisor in S, then a1,a2 is a zero-divisor in R⊕S. Hence, a1,a2 is a zero divisor in R⊕S if and only if a1 is a zero divisor in R or a2 is a zero divisor in S, that is, a1,a2 is not a zero divisor in R⊕S if and only if a1 is not a zero divisor in R and a2 is not a zero divisor in S. Hence R⊕S is an integral domain if and only if both the rings R and S are integral domains.

Answered: 8.32. Let R and S be rings. Under…

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8.32. Let R and S be rings. Under precisely what circumstances is ROS an integral domain?

Advanced Engineering Mathematics

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ISBN:9780470458365

Author:Erwin Kreyszig

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Could you explain how to show 8.32 in detail? I also attached definitions and theorems in my textbook.

**8.28.** For each of the following rings, which elements are units? Which are zero divisors?
1. $ \mathbb{Z}_{18} $
2. $ \mathbb{Z}_3 \oplus \mathbb{Z}_9 $

**8.32.** Let $ R $ and $ S $ be rings. Under precisely what circumstances is $ R \oplus S $ an integral domain?

**8.33.** Let $ F $ be a field with subfields $ K $ and $ L $. Show that $ K \cap L $ is a subfield of $ F $. Extend this to show that the intersection of any collection of subfields is a subfield.

**8.34.** Let $ p $ be a prime and $ F $ a field with $ p^2 $ elements. Show that $ F $ cannot have more than one proper subfield.

**Definition 8.8.** Let $ R $ be a commutative ring. Then a nonzero element $ a \in R $ is said to be a **zero divisor** if there exists a nonzero $ b \in R $ such that $ ab = 0 $.

**Example 8.19.** In $ \mathbb{Z}_6 $, we note that 4 is a zero divisor, as $ 4 \cdot 3 = 0 $. On the other hand, 5 is not a zero divisor.

**Example 8.20.** The ring of integers has no zero divisors.

**Definition 8.9.** An **integral domain** is a commutative ring $ R $ with identity $ 1 \neq 0 $ having no zero divisors.

**Example 8.21.** The rings $ \mathbb{Z}, \mathbb{Q}, \mathbb{R} $ and $ \mathbb{C} $ are all integral domains.

**Example 8.22.** The polynomial ring $ \mathbb{R}[x] $ is an integral domain. Indeed, we know that it is a commutative ring with identity. Also, if $ f(x) = a_0 + a_1x + \cdots + a_nx^n $ and $ g(x) = b_0 + b_1x + \cdots + b_mx^m $, with $ a_i, b_i \in R $ and $ a_n \neq 0 \neq b_m $, then the unique term of highest degree in $ f(x)g(x) $ is $ a_nb_mx^{m+n} $. As $ R $ is an integral domain, $ a_nb_m \neq 0 $. Thus, $ f(x)g(x) $ is not the zero polynomial.

**Example 8.23.** The rings $ \mathbb{Z}\mathbb{Z}, \mathbb{Z}_6 $ and $ M_2(\mathbb{R}) $ all fail to be integral domains. The first lacks an identity, the second has zero divisors and the third is not commutative.

**Theorem 8.7 (Cancellation Law).** Let \( R \

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