Commutative Ring Check E79071

1. **Determine whether $(\mathbb{Z}, +, \cdot)$ with $x \cdot y = 0$ for all $x,y \in \mathbb{Z}$ is a commutative ring.** 2. **Show that a ring $D$ is an integral domain if and only if it obeys cancellation laws.** 3. **Prove that an ideal of a ring is a subring of the ring.** 4. **Show that $S = \left\{ \begin{pmatrix}0 & x \\ 0 & y \end{pmatrix} : x,y \in \mathbb{Z} \right\}$ is a subring of $M_2(\mathbb{Z})$, determine if it is unital, and if it is an ideal of $M_2(\mathbb{Z})$.** 5. **Explain why zero divisors, nilpotent, units, and idempotent elements are preserved under a ring isomorphism $f : R \to R$.** --- ### 1. Commutative ring check for $(\mathbb{Z}, +, \cdot)$ with $x \cdot y = 0$ 1. The problem: Define multiplication by $x \cdot y = 0$ for all $x,y \in \mathbb{Z}$. Is $(\mathbb{Z}, +, \cdot)$ a commutative ring? 2. Recall the ring axioms: $(\mathbb{Z}, +)$ is an abelian group, multiplication must be associative, distributive over addition, and commutative for a commutative ring. 3. Addition is usual integer addition, so $(\mathbb{Z}, +)$ is an abelian group. 4. Check multiplication: - Multiplication is defined as $x \cdot y = 0$ for all $x,y$. - Multiplication is commutative trivially since $x \cdot y = 0 = y \cdot x$. - Multiplication is associative: $$ (x \cdot y) \cdot z = 0 \cdot z = 0, \quad x \cdot (y \cdot z) = x \cdot 0 = 0 $$ - Distributive laws: $$ x \cdot (y + z) = 0, \quad x \cdot y + x \cdot z = 0 + 0 = 0 $$ Similarly for right distributivity. 5. There is a multiplicative identity? No, since $1 \cdot y = 0 \neq y$ unless $y=0$. 6. Conclusion: $(\mathbb{Z}, +, \cdot)$ with this multiplication is a commutative ring without unity. --- ### 2. Integral domain and cancellation laws 1. Problem: Show $D$ is an integral domain iff it obeys cancellation laws. 2. Definitions: - Integral domain: commutative ring with unity and no zero divisors. - Cancellation law: For $a,b,c \in D$, if $a \neq 0$ and $ab = ac$, then $b = c$. 3. Proof: - $(\Rightarrow)$ Assume $D$ is integral domain. If $ab = ac$, then $a(b-c) = 0$. Since no zero divisors and $a \neq 0$, $b-c=0 \Rightarrow b=c$. - $(\Leftarrow)$ Assume cancellation laws hold. If $ab=0$ and $a \neq 0$, then $ab = a0$. By cancellation, $b=0$. So no zero divisors. Hence $D$ is integral domain. --- ### 3. Ideal is a subring 1. Problem: Prove an ideal $I$ of ring $R$ is a subring. 2. Recall: - Ideal $I$ is a nonempty subset of $R$ closed under subtraction and absorbs multiplication by elements of $R$. 3. To show $I$ is a subring: - $I$ is nonempty. - Closed under subtraction: if $a,b \in I$, then $a-b \in I$. - Closed under multiplication: if $a,b \in I$, then $ab \in I$ because $I$ absorbs multiplication by $R$ and $b \in I \subseteq R$. 4. Hence $I$ is a subring. --- ### 4. Subring and ideal properties of $S \subseteq M_2(\mathbb{Z})$ 1. Problem: Show $S = \left\{ \begin{pmatrix}0 & x \\ 0 & y \end{pmatrix} : x,y \in \mathbb{Z} \right\}$ is a subring of $M_2(\mathbb{Z})$, if it is unital, and if it is an ideal. 2. Check closure under addition and multiplication: - Addition: $$ \begin{pmatrix}0 & x \\ 0 & y \end{pmatrix} + \begin{pmatrix}0 & x' \\ 0 & y' \end{pmatrix} = \begin{pmatrix}0 & x+x' \\ 0 & y+y' \end{pmatrix} \in S $$ - Additive identity is zero matrix in $S$. - Additive inverses exist since $-x, -y \in \mathbb{Z}$. - Multiplication: $$ \begin{pmatrix}0 & x \\ 0 & y \end{pmatrix} \begin{pmatrix}0 & x' \\ 0 & y' \end{pmatrix} = \begin{pmatrix}0 & xy' \\ 0 & yy' \end{pmatrix} \in S $$ 3. So $S$ is a subring. 4. Is $S$ unital? The identity in $M_2(\mathbb{Z})$ is $I = \begin{pmatrix}1 & 0 \\ 0 & 1 \end{pmatrix}$ which is not in $S$ (top-left entry is 1, not 0). So $S$ is not unital. 5. Is $S$ an ideal? - For $A \in M_2(\mathbb{Z})$, $S$ is a left ideal if $A S \subseteq S$ and right ideal if $S A \subseteq S$. - Take arbitrary $A = \begin{pmatrix}a & b \\ c & d \end{pmatrix}$ and $S$ element $M = \begin{pmatrix}0 & x \\ 0 & y \end{pmatrix}$. - Compute $A M$: $$ A M = \begin{pmatrix}a & b \\ c & d \end{pmatrix} \begin{pmatrix}0 & x \\ 0 & y \end{pmatrix} = \begin{pmatrix}0 & a x + b y \\ 0 & c x + d y \end{pmatrix} \in S $$ - Compute $M A$: $$ M A = \begin{pmatrix}0 & x \\ 0 & y \end{pmatrix} \begin{pmatrix}a & b \\ c & d \end{pmatrix} = \begin{pmatrix}0 \cdot a + x \cdot c & 0 \cdot b + x \cdot d \\ 0 \cdot a + y \cdot c & 0 \cdot b + y \cdot d \end{pmatrix} = \begin{pmatrix}x c & x d \\ y c & y d \end{pmatrix} $$ - The matrix $M A$ is not in $S$ because the top-left entry is $x c$ which may not be zero. - Therefore, $S$ is a left ideal but not a right ideal, so not a two-sided ideal. --- ### 5. Preservation of properties under ring isomorphism $f : R \to R$ 1. Problem: Show zero divisors, nilpotent, units, and idempotents are preserved under $f$. 2. Since $f$ is a ring isomorphism, it is bijective and respects addition and multiplication. 3. Zero divisor: If $a$ is zero divisor, $a b = 0$ for some $b \neq 0$. Apply $f$: $$ f(a) f(b) = f(ab) = f(0) = 0 $$ Since $f$ is bijection, $f(b) \neq 0$, so $f(a)$ is zero divisor. 4. Nilpotent: If $a^n = 0$, then $$ f(a)^n = f(a^n) = f(0) = 0 $$ So $f(a)$ is nilpotent. 5. Unit: If $a$ is unit, $a a^{-1} = 1$. Apply $f$: $$ f(a) f(a^{-1}) = f(1) = 1 $$ So $f(a)$ is unit with inverse $f(a^{-1})$. 6. Idempotent: If $a^2 = a$, then $$ f(a)^2 = f(a^2) = f(a) $$ So $f(a)$ is idempotent. --- **Final answers:** - $(\mathbb{Z}, +, \cdot)$ with $x \cdot y = 0$ is a commutative ring without unity. - $D$ is integral domain iff cancellation laws hold. - Every ideal is a subring. - $S$ is a subring of $M_2(\mathbb{Z})$, not unital, left ideal but not two-sided ideal. - Ring isomorphisms preserve zero divisors, nilpotents, units, and idempotents.